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== 7.2#h(0.6em)Factor Trinomials of the Form x#super[2]+bx+c

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Multiply: #math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x plus 5 close parenthesis .")[$( x + 4 ) ( x + 5 ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x squared plus 9 x plus 20")[$x^(2) + 9 x + 20$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: ⓐ #math.equation(block: false, alt: "−9 plus open parenthesis −6 close parenthesis")[$−9 + ( −6 )$] ⓑ #math.equation(block: false, alt: "−9 plus 6 .")[$−9 + 6 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
ⓐ −15 ⓑ −3
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: ⓐ #math.equation(block: false, alt: "−9 open parenthesis 6 close parenthesis")[$−9 ( 6 )$] ⓑ #math.equation(block: false, alt: "−9 open parenthesis −6 close parenthesis .")[$−9 ( −6 ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
ⓐ −54 ⓑ 54
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: ⓐ #math.equation(block: false, alt: "| −5 |")[$| −5 |$] ⓑ #math.equation(block: false, alt: "| 3 | .")[$| 3 | .$] #linebreak() 
If you missed this problem, review .
]


=== Factor Trinomials of the Form #emph[x]#super[2] + #emph[bx] + #emph[c]

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication—to start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

#figure(figph[This figure shows the steps of multiplying the factors (x + 2) times (x + 3). The multiplying is completed using FOIL to demonstrate. The first term is x squared and is below F. The second term is 3 x below “O”. The third term is 2 x below “I”. The fourth term is 6 below L. The simplified product is then given as x 2 plus 5 x + 6.], alt: "This figure shows the steps of multiplying the factors (x + 2) times (x + 3). The multiplying is completed using FOIL to demonstrate. The first term is x squared and is below F. The second term is 3 x below “O”. The third term is 2 x below “I”. The fourth term is 6 below L. The simplified product is then given as x 2 plus 5 x + 6.", caption: none)

To factor the trinomial means to start with the product, #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$], and end with the factors, #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 2 ) ( x + 3 )$]. You need to think about where each of the terms in the trinomial came from.

The #emph[first term] came from multiplying the first term in each binomial. So to get #math.equation(block: false, alt: "x squared")[$x^(2)$] in the product, each binomial must start with an #emph[x].

#math.equation(block: true, alt: "x squared plus 5 x plus 6; open parenthesis x close parenthesis open parenthesis x close parenthesis")[$x^(2) + 5 x + 6 \ ( x #h(1em) ) ( x #h(1em) )$]The #emph[last term] in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the #emph[middle term]. It came from adding the outer and inner terms.

So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in —the table will be very helpful when you work with numbers that can be factored in many different ways.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "6")[$6$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , 6")[$1 , 6$]], [#math.equation(block: false, alt: "1 plus 6 equals 7")[$1 + 6 = 7$]],
  [#math.equation(block: false, alt: "2 , 3")[$2 , 3$]], [#math.equation(block: false, alt: "2 plus 3 equals 5")[$2 + 3 = 5$]],
))

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. So we have the factors of #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$]. They are #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 2 ) ( x + 3 )$].

#math.equation(block: true, alt: "x squared plus 5 x plus 6, product; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis, factors")[$x^(2) + 5 x + 6 &  &  & "product" \ ( x + 2 ) ( x + 3 ) &  &  & "factors"$]You should check this by multiplying.

Looking back, we started with #math.equation(block: false, alt: "x squared plus 5 x plus 6")[$x^(2) + 5 x + 6$], which is of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$], where #math.equation(block: false, alt: "b equals 5")[$b = 5$] and #math.equation(block: false, alt: "c equals 6")[$c = 6$]. We factored it into two binomials of the form #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis and open parenthesis x plus n close parenthesis")[$( x + m ) #h(0.2em) "and" #h(0.2em) ( x + n )$].

#math.equation(block: true, alt: "x squared plus 5 x plus 6, x squared plus b x plus c; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis, open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$x^(2) + 5 x + 6 &  &  & x^(2) + b x + c \ ( x + 2 ) ( x + 3 ) &  &  & ( x + m ) ( x + n )$]To get the correct factors, we found two numbers #emph[m] and #emph[n] whose product is #emph[c] and sum is #emph[b].

#examplebox("Example 1")[How to Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]][
Factor: #math.equation(block: false, alt: "x squared plus 7 x plus 12")[$x^(2) + 7 x + 12$].

#solutionbox[
#figure(figph[This table gives the steps for factoring x squared + 7 x + 12. The first row states the first step “write the factors as two binomials with first terms x”. In the second column of the first row it states, “write two sets of parentheses and put x as the first term”. In the third column, it has the expression x squared + 7 x +12. Below the expression are two sets of parentheses with x as the first term.], alt: "This table gives the steps for factoring x squared + 7 x + 12. The first row states the first step “write the factors as two binomials with first terms x”. In the second column of the first row it states, “write two sets of parentheses and put x as the first term”. In the third column, it has the expression x squared + 7 x +12. Below the expression are two sets of parentheses with x as the first term.", caption: none)

#figure(figph[The second row states the second step “find two numbers m and n that multiply to c, m times n = c and add to b, m + n = b”. In the second column of the second row are the factors of 12 and their sums. 1,12 with sum 1 + 12 = 13. 2, 6 with sum 2 + 6 =8. 3, 4 with sum 3 + 4 = 7.], alt: "The second row states the second step “find two numbers m and n that multiply to c, m times n = c and add to b, m + n = b”. In the second column of the second row are the factors of 12 and their sums. 1,12 with sum 1 + 12 = 13. 2, 6 with sum 2 + 6 =8. 3, 4 with sum 3 + 4 = 7.", caption: none)

#figure(figph[The third row states “use m and n as the last terms of the factors”. The second column states “use 3 and 4 as the last terms of the binomials”. The third column in this row has the product (x + 3)(x + 4).], alt: "The third row states “use m and n as the last terms of the factors”. The second column states “use 3 and 4 as the last terms of the binomials”. The third column in this row has the product (x + 3)(x + 4).", caption: none)

#figure(figph[In the fourth row the statement is “check by multiplying the factors”. The product of (x + 3)(x +4) is shown to be x 2 + 7 x + 12.], alt: "In the fourth row the statement is “check by multiplying the factors”. The product of (x + 3)(x +4) is shown to be x 2 + 7 x + 12.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared plus 6 x plus 8")[$x^(2) + 6 x + 8$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x plus 4 close parenthesis")[$( x + 2 ) ( x + 4 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "y squared plus 8 y plus 15")[$y^(2) + 8 y + 15$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y plus 3 close parenthesis open parenthesis y plus 5 close parenthesis")[$( y + 3 ) ( y + 5 )$]
]
]

Let’s summarize the steps we used to find the factors.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$].]

+ Write the factors as two binomials with first terms #emph[x]: #math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$#h(1em) ( x #h(1.5em) ) ( x #h(1.5em) )$].
+ Find two numbers #emph[m] and #emph[n] that #linebreak() 
 Multiply to #emph[c], #math.equation(block: false, alt: "m times n equals c")[$m · n = c$] #linebreak() 
 Add to #emph[b], #math.equation(block: false, alt: "m plus n equals b")[$m + n = b$]
+ Use #emph[m] and #emph[n] as the last terms of the factors: #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$#h(4em) ( x + m ) ( x + n )$].
+ Check by multiplying the factors.
]

#examplebox("Example 2")[][
Factor: #math.equation(block: false, alt: "u squared plus 11 u plus 24")[$u^(2) + 11 u + 24$].

#solutionbox[
Notice that the variable is #emph[u], so the factors will have first terms #emph[u].

#math.equation(block: true, alt: "u squared plus 11 u plus 24; Write the factors as two binomials with first terms u ., open parenthesis u close parenthesis open parenthesis u close parenthesis")[$&  &  & u^(2) + 11 u + 24 \ "Write the factors as two binomials with first terms" #h(0.2em) u . &  &  & ( u #h(1.5em) ) ( u #h(1.5em) )$]

Find two numbers that: multiply to 24 and add to 11.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "24")[$24$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , 24")[$#h(0.4em) 1 , 24$]], [#math.equation(block: false, alt: "1 plus 24 equals 25")[$1 + 24 = 25$]],
  [#math.equation(block: false, alt: "2 , 12")[$#h(0.4em) 2 , 12$]], [#math.equation(block: false, alt: "2 plus 12 equals 14")[$2 + 12 = 14$]],
  [#math.equation(block: false, alt: "3 , 8")[$3 , 8$]], [#math.equation(block: false, alt: "3 plus 8 equals 11 *")[$#h(0.8em) 3 + 8 = 11 "*"$]],
  [#math.equation(block: false, alt: "4 , 6")[$4 , 6$]], [#math.equation(block: false, alt: "4 plus 6 equals 10")[$#h(0.4em) 4 + 6 = 10$]],
))

#math.equation(block: true, alt: "Use 3 and 8 as the last terms of the binomials., open parenthesis u plus 3 close parenthesis open parenthesis u plus 8 close parenthesis; Check.; open parenthesis u plus 3 close parenthesis open parenthesis u plus 8 close parenthesis; u squared plus 3 u plus 8 u plus 24; u squared plus 11 u plus 24 ✓")[$"Use 3 and 8 as the last terms of the binomials." &  &  & #h(2.8em) ( u + 3 ) ( u + 8 ) \  \ "Check." &  &  &  \  \  \ #h(2.5em) ( u + 3 ) ( u + 8 ) &  &  &  \ #h(2.5em) u^(2) + 3 u + 8 u + 24 &  &  &  \ #h(2.5em) u^(2) + 11 u + 24 #h(0.2em) ✓ &  &  &$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "q squared plus 10 q plus 24")[$q^(2) + 10 q + 24$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis q plus 4 close parenthesis open parenthesis q plus 6 close parenthesis")[$( q + 4 ) ( q + 6 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "t squared plus 14 t plus 24")[$t^(2) + 14 t + 24$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis t plus 2 close parenthesis open parenthesis t plus 12 close parenthesis")[$( t + 2 ) ( t + 12 )$]
]
]

#examplebox("Example 3")[][
Factor: #math.equation(block: false, alt: "y squared plus 17 y plus 60")[$y^(2) + 17 y + 60$].

#solutionbox[
#math.equation(block: true, alt: "y squared plus 17 y plus 60; Write the factors as two binomials with first terms y ., open parenthesis y close parenthesis open parenthesis y close parenthesis")[$&  &  & y^(2) + 17 y + 60 \ "Write the factors as two binomials with first terms" #h(0.2em) y . &  &  & ( y #h(1.5em) ) ( y #h(1.5em) )$]

Find two numbers that multiply to 60 and add to 17.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "60")[$60$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , 60")[$1 , 60$]], [#math.equation(block: false, alt: "1 plus 60 equals 61")[$1 + 60 = 61$]],
  [#math.equation(block: false, alt: "2 , 30")[$2 , 30$]], [#math.equation(block: false, alt: "2 plus 30 equals 32")[$2 + 30 = 32$]],
  [#math.equation(block: false, alt: "3 , 20")[$3 , 20$]], [#math.equation(block: false, alt: "3 plus 20 equals 23")[$3 + 20 = 23$]],
  [#math.equation(block: false, alt: "4 , 15")[$4 , 15$]], [#math.equation(block: false, alt: "4 plus 15 equals 19")[$4 + 15 = 19$]],
  [#math.equation(block: false, alt: "5 , 12")[$5 , 12$]], [#math.equation(block: false, alt: "5 plus 12 equals 17 *")[$#h(0.4em) 5 + 12 = 17 "*"$]],
  [#math.equation(block: false, alt: "6 , 10")[$6 , 10$]], [#math.equation(block: false, alt: "6 plus 10 equals 16")[$6 + 10 = 16$]],
))

#math.equation(block: true, alt: "Use 5 and 12 as the last terms., open parenthesis y plus 5 close parenthesis open parenthesis y plus 12 close parenthesis; Check.; open parenthesis y plus 5 close parenthesis open parenthesis y plus 12 close parenthesis; open parenthesis y squared plus 12 y plus 5 y plus 60 close parenthesis; open parenthesis y squared plus 17 y plus 60 close parenthesis ✓")[$"Use" #h(0.2em) 5 #h(0.2em) "and" #h(0.2em) 12 #h(0.2em) "as the last terms." &  &  & #h(9em) ( y + 5 ) ( y + 12 ) \  \ "Check." &  &  &  \  \  \ #h(2.5em) ( y + 5 ) ( y + 12 ) &  &  &  \ #h(2.5em) ( y^(2) + 12 y + 5 y + 60 ) &  &  &  \ #h(2.5em) ( y^(2) + 17 y + 60 ) ✓ &  &  &$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared plus 19 x plus 60")[$x^(2) + 19 x + 60$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x plus 15 close parenthesis")[$( x + 4 ) ( x + 15 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "v squared plus 23 v plus 60")[$v^(2) + 23 v + 60$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis v plus 3 close parenthesis open parenthesis v plus 20 close parenthesis")[$( v + 3 ) ( v + 20 )$]
]
]


==== Factor Trinomials of the Form #emph[x]#super[2] + #emph[bx] + #emph[c] with #emph[b] Negative, #emph[c] Positive

In the examples so far, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

Remember: To get a negative sum and a positive product, the numbers must both be negative.

Again, think about FOIL and where each term in the trinomial came from. Just as before,

- the first term, #math.equation(block: false, alt: "x squared")[$x^(2)$], comes from the product of the two first terms in each binomial factor, #emph[x] and #emph[y];
- the positive last term is the product of the two last terms
- the negative middle term is the sum of the outer and inner terms.

How do you get a #emph[positive product] and a #emph[negative sum]? With two negative numbers.

#examplebox("Example 4")[][
Factor: #math.equation(block: false, alt: "t squared minus 11 t plus 28")[$t^(2) − 11 t + 28$].

#solutionbox[
Again, with the positive last term, 28, and the negative middle term, #math.equation(block: false, alt: "−11 t")[$−11 t$], we need two negative factors. Find two numbers that multiply 28 and add to #math.equation(block: false, alt: "−11")[$−11$].

#math.equation(block: true, alt: "t squared minus 11 t plus 28; Write the factors as two binomials with first terms t ., open parenthesis t close parenthesis open parenthesis t close parenthesis")[$&  &  & t^(2) − 11 t + 28 \  \ "Write the factors as two binomials with first terms" #h(0.2em) t . &  &  & ( t #h(1.5em) ) ( t #h(1.5em) )$]

Find two numbers that: multiply to 28 and add to #math.equation(block: false, alt: "−11")[$−11$].

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "28")[$28$]], [Sum of factors]),
  [#math.equation(block: false, alt: "−1 , −28")[$#h(0.4em) −1 , −28$]], [#math.equation(block: false, alt: "−1 plus open parenthesis −28 close parenthesis equals −29")[$−1 + ( −28 ) = −29$]],
  [#math.equation(block: false, alt: "−2 , −14")[$#h(0.4em) −2 , −14$]], [#math.equation(block: false, alt: "−2 plus open parenthesis −14 close parenthesis equals −16")[$−2 + ( −14 ) = −16$]],
  [#math.equation(block: false, alt: "−4 , −7")[$−4 , −7$]], [#math.equation(block: false, alt: "−4 plus open parenthesis −7 close parenthesis equals −11 *")[$#h(0.8em) −4 + ( −7 ) = −11 "*"$]],
))

#math.equation(block: true, alt: "Use −4 , −7 as the last terms of the binomials., open parenthesis t minus 4 close parenthesis open parenthesis t minus 7 close parenthesis; Check.; open parenthesis t minus 4 close parenthesis open parenthesis t minus 7 close parenthesis; t squared minus 7 t minus 4 t plus 28; t squared minus 11 t plus 28 ✓")[$"Use" #h(0.2em) −4 , −7 #h(0.2em) "as the last terms of the binomials." &  &  & #h(2.8em) ( t − 4 ) ( t − 7 ) \ "Check." &  &  &  \  \  \ #h(2.5em) ( t − 4 ) ( t − 7 ) &  &  &  \ #h(2.5em) t^(2) − 7 t − 4 t + 28 &  &  &  \ #h(2.5em) t^(2) − 11 t + 28 #h(0.2em) ✓ &  &  &$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "u squared minus 9 u plus 18")[$u^(2) − 9 u + 18$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis u minus 3 close parenthesis open parenthesis u minus 6 close parenthesis")[$( u − 3 ) ( u − 6 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "y squared minus 16 y plus 63")[$y^(2) − 16 y + 63$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y minus 7 close parenthesis open parenthesis y minus 9 close parenthesis")[$( y − 7 ) ( y − 9 )$]
]
]


==== Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] with #emph[c] Negative

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

Remember: To get a negative product, the numbers must have different signs.

#examplebox("Example 5")[][
Factor: #math.equation(block: false, alt: "z squared plus 4 z minus 5")[$z^(2) + 4 z − 5$].

#solutionbox[
To get a negative last term, multiply one positive and one negative. We need factors of #math.equation(block: false, alt: "−5")[$−5$] that add to positive 4.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−5")[$−5$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , −5")[$#h(1.2em) 1 , −5$]], [#math.equation(block: false, alt: "1 plus open parenthesis −5 close parenthesis equals −4")[$1 + ( −5 ) = −4$]],
  [#math.equation(block: false, alt: "−1 , 5")[$−1 , 5$]], [#math.equation(block: false, alt: "−1 plus 5 equals 4 *")[$#h(0.5em) −1 + 5 = 4 "*"$]],
))

Notice: We listed both #math.equation(block: false, alt: "1 , −5 and minus 1 , 5")[$1 , −5 #h(0.2em) "and" #h(0.2em) − 1 , 5$] to make sure we got the sign of the middle term correct.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "z squared plus 4 z minus 5")[$z^(2) + 4 z − 5$]]),
  [Factors will be two binomials with first terms #emph[z].], [#math.equation(block: false, alt: "open parenthesis z close parenthesis open parenthesis z close parenthesis")[$( z #h(1.5em) ) ( z #h(1.5em) )$]],
  [Use −1, 5 as the last terms of the binomials.], [#math.equation(block: false, alt: "open parenthesis z minus 1 close parenthesis open parenthesis z plus 5 close parenthesis")[$( z − 1 ) ( z + 5 )$]],
  [Check. #linebreak() #math.equation(block: false, alt: "open parenthesis z minus 1 close parenthesis open parenthesis z plus 5 close parenthesis; z squared plus 5 z minus 1 z minus 5; z squared plus 4 z minus 5 ✓")[$#h(2.5em) ( z − 1 ) ( z + 5 ) \ #h(2.5em) z^(2) + 5 z − 1 z − 5 \ #h(2.5em) z^(2) + 4 z − 5 #h(0.2em) ✓$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "h squared plus 4 h minus 12")[$h^(2) + 4 h − 12$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis h minus 2 close parenthesis open parenthesis h plus 6 close parenthesis")[$( h − 2 ) ( h + 6 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "k squared plus k minus 20")[$k^(2) + k − 20$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis k minus 4 close parenthesis open parenthesis k plus 5 close parenthesis")[$( k − 4 ) ( k + 5 )$]
]
]

Let’s make a minor change to the last trinomial and see what effect it has on the factors.

#examplebox("Example 6")[][
Factor: #math.equation(block: false, alt: "z squared minus 4 z minus 5")[$z^(2) − 4 z − 5$].

#solutionbox[
This time, we need factors of #math.equation(block: false, alt: "−5")[$−5$] that add to #math.equation(block: false, alt: "−4")[$−4$].

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−5")[$−5$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , −5")[$#h(1.2em) 1 , −5$]], [#math.equation(block: false, alt: "1 plus open parenthesis −5 close parenthesis equals −4 *")[$#h(0.4em) 1 + ( −5 ) = −4 "*"$]],
  [#math.equation(block: false, alt: "−1 , 5")[$−1 , 5$]], [#math.equation(block: false, alt: "−1 plus 5 equals 4")[$−1 + 5 = 4$]],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "z squared minus 4 z minus 5")[$z^(2) − 4 z − 5$]]),
  [Factors will be two binomials with first terms #emph[z].], [#math.equation(block: false, alt: "open parenthesis z close parenthesis open parenthesis z close parenthesis")[$( z #h(1.5em) ) ( z #h(1.5em) )$]],
  [Use 1, −5 as the last terms of the binomials.], [#math.equation(block: false, alt: "open parenthesis z plus 1 close parenthesis open parenthesis z minus 5 close parenthesis")[$( z + 1 ) ( z − 5 )$]],
  [Check. #linebreak() #math.equation(block: false, alt: "open parenthesis z plus 1 close parenthesis open parenthesis z minus 5 close parenthesis; z squared minus 5 z plus 1 z minus 5; z squared minus 4 z minus 5 ✓")[$#h(2.5em) ( z + 1 ) ( z − 5 ) \ #h(2.5em) z^(2) − 5 z + 1 z − 5 \ #h(2.5em) z^(2) − 4 z − 5 #h(0.2em) ✓$]], [],
))

Notice that the factors of #math.equation(block: false, alt: "z squared minus 4 z minus 5")[$z^(2) − 4 z − 5$] are very similar to the factors of #math.equation(block: false, alt: "z squared plus 4 z minus 5")[$z^(2) + 4 z − 5$]. It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared minus 4 x minus 12")[$x^(2) − 4 x − 12$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x minus 6 close parenthesis")[$( x + 2 ) ( x − 6 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "y squared minus y minus 20")[$y^(2) − y − 20$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y plus 4 close parenthesis open parenthesis y minus 5 close parenthesis")[$( y + 4 ) ( y − 5 )$]
]
]

#examplebox("Example 7")[][
Factor: #math.equation(block: false, alt: "q squared minus 2 q minus 15")[$q^(2) − 2 q − 15$].

#solutionbox[
#math.equation(block: true, alt: "q squared minus 2 q minus 15; Factors will be two binomials with first terms q ., open parenthesis q close parenthesis open parenthesis q close parenthesis; You can use 3 , −5 as the last terms of the, open parenthesis q plus 3 close parenthesis open parenthesis q minus 5 close parenthesis; binomials.")[$&  &  & q^(2) − 2 q − 15 \ "Factors will be two binomials with first terms" #h(0.2em) q . &  &  & ( q #h(1.5em) ) ( q #h(1.5em) ) \ "You can use" #h(0.2em) 3 , −5 #h(0.2em) "as the last terms of the" &  &  & ( q + 3 ) ( q − 5 ) \ "binomials." &  &  &$]

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−15")[$−15$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , −15")[$#h(0.6em) 1 , −15$]], [#math.equation(block: false, alt: "1 plus open parenthesis −15 close parenthesis equals −14")[$1 + ( −15 ) = −14$]],
  [#math.equation(block: false, alt: "−1 , 15")[$−1 , 15$]], [#math.equation(block: false, alt: "−1 plus 15 equals 14")[$#h(0.7em) −1 + 15 = 14$]],
  [#math.equation(block: false, alt: "3 , −5")[$#h(0.6em) 3 , −5$]], [#math.equation(block: false, alt: "3 plus open parenthesis −5 close parenthesis equals −2 *")[$#h(0.6em) 3 + ( −5 ) = −2 "*"$]],
  [#math.equation(block: false, alt: "−3 , 5")[$−3 , 5$]], [#math.equation(block: false, alt: "−3 plus 5 equals 2")[$#h(1.2em) −3 + 5 = 2$]],
))

Check.

#math.equation(block: true, alt: "open parenthesis q plus 3 close parenthesis open parenthesis q minus 5 close parenthesis")[$#h(2.5em) ( q + 3 ) ( q − 5 )$]

#math.equation(block: true, alt: "q squared minus 5 q plus 3 q minus 15")[$#h(2.5em) q^(2) − 5 q + 3 q − 15$]

#math.equation(block: true, alt: "q squared minus 2 q minus 15 ✓")[$#h(2.5em) q^(2) − 2 q − 15 #h(0.2em) ✓$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "r squared minus 3 r minus 40")[$r^(2) − 3 r − 40$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis r plus 5 close parenthesis open parenthesis r minus 8 close parenthesis")[$( r + 5 ) ( r − 8 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "s squared minus 3 s minus 10")[$s^(2) − 3 s − 10$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis s plus 2 close parenthesis open parenthesis s minus 5 close parenthesis")[$( s + 2 ) ( s − 5 )$]
]
]

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

#examplebox("Example 8")[][
Factor: #math.equation(block: false, alt: "y squared minus 6 y plus 15")[$y^(2) − 6 y + 15$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "y squared minus 6 y plus 15")[$y^(2) − 6 y + 15$]]),
  [Factors will be two binomials with first terms #emph[y].], [#math.equation(block: false, alt: "open parenthesis y close parenthesis open parenthesis y close parenthesis")[$( y #h(1.5em) ) ( y #h(1.5em) )$]],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of 15], [Sum of factors]),
  [#math.equation(block: false, alt: "−1 , −15")[$−1 , −15$]], [#math.equation(block: false, alt: "−1 plus open parenthesis −15 close parenthesis equals −16")[$−1 + ( −15 ) = −16$]],
  [#math.equation(block: false, alt: "−3 , −5")[$−3 , −5$]], [#math.equation(block: false, alt: "−3 plus open parenthesis −5 close parenthesis equals −8")[$#h(0.4em) −3 + ( −5 ) = −8$]],
))

As shown in the table, none of the factors add to #math.equation(block: false, alt: "−6")[$−6$]; therefore, the expression is prime.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "m squared plus 4 m plus 18")[$m^(2) + 4 m + 18$].

#solutionbox[
prime
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "n squared minus 10 n plus 12")[$n^(2) − 10 n + 12$].

#solutionbox[
prime
]
]

#examplebox("Example 9")[][
Factor: #math.equation(block: false, alt: "2 x plus x squared minus 48")[$2 x + x^(2) − 48$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "2 x plus x squared minus 48")[$2 x + x^(2) − 48$]]),
  [First we put the terms in decreasing degree order.], [#math.equation(block: false, alt: "x squared plus 2 x minus 48")[$x^(2) + 2 x − 48$]],
  [Factors will be two binomials with first terms #emph[x].], [#math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1.5em) ) ( x #h(1.5em) )$]],
))

As shown in the table, you can use #math.equation(block: false, alt: "−6 , 8")[$−6 , 8$] as the last terms of the binomials.

#math.equation(block: true, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 8 close parenthesis")[$#h(1.6em) ( x − 6 ) ( x + 8 )$]#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−48")[$−48$]], [Sum of factors]),
  [#math.equation(block: false, alt: "−1 , 48")[$−1 , 48$]], [#math.equation(block: false, alt: "−1 plus 48 equals 47")[$−1 + 48 = 47$]],
  [#math.equation(block: false, alt: "−2 , 24")[$−2 , 24$] #linebreak() 
#math.equation(block: false, alt: "−3 , 16")[$−3 , 16$] #linebreak() 
#math.equation(block: false, alt: "−4 , 12")[$−4 , 12$] #linebreak() 
#math.equation(block: false, alt: "−6 , 8")[$−6 , 8$]], [#math.equation(block: false, alt: "−2 plus 24 equals 22")[$−2 + 24 = 22$] #linebreak() 
#math.equation(block: false, alt: "−3 plus 16 equals 13")[$−3 + 16 = 13$] #linebreak() 
#math.equation(block: false, alt: "−4 plus 12 equals 8")[$−4 + 12 = 8$] #linebreak() 
#math.equation(block: false, alt: "−6 plus 8 equals 2")[$#h(0.5em) −6 + 8 = 2$]],
))

Check.

#math.equation(block: true, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 8 close parenthesis")[$#h(2.5em) ( x − 6 ) ( x + 8 )$]

#math.equation(block: true, alt: "x squared minus 6 x plus 8 x minus 48")[$#h(2.5em) x^(2) − 6 x + 8 x − 48$]

#math.equation(block: true, alt: "x squared plus 2 x minus 48 ✓")[$#h(2.5em) x^(2) + 2 x − 48 #h(0.2em) ✓$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "9 m plus m squared plus 18")[$9 m + m^(2) + 18$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m plus 3 close parenthesis open parenthesis m plus 6 close parenthesis")[$( m + 3 ) ( m + 6 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "−7 n plus 12 plus n squared")[$−7 n + 12 + n^(2)$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis n minus 3 close parenthesis open parenthesis n minus 4 close parenthesis")[$( n − 3 ) ( n − 4 )$]
]
]

Let’s summarize the method we just developed to factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Factor trinomials.]

When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.

#math.equation(block: true, alt: "x squared plus b x plus c; open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$x^(2) + b x + c \ ( x + m ) ( x + n )$]#strong[When #emph[c] is positive, #emph[m] and #emph[n] have the same sign.]

#math.equation(block: true, alt: "b positive, b negative; m , n positive, m , n negative; x squared plus 5 x plus 6, x squared minus 6 x plus 8; open parenthesis x plus 2 close parenthesis open parenthesis x plus 3 close parenthesis, open parenthesis x minus 4 close parenthesis open parenthesis x minus 2 close parenthesis; same signs, same signs")[$b #h(0.2em) "positive" &  &  & #h(2em) b #h(0.2em) "negative" \ m , n #h(0.2em) "positive" &  &  & #h(2em) m , n #h(0.2em) "negative" \ x^(2) + 5 x + 6 &  &  & #h(2em) x^(2) − 6 x + 8 \ ( x + 2 ) ( x + 3 ) &  &  & #h(2em) ( x − 4 ) ( x − 2 ) \ "same signs" &  &  & #h(2em) "same signs"$]#strong[When #emph[c] is negative, #emph[m] and #emph[n] have opposite signs.]

#math.equation(block: true, alt: "x squared plus x minus 12, x squared minus 2 x minus 15; open parenthesis x plus 4 close parenthesis open parenthesis x minus 3 close parenthesis, open parenthesis x minus 5 close parenthesis open parenthesis x plus 3 close parenthesis; opposite signs, opposite signs")[$x^(2) + x − 12 &  &  & #h(2em) x^(2) − 2 x − 15 \ ( x + 4 ) ( x − 3 ) &  &  & #h(2em) ( x − 5 ) ( x + 3 ) \ "opposite signs" &  &  & #h(2em) "opposite signs"$]Notice that, in the case when #emph[m] and #emph[n] have opposite signs, the sign of the one with the larger absolute value matches the sign of #emph[b].
]


=== Factor Trinomials of the Form #emph[x]#super[2] + #emph[bxy] + #emph[cy]#super[2]

Sometimes you’ll need to factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x y plus c y squared")[$x^(2) + b x y + c y^(2)$] with two variables, such as #math.equation(block: false, alt: "x squared plus 12 x y plus 36 y squared .")[$x^(2) + 12 x y + 36 y^(2) .$] The first term, #math.equation(block: false, alt: "x squared")[$x^(2)$], is the product of the first terms of the binomial factors, #math.equation(block: false, alt: "x times x")[$x · x$]. The #math.equation(block: false, alt: "y squared")[$y^(2)$] in the last term means that the second terms of the binomial factors must each contain #emph[y]. To get the coefficients #emph[b] and #emph[c], you use the same process summarized in the previous objective.

#examplebox("Example 10")[][
Factor: #math.equation(block: false, alt: "x squared plus 12 x y plus 36 y squared")[$x^(2) + 12 x y + 36 y^(2)$].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "x squared plus 12 x y plus 36 y squared")[$x^(2) + 12 x y + 36 y^(2)$]]),
  [Note that the first terms are #emph[x], last terms contain #emph[y].], [#math.equation(block: false, alt: "open parenthesis x _ y close parenthesis open parenthesis x _ y close parenthesis")[$( x \_ y ) ( x \_ y )$]],
))

Find the numbers that multiply to 36 and add to 12.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "36")[$36$]], [Sum of factors]),
  [1, 36], [#math.equation(block: false, alt: "1 plus 36 equals 37")[$1 + 36 = 37$]],
  [2, 18], [#math.equation(block: false, alt: "2 plus 18 equals 20")[$2 + 18 = 20$]],
  [3, 12], [#math.equation(block: false, alt: "3 plus 12 equals 15")[$3 + 12 = 15$]],
  [4, 9], [#math.equation(block: false, alt: "4 plus 9 equals 13")[$4 + 9 = 13$]],
  [6, 6], [#math.equation(block: false, alt: "6 plus 6 equals 12 *")[$6 + 6 = 12 "*"$]],
))

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use 6 and 6 as the coefficients of the last terms.], [#math.equation(block: false, alt: "open parenthesis x plus 6 y close parenthesis open parenthesis x plus 6 y close parenthesis")[$( x + 6 y ) ( x + 6 y )$]]),
  [Check your answer. #linebreak() #math.equation(block: false, alt: "open parenthesis x plus 6 y close parenthesis open parenthesis x plus 6 y close parenthesis; x squared plus 6 x y plus 6 x y plus 36 y squared; x squared plus 12 x y plus 36 y squared ✓")[$#h(3.5em) ( x + 6 y ) ( x + 6 y ) \ #h(2.5em) x^(2) + 6 x y + 6 x y + 36 y^(2) \ #h(2.5em) x^(2) + 12 x y + 36 y^(2) ✓$]], [],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "u squared plus 11 u v plus 28 v squared")[$u^(2) + 11 u v + 28 v^(2)$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis u plus 4 v close parenthesis open parenthesis u plus 7 v close parenthesis")[$( u + 4 v ) ( u + 7 v )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared plus 13 x y plus 42 y squared")[$x^(2) + 13 x y + 42 y^(2)$]. #linebreak()  #linebreak()

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 6 y close parenthesis open parenthesis x plus 7 y close parenthesis")[$( x + 6 y ) ( x + 7 y )$]
]
]

#examplebox("Example 11")[][
Factor: #math.equation(block: false, alt: "r squared minus 8 r s minus 9 s squared")[$r^(2) − 8 r s − 9 s^(2)$].

#solutionbox[
We need #math.equation(block: false, alt: "r")[$r$] in the first term of each binomial and #math.equation(block: false, alt: "s")[$s$] in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

#math.equation(block: true, alt: "r squared minus 8 r s minus 9 s squared; Note that the first terms are r , last terms contain s ., open parenthesis r _ s close parenthesis open parenthesis r _ s close parenthesis")[$&  &  & r^(2) − 8 r s − 9 s^(2) \ "Note that the first terms are" #h(0.2em) r , #h(0.2em) "last terms contain" #h(0.2em) s . &  &  & ( r \_ s ) ( r \_ s )$]

Find the numbers that multiply to #math.equation(block: false, alt: "−9")[$−9$] and add to #math.equation(block: false, alt: "−8")[$−8$].

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−9")[$−9$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , −9")[$#h(1em) 1 , −9$]], [#math.equation(block: false, alt: "1 plus open parenthesis −9 close parenthesis equals −8 *")[$#h(0.7em) 1 + ( −9 ) = −8 "*"$]],
  [#math.equation(block: false, alt: "−1 , 9")[$−1 , 9$]], [#math.equation(block: false, alt: "−1 plus 9 equals 8")[$#h(0.4em) −1 + 9 = 8$]],
  [#math.equation(block: false, alt: "3 , −3")[$#h(1em) 3 , −3$]], [#math.equation(block: false, alt: "3 plus open parenthesis −3 close parenthesis equals 0")[$3 + ( −3 ) = 0$]],
))

#math.equation(block: true, alt: "Use 1 , −9 as coefficients of the last terms., open parenthesis r plus s close parenthesis open parenthesis r minus 9 s close parenthesis; Check your answer.; open parenthesis r minus 9 s close parenthesis open parenthesis r plus s close parenthesis; r squared plus r s minus 9 r s minus 9 s squared; r squared minus 8 r s minus 9 s squared ✓")[$"Use" #h(0.2em) 1 , −9 #h(0.2em) "as coefficients of the last terms." &  &  & #h(3.4em) ( r + s ) ( r − 9 s ) \ "Check your answer." &  &  &  \  \  \ #h(2.5em) ( r − 9 s ) ( r + s ) &  &  &  \ #h(2.5em) r^(2) + r s − 9 r s − 9 s^(2) &  &  &  \ #h(2.5em) r^(2) − 8 r s − 9 s^(2) #h(0.2em) ✓ &  &  &$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "a squared minus 11 a b plus 10 b squared")[$a^(2) − 11 a b + 10 b^(2)$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a minus b close parenthesis open parenthesis a minus 10 b close parenthesis")[$( a − b ) ( a − 10 b )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "m squared minus 13 m n plus 12 n squared")[$m^(2) − 13 m n + 12 n^(2)$].

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m minus n close parenthesis open parenthesis m minus 12 n close parenthesis")[$( m − n ) ( m − 12 n )$]
]
]

#linebreak()

#examplebox("Example 12")[][
Factor: #math.equation(block: false, alt: "u squared minus 9 u v minus 12 v squared")[$u^(2) − 9 u v − 12 v^(2)$].

#solutionbox[
We need #emph[u] in the first term of each binomial and #math.equation(block: false, alt: "v")[$v$] in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "u squared minus 9 u v minus 12 v squared")[$u^(2) − 9 u v − 12 v^(2)$]]),
  [Note that the first terms are #emph[u], last terms contain #emph[v].], [#math.equation(block: false, alt: "open parenthesis u _ v close parenthesis open parenthesis u _ v close parenthesis")[$( u \_ v ) ( u \_ v )$]],
))

Find the numbers that multiply to #math.equation(block: false, alt: "−12")[$−12$] and add to #math.equation(block: false, alt: "−9")[$−9$].

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Factors of #math.equation(block: false, alt: "−12")[$−12$]], [Sum of factors]),
  [#math.equation(block: false, alt: "1 , −12")[$#h(2em) 1 , −12$]], [#math.equation(block: false, alt: "1 plus open parenthesis −12 close parenthesis equals −11")[$1 + ( −12 ) = −11$]],
  [#math.equation(block: false, alt: "−1 , 12")[$#h(0.5em) −1 , 12$]], [#math.equation(block: false, alt: "−1 plus 12 equals 11")[$−1 + 12 = 11$]],
  [#math.equation(block: false, alt: "2 , −6")[$#h(1.2em) 2 , −6$]], [#math.equation(block: false, alt: "2 plus open parenthesis −6 close parenthesis equals −4")[$2 + ( −6 ) = −4$]],
  [#math.equation(block: false, alt: "−2 , 6")[$−2 , 6$]], [#math.equation(block: false, alt: "−2 plus 6 equals 4")[$−2 + 6 = 4$]],
  [#math.equation(block: false, alt: "3 , −4")[$#h(1.2em) 3 , −4$]], [#math.equation(block: false, alt: "3 plus open parenthesis −4 close parenthesis equals −1")[$3 + ( −4 ) = −1$]],
  [#math.equation(block: false, alt: "−3 , 4")[$−3 , 4$]], [#math.equation(block: false, alt: "−3 plus 4 equals 1")[$−3 + 4 = 1$]],
))

Note there are no factor pairs that give us #math.equation(block: false, alt: "−9")[$−9$] as a sum. The trinomial is prime.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "x squared minus 7 x y minus 10 y squared")[$x^(2) − 7 x y − 10 y^(2)$].

#solutionbox[
prime
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Factor: #math.equation(block: false, alt: "p squared plus 15 p q plus 20 q squared")[$p^(2) + 15 p q + 20 q^(2)$].

#solutionbox[
prime
]
]


=== Key Concepts

- #strong[Factor trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]]
+ Write the factors as two binomials with first terms #emph[x]: #math.equation(block: false, alt: "open parenthesis x close parenthesis open parenthesis x close parenthesis")[$( x #h(1em) ) ( x #h(1em) )$].
+ Find two numbers #emph[m] and #emph[n] that #linebreak() 
Multiply to #emph[c], #math.equation(block: false, alt: "m times n equals c")[$m · n = c$] #linebreak() 
Add to #emph[b], #math.equation(block: false, alt: "m plus n equals b")[$m + n = b$]
+ Use #emph[m] and #emph[n] as the last terms of the factors: #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$( x + m ) ( x + n )$].
+ Check by multiplying the factors.


==== Practice Makes Perfect

#strong[Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$]]

In the following exercises, factor each trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$].

#math.equation(block: true, alt: "x squared plus 4 x plus 3")[$x^(2) + 4 x + 3$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 1 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x + 1 ) ( x + 3 )$]
]

#math.equation(block: true, alt: "y squared plus 8 y plus 7")[$y^(2) + 8 y + 7$]

#math.equation(block: true, alt: "m squared plus 12 m plus 11")[$m^(2) + 12 m + 11$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m plus 1 close parenthesis open parenthesis m plus 11 close parenthesis")[$( m + 1 ) ( m + 11 )$]
]

#math.equation(block: true, alt: "b squared plus 14 b plus 13")[$b^(2) + 14 b + 13$]

#math.equation(block: true, alt: "a squared plus 9 a plus 20")[$a^(2) + 9 a + 20$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a plus 4 close parenthesis open parenthesis a plus 5 close parenthesis")[$( a + 4 ) ( a + 5 )$]
]

#math.equation(block: true, alt: "m squared plus 7 m plus 12")[$m^(2) + 7 m + 12$]

#math.equation(block: true, alt: "p squared plus 11 p plus 30")[$p^(2) + 11 p + 30$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis p plus 5 close parenthesis open parenthesis p plus 6 close parenthesis")[$( p + 5 ) ( p + 6 )$]
]

#math.equation(block: true, alt: "w squared plus 10 w plus 21")[$w^(2) + 10 w + 21$]

#math.equation(block: true, alt: "n squared plus 19 n plus 48")[$n^(2) + 19 n + 48$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis n plus 3 close parenthesis open parenthesis n plus 16 close parenthesis")[$( n + 3 ) ( n + 16 )$]
]

#math.equation(block: true, alt: "b squared plus 14 b plus 48")[$b^(2) + 14 b + 48$]

#math.equation(block: true, alt: "a squared plus 25 a plus 100")[$a^(2) + 25 a + 100$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a plus 5 close parenthesis open parenthesis a plus 20 close parenthesis")[$( a + 5 ) ( a + 20 )$]
]

#math.equation(block: true, alt: "u squared plus 101 u plus 100")[$u^(2) + 101 u + 100$]

#math.equation(block: true, alt: "x squared minus 8 x plus 12")[$x^(2) − 8 x + 12$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x minus 6 close parenthesis")[$( x − 2 ) ( x − 6 )$]
]

#math.equation(block: true, alt: "q squared minus 13 q plus 36")[$q^(2) − 13 q + 36$]

#math.equation(block: true, alt: "y squared minus 18 y plus 45")[$y^(2) − 18 y + 45$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y minus 3 close parenthesis open parenthesis y minus 15 close parenthesis")[$( y − 3 ) ( y − 15 )$]
]

#math.equation(block: true, alt: "m squared minus 13 m plus 30")[$m^(2) − 13 m + 30$]

#math.equation(block: true, alt: "x squared minus 8 x plus 7")[$x^(2) − 8 x + 7$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis open parenthesis x minus 7 close parenthesis")[$( x − 1 ) ( x − 7 )$]
]

#math.equation(block: true, alt: "y squared minus 5 y plus 6")[$y^(2) − 5 y + 6$]

#math.equation(block: true, alt: "p squared plus 5 p minus 6")[$p^(2) + 5 p − 6$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis p minus 1 close parenthesis open parenthesis p plus 6 close parenthesis")[$( p − 1 ) ( p + 6 )$]
]

#math.equation(block: true, alt: "n squared plus 6 n minus 7")[$n^(2) + 6 n − 7$]

#math.equation(block: true, alt: "y squared minus 6 y minus 7")[$y^(2) − 6 y − 7$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis y plus 1 close parenthesis open parenthesis y minus 7 close parenthesis")[$( y + 1 ) ( y − 7 )$]
]

#math.equation(block: true, alt: "v squared minus 2 v minus 3")[$v^(2) − 2 v − 3$]

#math.equation(block: true, alt: "x squared minus x minus 12")[$x^(2) − x − 12$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 4 close parenthesis open parenthesis x plus 3 close parenthesis")[$( x − 4 ) ( x + 3 )$]
]

#math.equation(block: true, alt: "r squared minus 2 r minus 8")[$r^(2) − 2 r − 8$]

#math.equation(block: true, alt: "a squared minus 3 a minus 28")[$a^(2) − 3 a − 28$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a minus 7 close parenthesis open parenthesis a plus 4 close parenthesis")[$( a − 7 ) ( a + 4 )$]
]

#math.equation(block: true, alt: "b squared minus 13 b minus 30")[$b^(2) − 13 b − 30$]

#math.equation(block: true, alt: "w squared minus 5 w minus 36")[$w^(2) − 5 w − 36$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis w minus 9 close parenthesis open parenthesis w plus 4 close parenthesis")[$( w − 9 ) ( w + 4 )$]
]

#math.equation(block: true, alt: "t squared minus 3 t minus 54")[$t^(2) − 3 t − 54$]

#math.equation(block: true, alt: "x squared plus x plus 5")[$x^(2) + x + 5$]

#solutionbox[
prime
]

#math.equation(block: true, alt: "x squared minus 3 x minus 9")[$x^(2) − 3 x − 9$]

#math.equation(block: true, alt: "8 minus 6 x plus x squared")[$8 − 6 x + x^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 4 close parenthesis open parenthesis x minus 2 close parenthesis")[$( x − 4 ) ( x − 2 )$]
]

#math.equation(block: true, alt: "7 x plus x squared plus 6")[$7 x + x^(2) + 6$]

#math.equation(block: true, alt: "x squared minus 12 minus 11 x")[$x^(2) − 12 − 11 x$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x minus 12 close parenthesis open parenthesis x plus 1 close parenthesis")[$( x − 12 ) ( x + 1 )$]
]

#math.equation(block: true, alt: "−11 minus 10 x plus x squared")[$−11 − 10 x + x^(2)$]

#strong[Factor Trinomials of the Form #math.equation(block: false, alt: "x squared plus b x y plus c y squared")[$x^(2) + b x y + c y^(2)$]]

In the following exercises, factor each trinomial of the form #math.equation(block: false, alt: "x squared plus b x y plus c y squared")[$x^(2) + b x y + c y^(2)$]. If the trinomial cannot be factored, answer “Prime.”

#math.equation(block: true, alt: "p squared plus 3 p q plus 2 q squared")[$p^(2) + 3 p q + 2 q^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis p plus q close parenthesis open parenthesis p plus 2 q close parenthesis")[$( p + q ) ( p + 2 q )$]
]

#math.equation(block: true, alt: "m squared plus 6 m n plus 5 n squared")[$m^(2) + 6 m n + 5 n^(2)$]

#math.equation(block: true, alt: "r squared plus 15 r s plus 36 s squared")[$r^(2) + 15 r s + 36 s^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis r plus 3 s close parenthesis open parenthesis r plus 12 s close parenthesis")[$( r + 3 s ) ( r + 12 s )$]
]

#math.equation(block: true, alt: "u squared plus 10 u v plus 24 v squared")[$u^(2) + 10 u v + 24 v^(2)$]

#math.equation(block: true, alt: "m squared minus 12 m n plus 20 n squared")[$m^(2) − 12 m n + 20 n^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m minus 2 n close parenthesis open parenthesis m minus 10 n close parenthesis")[$( m − 2 n ) ( m − 10 n )$]
]

#math.equation(block: true, alt: "p squared minus 16 p q plus 63 q squared")[$p^(2) − 16 p q + 63 q^(2)$]

#math.equation(block: true, alt: "x squared minus 2 x y minus 80 y squared")[$x^(2) − 2 x y − 80 y^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 8 y close parenthesis open parenthesis x minus 10 y close parenthesis")[$( x + 8 y ) ( x − 10 y )$]
]

#math.equation(block: true, alt: "p squared minus 8 p q minus 65 q squared")[$p^(2) − 8 p q − 65 q^(2)$]

#math.equation(block: true, alt: "m squared minus 64 m n minus 65 n squared")[$m^(2) − 64 m n − 65 n^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m plus n close parenthesis open parenthesis m minus 65 n close parenthesis")[$( m + n ) ( m − 65 n )$]
]

#math.equation(block: true, alt: "p squared minus 2 p q minus 35 q squared")[$p^(2) − 2 p q − 35 q^(2)$]

#math.equation(block: true, alt: "a squared plus 5 a b minus 24 b squared")[$a^(2) + 5 a b − 24 b^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis a plus 8 b close parenthesis open parenthesis a minus 3 b close parenthesis")[$( a + 8 b ) ( a − 3 b )$]
]

#math.equation(block: true, alt: "r squared plus 3 r s minus 28 s squared")[$r^(2) + 3 r s − 28 s^(2)$]

#math.equation(block: true, alt: "x squared minus 3 x y minus 14 y squared")[$x^(2) − 3 x y − 14 y^(2)$]

#solutionbox[
prime
]

#math.equation(block: true, alt: "u squared minus 8 u v minus 24 v squared")[$u^(2) − 8 u v − 24 v^(2)$]

#math.equation(block: true, alt: "m squared minus 5 m n plus 30 n squared")[$m^(2) − 5 m n + 30 n^(2)$]

#solutionbox[
prime
]

#math.equation(block: true, alt: "c squared minus 7 c d plus 18 d squared")[$c^(2) − 7 c d + 18 d^(2)$]

#strong[Mixed Practice]

In the following exercises, factor each expression.

#math.equation(block: true, alt: "u squared minus 12 u plus 36")[$u^(2) − 12 u + 36$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis u minus 6 close parenthesis open parenthesis u minus 6 close parenthesis")[$( u − 6 ) ( u − 6 )$]
]

#math.equation(block: true, alt: "w squared plus 4 w minus 32")[$w^(2) + 4 w − 32$]

#math.equation(block: true, alt: "x squared minus 14 x minus 32")[$x^(2) − 14 x − 32$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x minus 16 close parenthesis")[$( x + 2 ) ( x − 16 )$]
]

#math.equation(block: true, alt: "y squared plus 41 y plus 40")[$y^(2) + 41 y + 40$]

#math.equation(block: true, alt: "r squared minus 20 r s plus 64 s squared")[$r^(2) − 20 r s + 64 s^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis r minus 4 s close parenthesis open parenthesis r minus 16 s close parenthesis")[$( r − 4 s ) ( r − 16 s )$]
]

#math.equation(block: true, alt: "x squared minus 16 x y plus 64 y squared")[$x^(2) − 16 x y + 64 y^(2)$]

#math.equation(block: true, alt: "k squared plus 34 k plus 120")[$k^(2) + 34 k + 120$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis k plus 4 close parenthesis open parenthesis k plus 30 close parenthesis")[$( k + 4 ) ( k + 30 )$]
]

#math.equation(block: true, alt: "m squared plus 29 m plus 120")[$m^(2) + 29 m + 120$]

#math.equation(block: true, alt: "y squared plus 10 y plus 15")[$y^(2) + 10 y + 15$]

#solutionbox[
prime
]

#math.equation(block: true, alt: "z squared minus 3 z plus 28")[$z^(2) − 3 z + 28$]

#math.equation(block: true, alt: "m squared plus m n minus 56 n squared")[$m^(2) + m n − 56 n^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis m plus 8 n close parenthesis open parenthesis m minus 7 n close parenthesis")[$( m + 8 n ) ( m − 7 n )$]
]

#math.equation(block: true, alt: "q squared minus 29 q r minus 96 r squared")[$q^(2) − 29 q r − 96 r^(2)$]

#math.equation(block: true, alt: "u squared minus 17 u v plus 30 v squared")[$u^(2) − 17 u v + 30 v^(2)$]

#solutionbox[
#math.equation(block: true, alt: "open parenthesis u minus 15 v close parenthesis open parenthesis u minus 2 v close parenthesis")[$( u − 15 v ) ( u − 2 v )$]
]

#math.equation(block: true, alt: "m squared minus 31 m n plus 30 n squared")[$m^(2) − 31 m n + 30 n^(2)$]

#math.equation(block: true, alt: "c squared minus 8 c d plus 26 d squared")[$c^(2) − 8 c d + 26 d^(2)$]

#solutionbox[
prime
]

#math.equation(block: true, alt: "r squared plus 11 r s plus 36 s squared")[$r^(2) + 11 r s + 36 s^(2)$]


==== Everyday Math

#strong[Consecutive integers] Deirdre is thinking of two consecutive integers whose product is 56. The trinomial #math.equation(block: false, alt: "x squared plus x minus 56")[$x^(2) + x − 56$] describes how these numbers are related. Factor the trinomial.

#solutionbox[
#math.equation(block: true, alt: "open parenthesis x plus 8 close parenthesis open parenthesis x minus 7 close parenthesis")[$( x + 8 ) ( x − 7 )$]
]

#strong[Consecutive integers] Deshawn is thinking of two consecutive integers whose product is 182. The trinomial #math.equation(block: false, alt: "x squared plus x minus 182")[$x^(2) + x − 182$] describes how these numbers are related. Factor the trinomial.


==== Writing Exercises

Many trinomials of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] factor into the product of two binomials #math.equation(block: false, alt: "open parenthesis x plus m close parenthesis open parenthesis x plus n close parenthesis")[$( x + m ) ( x + n )$]. Explain how you find the values of #emph[m] and #emph[n].

#solutionbox[
Answers may vary
]

How do you determine whether to use plus or minus signs in the binomial factors of a trinomial of the form #math.equation(block: false, alt: "x squared plus b x plus c")[$x^(2) + b x + c$] where #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "c")[$c$] may be positive or negative numbers?

Will factored #math.equation(block: false, alt: "x squared minus x minus 20")[$x^(2) − x − 20$] as #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis open parenthesis x minus 4 close parenthesis")[$( x + 5 ) ( x − 4 )$]. Bill factored it as #math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis open parenthesis x minus 5 close parenthesis")[$( x + 4 ) ( x − 5 )$]. Phil factored it as #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis open parenthesis x minus 4 close parenthesis")[$( x − 5 ) ( x − 4 )$]. Who is correct? Explain why the other two are wrong.

#solutionbox[
Answers may vary
]

Look at , where we factored #math.equation(block: false, alt: "y squared plus 17 y plus 60")[$y^(2) + 17 y + 60$]. We made a table listing all pairs of factors of 60 and their sums. Do you find this kind of table helpful? Why or why not?


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has the following statements all to be preceded by “I can…”. The first is “factor trinomials of the form x ^ 2 +b x + c”. The second is “factor trinomials of the form x^2 + b x y + c y ^ 2”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.], alt: "This table has the following statements all to be preceded by “I can…”. The first is “factor trinomials of the form x ^ 2 +b x + c”. The second is “factor trinomials of the form x^2 + b x y + c y ^ 2”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.", caption: none)

ⓑ After reviewing this checklist, what will you do to become confident for all goals?