5.3 Multiply Polynomials

Multiply Monomials

We are ready to perform operations on polynomials. Since monomials are algebraic expressions, we can use the properties of exponents to multiply monomials.

Example 1

Multiply: ⓐ $open parenthesis 3 x squared close parenthesis open parenthesis −4 x cubed close parenthesis$ ⓑ $(\frac{5}{6} x^{3} y) (12 x y^{2}) .$

ⓐ

	$open parenthesis 3 x squared close parenthesis open parenthesis −4 x cubed close parenthesis$
Use the Commutative Property to rearrange the terms.	$3 \cdot (−4) \cdot x^{2} \cdot x^{3}$
Multiply.	$−12 x to the power 5$

ⓑ

	$(\frac{5}{6} x^{3} y) (12 x y^{2})$
Use the Commutative Property to rearrange the terms.	$\frac{5}{6} \cdot 12 \cdot x^{3} \cdot x \cdot y \cdot y^{2}$
Multiply.	$10 x to the power 4 y cubed$

Multiply a Polynomial by a Monomial

Multiplying a polynomial by a monomial is really just applying the Distributive Property.

Multiply a Binomial by a Binomial

Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

If you multiply binomials often enough you may notice a pattern. Notice that the first term in the result is the product of the first terms in each binomial. The second and third terms are the product of multiplying the two outer terms and then the two inner terms. And the last term results from multiplying the two last terms,

We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘First, Outer, Inner, Last’. We use this as another method of multiplying binomials. The word FOIL is easy to remember and ensures we find all four products.

Let’s multiply $(x + 3) (x + 7)$ using both methods.

The figure shows how four terms in the product of two binomials can be remembered according to the mnemonic acronym FOIL. The example is the quantity x plus 3 in parentheses times the quantity x plus 7 in parentheses. The expression is expanded as in the previous examples by using the distributive property twice. After distributing the quantity x plus 7 in parentheses the result is x times the quantity x plus 7 in parentheses plus 3 times the quantity x plus 7 in parentheses. Then the x is distributed the x plus 7 and the 3 is distributed to the x plus 7 to get x squared plus 7 x plus 3 x plus 21. The letter F is written under the term x squared since it was the product of the first terms in the binomials. The letter O is written under the 7 x term sine it was the product of the outer terms in the binomials. The letter I is written under the 3 x term since it was the product of the inner terms in the binomials. The letter L is written under the 21 since it was the product of the last terms in the binomial. The original expression is shown again with four arrows connecting the first, outer, inner, and last terms in the binomials showing how the four terms can be determined directly from the factored form.

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

When you multiply by the FOIL method, drawing the lines will help your brain focus on the pattern and make it easier to apply.

Now we will do an example where we use the FOIL pattern to multiply two binomials.

The final products in the last example were trinomials because we could combine the two middle terms. This is not always the case.

Example 5

Multiply: ⓐ $(n^{2} + 4) (n - 1)$ ⓑ $(3 p q + 5) (6 p q - 11) .$

ⓐ



Step 1. Multiply the First terms.
Step 2. Multiply the Outer terms.
Step 3. Multiply the Inner terms.
Step 4. Multiply the Last terms.
Step 5. Combine like terms—there are none.

ⓑ



Step 1. Multiply the First terms.
Step 2. Multiply the Outer terms.
Step 3. Multiply the Inner terms.
Step 4. Multiply the Last terms.
Step 5. Combine like terms.

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.

This figure shows the vertical multiplication of 23 and 46. The number 23 is above the number 46. Below this, there is the partial product 138 over the partial product 92. The final product is at the bottom and is 1058. Text on the right side of the image says “You start by multiplying 23 by 6 to get 138. Then you multiply 23 by 4, lining up the partial product in the correct columns. Last, you add the partial products.”

Now we’ll apply this same method to multiply two binomials.

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The methods are listed here all together, to help you remember them.

Multiply a Polynomial by a Polynomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we’re ready to multiply a polynomial by a polynomial. Remember, FOIL will not work in this case, but we can use either the Distributive Property or the Vertical Method.

Example 7

Multiply $(b + 3) (2 b^{2} - 5 b + 8)$ using ⓐ the Distributive Property and ⓑ the Vertical Method.

ⓐ


Distribute.
Multiply.
Combine like terms.

ⓑ It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

Multiply $(2 b^{2} - 5 b + 8)$ by 3. Multiply $(2 b^{2} - 5 b + 8)$ by $b$ .
Add like terms.

We have now seen two methods you can use to multiply a polynomial by a polynomial. After you practice each method, you’ll probably find you prefer one way over the other. We list both methods are listed here, for easy reference.

Multiply Special Products

Mathematicians like to look for patterns that will make their work easier. A good example of this is squaring binomials. While you can always get the product by writing the binomial twice and multiplying them, there is less work to do if you learn to use a pattern. Let’s start by looking at three examples and look for a pattern.

Look at these results. Do you see any patterns?

The figure shows three examples of squaring a binomial. In the first example x plus 9 is squared to get x plus 9 times x plus 9 which is x squared plus 9 x plus 9 x plus 81 which simplifies to x squared plus 18 x plus 81. Colors show that x squared comes from the square of the x in the original binomial and 81 comes from the square of the 9 in the original binomial. In the second example y minus 7 is squared to get y minus y times y minus 7 which is y squared minus 7 y minus 7 y plus 49 which simplifies to y squared minus 14 y plus 49. Colors show that y squared comes from the square of the y in the original binomial and 49 comes from the square of the negative 7 in the original binomial. In the third example 2 x plus 3 is squared to get 2 x plus 3 times 2 x plus 3 which is 4 x squared plus 6 x plus 6 x plus 9 which simplifies to 4 x squared plus 12 x plus 9. Colors show that 4 x squared comes from the square of the 2 x in the original binomial and 9 comes from the square of the 3 in the original binomial.

What about the number of terms? In each example we squared a binomial and the result was a trinomial.

${(a + b)}^{2} =___+___+___$

Now look at the first term in each result. Where did it come from?

The first term is the product of the first terms of each binomial. Since the binomials are identical, it is just the square of the first term!

${(a + b)}^{2} = a^{2} +___+___$

To get the first term of the product, square the first term.

Where did the last term come from? Look at the examples and find the pattern.

The last term is the product of the last terms, which is the square of the last term.

${(a + b)}^{2} =___+___+ b^{2}$

To get the last term of the product, square the last term.

Finally, look at the middle term. Notice it came from adding the “outer” and the “inner” terms—which are both the same! So the middle term is double the product of the two terms of the binomial.

$\begin{matrix} {(a + b)}^{2} =___+ 2 a b +___ \\ {(a - b)}^{2} =___- 2 a b +___ \end{matrix}$

To get the middle term of the product, multiply the terms and double their product.

Putting it all together:

Example 8

Multiply: ⓐ $open parenthesis x plus 5 close parenthesis squared$ ⓑ $open parenthesis 2 x minus 3 y close parenthesis squared.$

ⓐ


Square the first term.
Square the last term.
Double their product.
Simplify.

ⓑ


Use the pattern.
Simplify.

We just saw a pattern for squaring binomials that we can use to make multiplying some binomials easier. Similarly, there is a pattern for another product of binomials. But before we get to it, we need to introduce some vocabulary.

A pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form $(a - b), (a + b) .$

There is a nice pattern for finding the product of conjugates. You could, of course, simply FOIL to get the product, but using the pattern makes your work easier. Let’s look for the pattern by using FOIL to multiply some conjugate pairs.

The figure shows three examples of multiplying a binomial with its conjugate. In the first example x plus 9 is multiplied with x minus 9 to get x squared minus 9 x plus 9 x minus 81 which simplifies to x squared minus 81. Colors show that x squared comes from the square of the x in the original binomial and 81 comes from the square of the 9 in the original binomial. In the second example y minus 8 is multiplied with y plus 8 to get y squared plus 8 y minus 8 y minus 64 which simplifies to y squared minus 64. Colors show that y squared comes from the square of the y in the original binomial and 64 comes from the square of the 8 in the original binomial. In the third example 2 x minus 5 is multiplied with 2 x plus 5 to get 4 x squared plus 10 x minus 10 x minus 25 which simplifies to 4 x squared minus 25. Colors show that 4 x squared comes from the square of the 2 x in the original binomial and 25 comes from the square of the 5 in the original binomial.

What do you observe about the products?

The product of the two binomials is also a binomial! Most of the products resulting from FOIL have been trinomials.

Each first term is the product of the first terms of the binomials, and since they are identical it is the square of the first term.

$(a + b) (a - b) = a^{2} -___$

To get the first term, square the first term.

The last term came from multiplying the last terms, the square of the last term.

$(a + b) (a - b) = a^{2} - b^{2}$

To get the last term, square the last term.

Why is there no middle term? Notice the two middle terms you get from FOIL combine to 0 in every case, the result of one addition and one subtraction.

The product of conjugates is always of the form $a^{2} - b^{2} .$ This is called a difference of squares.

This leads to the pattern:

Example 9

Multiply using the product of conjugates pattern: ⓐ $(2 x + 5) (2 x - 5)$ ⓑ $(5 m - 9 n) (5 m + 9 n) .$

ⓐ

Are the binomials conjugates?
It is the product of conjugates.
Square the first term, $2 x.$
Square the last term, $5 .$
Simplify. The product is a difference of squares.

ⓑ


This fits the pattern.
Use the pattern.
Simplify.

We just developed special product patterns for Binomial Squares and for the Product of Conjugates. The products look similar, so it is important to recognize when it is appropriate to use each of these patterns and to notice how they differ. Look at the two patterns together and note their similarities and differences.

Note

Comparing the Special Product Patterns

Binomial Squares	Product of Conjugates
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$	$(a - b) (a + b) = a^{2} - b^{2}$
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
⋅ Squaring a binomial	⋅ Multiplying conjugates
⋅ Product is a trinomial	⋅ Product is a binomial.
⋅ Inner and outer terms with FOIL are the same.	⋅ Inner and outer terms with FOIL are opposites.
⋅ Middle term is double the product of the terms	⋅ There is no middle term.

Example 10

Choose the appropriate pattern and use it to find the product:

ⓐ $(2 x - 3) (2 x + 3)$ ⓑ $open parenthesis 8 x minus 5 close parenthesis squared$ ⓒ $open parenthesis 6 m plus 7 close parenthesis squared$ ⓓ $(5 x - 6) (6 x + 5) .$

ⓐ $(2 x - 3) (2 x + 3)$

These are conjugates. They have the same first numbers, and the same last numbers, and one binomial is a sum and the other is a difference. It fits the Product of Conjugates pattern.


Use the pattern.
Simplify.

ⓑ $open parenthesis 8 x minus 5 close parenthesis squared$

We are asked to square a binomial. It fits the binomial squares pattern.


Use the pattern.
Simplify.

Again, we will square a binomial so we use the binomial squares pattern.

	$A mathematical expression displaying the fraction (a+b)/(6m+7) raised to the power of 2, with 'a' and 'b' highlighted in red.$
Use the pattern.
Simplify.

ⓓ $(5 x - 6) (6 x + 5)$

This product does not fit the patterns, so we will use FOIL.

$\begin{matrix} (5 x - 6) (6 x + 5) \\ Use FOIL. & 30 x^{2} + 25 x - 36 x - 30 \\ Simplify. & 30 x^{2} - 11 x - 30 \end{matrix}$

Multiply Polynomial Functions

Just as polynomials can be multiplied, polynomial functions can also be multiplied.

Key Concepts

How to use the FOIL method to multiply two binomials.
Multiplying Two Binomials: To multiply binomials, use the:
- Distributive Property
- FOIL Method
Multiplying a Polynomial by a Polynomial: To multiply a trinomial by a binomial, use the:
- Distributive Property
- Vertical Method
Binomial Squares Pattern
If a and b are real numbers,
Product of Conjugates Pattern
If $a, b$ are real numbers

The product is called a difference of squares.
To multiply conjugates, square the first term, square the last term, write it as a difference of squares.

Comparing the Special Product Patterns

Binomial Squares	Product of Conjugates
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$	$(a - b) (a + b) = a^{2} - b^{2}$
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
⋅ Squaring a binomial	⋅ Multiplying conjugates
⋅ Product is a trinomial	⋅ Product is a binomial.
⋅ Inner and outer terms with FOIL are the same.	⋅ Inner and outer terms with FOIL are opposites.
⋅ Middle term is double the product of the terms	⋅ There is no middle term.

Multiplication of Polynomial Functions:
- For functions $f (x)$ and $g (x),$
  
  $(f \cdot g) (x) = f (x) \cdot g (x)$

Practice Makes Perfect

Multiply Monomials

In the following exercises, multiply the monomials.

ⓐ $(6 y^{7}) (−3 y^{4})$
ⓑ $(\frac{4}{7} r s^{2}) (14 r s^{3})$

ⓐ $open parenthesis −10 x to the power 5 close parenthesis open parenthesis −3 x cubed close parenthesis$
ⓑ $(\frac{5}{8} x^{3} y) (24 x^{5} y)$

ⓐ $30 x to the power 8$ ⓑ $15 x to the power 8 y squared$

ⓐ $open parenthesis −8 u to the power 6 close parenthesis open parenthesis −9 u close parenthesis$
ⓑ $(\frac{2}{3} x^{2} y) (\frac{3}{4} x y^{2})$

ⓐ $open parenthesis −6 c to the power 4 close parenthesis open parenthesis −12 c close parenthesis$
ⓑ $(\frac{3}{5} m^{3} n^{2}) (\frac{5}{9} m^{2} n^{3})$

ⓐ $72 c to the power 5$ ⓑ $\frac{1}{3} m^{5} n^{5}$

Multiply a Polynomial by a Monomial

In the following exercises, multiply.

ⓐ $−8 x (x^{2} + 2 x - 15)$
ⓑ $5 p q^{3} (p^{2} - 2 p q + 6 q^{2})$

ⓐ $−5 t (t^{2} + 3 t - 18);$
ⓑ $9 r^{3} s (r^{2} - 3 r s + 5 s^{2})$

ⓐ $−5 t^{3} - 15 t^{2} + 90 t$
ⓑ $9 s r^{5} - 27 s^{2} r^{4} + 45 s^{3} r^{3}$

ⓐ $−8 y (y^{2} + 2 y - 15)$
ⓑ $−4 y^{2} z^{2} (3 y^{2} + 12 y z - z^{2})$

ⓐ $−5 m (m^{2} + 3 m - 18)$
ⓑ $−3 x^{2} y^{2} (7 x^{2} + 10 x y - y^{2})$

ⓐ $−5 m^{3} - 15 m^{2} + 90 m$
ⓑ $−21 x^{4} y^{2} - 30 x^{3} y^{3} + 3 x^{2} y^{4}$

Multiply a Binomial by a Binomial

In the following exercises, multiply the binomials using ⓐ the Distributive Property; ⓑ the FOIL method; ⓒ the Vertical Method.

$(w + 5) (w + 7)$

$(y + 9) (y + 3)$

$y^{2} + 12 y + 27$

$(4 p + 11) (5 p - 4)$

$(7 q + 4) (3 q - 8)$

$21 q^{2} - 44 q - 32$

In the following exercises, multiply the binomials. Use any method.

$(x + 8) (x + 3)$

$(y - 6) (y - 2)$

$y^{2} - 8 y + 12$

$(2 t - 9) (10 t + 1)$

$(6 p + 5) (p + 1)$

$6 p^{2} + 11 p + 5$

$(q - 5) (q + 8)$

$(m + 11) (m - 4)$

$m^{2} + 7 m - 44$

$(7 m + 1) (m - 3)$

$(3 r - 8) (11 r + 1)$

$33 r^{2} - 85 r - 8$

$(x^{2} + 3) (x + 2)$

$(y^{2} - 4) (y + 3)$

$y^{3} + 3 y^{2} - 4 y - 12$

$(5 a b - 1) (2 a b + 3)$

$(2 x y + 3) (3 x y + 2)$

$6 x^{2} y^{2} + 13 x y + 6$

$(x^{2} + 8) (x^{2} - 5)$

$(y^{2} - 7) (y^{2} - 4)$

$y^{4} - 11 y^{2} + 28$

$(6 p q - 3) (4 p q - 5)$

$(3 r s - 7) (3 r s - 4)$

$9 r^{2} s^{2} - 33 r s + 28$

Multiply a Polynomial by a Polynomial

In the following exercises, multiply using ⓐ the Distributive Property; ⓑ the Vertical Method.

$(x + 5) (x^{2} + 4 x + 3)$

$(u + 4) (u^{2} + 3 u + 2)$

$u^{3} + 7 u^{2} + 14 u + 8$

$(y + 8) (4 y^{2} + y - 7)$

$(a + 10) (3 a^{2} + a - 5)$

$3 a^{3} + 31 a^{2} + 5 a - 50$

$(y^{2} - 3 y + 8) (4 y^{2} + y - 7)$

$(2 a^{2} - 5 a + 10) (3 a^{2} + a - 5)$

$6 a^{4} - 13 a^{3} + 15 a^{2} + 35 a - 50$

Multiply Special Products

In the following exercises, multiply. Use either method.

$(w - 7) (w^{2} - 9 w + 10)$

$(p - 4) (p^{2} - 6 p + 9)$

$p^{3} - 10 p^{2} + 33 p - 36$

$(3 q + 1) (q^{2} - 4 q - 5)$

$(6 r + 1) (r^{2} - 7 r - 9)$

$6 r^{3} - 41 r^{2} - 61 r - 9$

In the following exercises, square each binomial using the Binomial Squares Pattern.

$open parenthesis w plus 4 close parenthesis squared$

$open parenthesis q plus 12 close parenthesis squared$

$q^{2} + 24 q + 144$

$open parenthesis 3 x minus y close parenthesis squared$

$open parenthesis 2 y minus 3 z close parenthesis squared$

$4 y^{2} - 12 y z + 9 z^{2}$

${(y + \frac{1}{4})}^{2}$

${(x + \frac{2}{3})}^{2}$

$x^{2} + \frac{4}{3} x + \frac{4}{9}$

${(\frac{1}{5} x - \frac{1}{7} y)}^{2}$

${(\frac{1}{8} x - \frac{1}{9} y)}^{2}$

$\frac{1}{64} x^{2} - \frac{1}{36} x y + \frac{1}{81} y^{2}$

$open parenthesis 3 x squared plus 2 close parenthesis squared$

$open parenthesis 5 u squared plus 9 close parenthesis squared$

$25 u^{4} + 90 u^{2} + 81$

$open parenthesis 4 y cubed minus 2 close parenthesis squared$

$open parenthesis 8 p cubed minus 3 close parenthesis squared$

$64 p^{6} - 48 p^{3} + 9$

In the following exercises, multiply each pair of conjugates using the Product of Conjugates Pattern.

$(5 k + 6) (5 k - 6)$

$(8 j + 4) (8 j - 4)$

$64 j squared minus 16$

$(11 k + 4) (11 k - 4)$

$(9 c + 5) (9 c - 5)$

$81 c squared minus 25$

$(9 c - 2 d) (9 c + 2 d)$

$(7 w + 10 x) (7 w - 10 x)$

$49 w squared minus 100 x squared$

$(m + \frac{2}{3} n) (m - \frac{2}{3} n)$

$(p + \frac{4}{5} q) (p - \frac{4}{5} q)$

$p^{2} - \frac{16}{25} q^{2}$

$(a b - 4) (a b + 4)$

$(x y - 9) (x y + 9)$

$x squared y squared minus 81$

$(12 p^{3} - 11 q^{2}) (12 p^{3} + 11 q^{2})$

$(15 m^{2} - 8 n^{4}) (15 m^{2} + 8 n^{4})$

$225 m^{4} - 64 n^{8}$

In the following exercises, find each product.

$(p - 3) (p + 3)$

$open parenthesis t minus 9 close parenthesis squared$

$t^{2} - 18 t + 81$

${(m + n)}^{2}$

$(2 x + y) (x - 2 y)$

$2 x^{2} - 3 x y - 2 y^{2}$

$open parenthesis 2 r plus 12 close parenthesis squared$

$(3 p + 8) (3 p - 8)$

$9 p squared minus 64$

$(7 a + b) (a - 7 b)$

$open parenthesis k minus 6 close parenthesis squared$

$k^{2} - 12 k + 36$

${(a^{5} - 7 b)}^{2}$

$(x^{2} + 8 y) (8 x - y^{2})$

$8 x^{3} - x^{2} y^{2} + 64 x y - 8 y^{3}$

$(r^{6} + s^{6}) (r^{6} - s^{6})$

${(y^{4} + 2 z)}^{2}$

$y^{8} + 4 y^{4} z + 4 z^{2}$

$(x^{5} + y^{5}) (x^{5} - y^{5})$

$open parenthesis m cubed minus 8 n close parenthesis squared$

$m^{6} - 16 m^{3} n + 64 n^{2}$

$open parenthesis 9 p plus 8 q close parenthesis squared$

$(r^{2} - s^{3}) (r^{3} + s^{2})$

$r^{5} + r^{2} s^{2} - r^{3} s^{3} - s^{5}$

Mixed Practice

$(10 y - 6) + (4 y - 7)$

$(15 p - 4) + (3 p - 5)$

$18 p minus 9$

$(x^{2} - 4 x - 34) - (x^{2} + 7 x - 6)$

$(j^{2} - 8 j - 27) - (j^{2} + 2 j - 12)$

$−10 j minus 15$

$(\frac{1}{5} f^{8}) (20 f^{3})$

$(\frac{1}{4} d^{5}) (36 d^{2})$

$9 d to the power 7$

$open parenthesis 4 a cubed b close parenthesis open parenthesis 9 a squared b to the power 6 close parenthesis$

$(6 m^{4} n^{3}) (7 m n^{5})$

$42 m^{5} n^{8}$

$−5 m (m^{2} + 3 m - 18)$

$5 q^{3} (q^{2} - 2 q + 6)$

$5 q^{5} - 10 q^{4} + 30 q^{3}$

$(s - 7) (s + 9)$

$(y^{2} - 2 y) (y + 1)$

$y^{3} - y^{2} - 2 y$

$(5 x - y) (x - 4)$

$(6 k - 1) (k^{2} + 2 k - 4)$

$6 k^{3} + 11 k^{2} - 26 k + 4$

$(3 x - 11 y) (3 x - 11 y)$

$(11 - b) (11 + b)$

$121 minus b squared$

$(r s - \frac{2}{7}) (r s + \frac{2}{7})$

$(2 x^{2} - 3 y^{4}) (2 x^{2} + 3 y^{4})$

$4 x^{4} - 9 y^{8}$

$open parenthesis m minus 15 close parenthesis squared$

$open parenthesis 3 d plus 1 close parenthesis squared$

$9 d^{2} + 6 d + 1$

$open parenthesis 4 a plus 10 close parenthesis squared$

${(3 z + \frac{1}{5})}^{2}$

$9 z^{2} + \frac{6}{5} z + \frac{1}{25}$

Multiply Polynomial Functions

For functions $f (x) = x + 2$ and $g (x) = 3 x^{2} - 2 x + 4,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis −1 close parenthesis$

For functions $f (x) = x - 1$ and $g (x) = 4 x^{2} + 3 x - 5,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis −2 close parenthesis$

ⓐ $(f \cdot g) (x) = 4 x^{3} - x^{2} - 8 x + 5$
ⓑ $(f \cdot g) (−2) = −15$

For functions $f (x) = 2 x - 7$ and $g (x) = 2 x + 7,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis −3 close parenthesis$

For functions $f (x) = 7 x - 8$ and $g (x) = 7 x + 8,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis −2 close parenthesis$

ⓐ $(f \cdot g) (x) = 49 x^{2} - 64$
ⓑ $(f \cdot g) (−2) = 132$

For functions $f (x) = x^{2} - 5 x + 2$ and $g (x) = x^{2} - 3 x - 1,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis −1 close parenthesis$

For functions $f (x) = x^{2} + 4 x - 3$ and $g (x) = x^{2} + 2 x + 4,$ find ⓐ $(f \cdot g) (x)$ ⓑ $open parenthesis f times g close parenthesis open parenthesis 1 close parenthesis$

ⓐ $(f \cdot g) (x) = x^{4} + 6 x^{3} + 9 x^{2} + 10 x - 12$ ⓑ $(f \cdot g) (1) = 14$

Writing Exercises

Which method do you prefer to use when multiplying two binomials: the Distributive Property or the FOIL method? Why? Which method do you prefer to use when multiplying a polynomial by a polynomial: the Distributive Property or the Vertical Method? Why?

Multiply the following:

$\begin{matrix} (x + 2) (x - 2) \\ (y + 7) (y - 7) \\ (w + 5) (w - 5) \end{matrix}$

Explain the pattern that you see in your answers.

Answers will vary.

Multiply the following:

$\begin{matrix} (p + 3) (p + 3) \\ (q + 6) (q + 6) \\ (r + 1) (r + 1) \end{matrix}$

Explain the pattern that you see in your answers.

Why does ${(a + b)}^{2}$ result in a trinomial, but $(a - b) (a + b)$ result in a binomial?

Answers will vary.

Self Check

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

A self-assessment chart for students to rate their ability to multiply various polynomial types (monomials, binomials, general polynomials, special products) using 'Confidently,' 'With some help,' or 'No-I don't get it!' categories.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

	$3 x^{3} y (x^{2} - 8 x y + y^{2})$
Distribute.	$3 x^{3} y \cdot x^{2} + (3 x^{3} y) \cdot (−8 x y) + (3 x^{3} y) \cdot y^{2}$
Multiply.	$3 x^{5} y - 24 x^{4} y^{2} + 3 x^{3} y^{3}$

	$(f \cdot g) (x) = f (x) \cdot g (x)$
Substitute for $f (x) and g (x) .$	$(f \cdot g) (x) = (x + 2) (x^{2} - 3 x - 4)$
Multiply the polynomials.	$(f \cdot g) (x) = x (x^{2} - 3 x - 4) + 2 (x^{2} - 3 x - 4)$
Distribute.	$(f \cdot g) (x) = x^{3} - 3 x^{2} - 4 x + 2 x^{2} - 6 x - 8$
Combine like terms.	$(f \cdot g) (x) = x^{3} - x^{2} - 10 x - 8$

	$(f \cdot g) (x) = x^{3} - x^{2} - 10 x - 8$
To find $open parenthesis f times g close parenthesis open parenthesis 2 close parenthesis,$ substitute $x equals 2.$	$(f \cdot g) (2) = 2^{3} - 2^{2} - 10 \cdot 2 - 8$
	$(f \cdot g) (2) = 8 - 4 - 20 - 8$
	$(f \cdot g) (2) = −24$


Distribute $open parenthesis y plus 8 close parenthesis.$
Distribute again.
Combine like terms.