6.4 General Strategy for Factoring Polynomials

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

You have now become acquainted with all the methods of factoring that you will need in this course. The following chart summarizes all the factoring methods we have covered, and outlines a strategy you should use when factoring polynomials.

Remember, a polynomial is completely factored if, other than monomials, its factors are prime!

Example 1

Factor completely: $7 x^{3} - 21 x^{2} - 70 x .$

	$7 x^{3} - 21 x^{2} - 70 x$
Is there a GCF? Yes, $7 x$ .
Factor out the GCF.	$7 x (x^{2} - 3 x - 10)$
In the parentheses, is it a binomial, trinomial, or are there more terms?
Trinomial with leading coefficient 1.
“Undo” FOIL.	$7 x open parenthesis x close parenthesis open parenthesis x close parenthesis$
	$7 x (x + 2) (x - 5)$
Is the expression factored completely? Yes.
Neither binomial can be factored.
Check your answer.
Multiply.
$7 x (x + 2) (x - 5)$
$7 x (x^{2} - 5 x + 2 x - 10)$
$7 x (x^{2} - 3 x - 10)$
$7 x^{3} - 21 x^{2} - 70 x ✓$

Be careful when you are asked to factor a binomial as there are several options!

Example 2

Factor completely: $24 y squared minus 150.$

	$24 y squared minus 150$
Is there a GCF? Yes, 6.
Factor out the GCF.	$6 open parenthesis 4 y squared minus 25 close parenthesis$
In the parentheses, is it a binomial, trinomial or are there more than three terms? Binomial.
Is it a sum? No.
Is it a difference? Of squares or cubes? Yes, squares.	$6 open parenthesis open parenthesis 2 y close parenthesis squared minus open parenthesis 5 close parenthesis squared close parenthesis$
Write as a product of conjugates.	$6 (2 y - 5) (2 y + 5)$
$Is the expression factored completely?$
$Neither binomial can be factored.$
Check:
$Multiply.$
$6 (2 y - 5) (2 y + 5)$
$6 open parenthesis 4 y squared minus 25 close parenthesis$
$24 y squared minus 150 ✓$

The next example can be factored using several methods. Recognizing the trinomial squares pattern will make your work easier.

Remember, sums of squares do not factor, but sums of cubes do!

When using the sum or difference of cubes pattern, being careful with the signs.

Example 5

Factor completely: $24 x cubed plus 81 y cubed.$

Is there a GCF? Yes, 3.
Factor it out.
In the parentheses, is it a binomial, trinomial, of are there more than three terms? Binomial.
Is it a sum or difference? Sum.
Of squares or cubes? Sum of cubes.
Write it using the sum of cubes pattern.
Is the expression factored completely? Yes.
Check by multiplying.

Example 6

Factor completely: $3 x^{5} y - 48 x y .$

	$3 x^{5} y - 48 x y$
Is there a GCF? Factor out $3 x y$	$3 x y (x^{4} - 16)$
Is the binomial a sum or difference? Of squares or cubes? Write it as a difference of squares.	$3 x y open parenthesis open parenthesis x squared close parenthesis squared minus open parenthesis 4 close parenthesis squared close parenthesis$
Factor it as a product of conjugates	$3 x y (x^{2} - 4) (x^{2} + 4)$
The first binomial is again a difference of squares.	$3 x y ({(x)}^{2} - {(2)}^{2}) (x^{2} + 4)$
Factor it as a product of conjugates.	$3 x y (x - 2) (x + 2) (x^{2} + 4)$
Is the expression factored completely? Yes.
Check your answer.
Multiply.
$3 x y (x - 2) (x + 2) (x^{2} + 4)$
$3 x y (x^{2} - 4) (x^{2} + 4)$
$3 x y (x^{4} - 16)$
$3 x^{5} y - 48 x y ✓$

Taking out the complete GCF in the first step will always make your work easier.

Example 8

Factor completely: $40 x^{2} y + 44 x y - 24 y .$

	$40 x^{2} y + 44 x y - 24 y$
Is there a GCF? Factor out the GCF, $4 y$ .	$4 y (10 x^{2} + 11 x - 6)$
Factor the trinomial with $a not equal to 1$ .	$4 y (10 x^{2} + 11 x - 6)$
	$4 y (5 x - 2) (2 x + 3)$
Is the expression factored completely? Yes.
Check your answer.
Multiply.
$4 y (5 x - 2) (2 x + 3)$
$4 y (10 x^{2} + 11 x - 6)$
$40 x^{2} y + 44 x y - 24 y ✓$

When we have factored a polynomial with four terms, most often we separated it into two groups of two terms. Remember that we can also separate it into a trinomial and then one term.

Example 9

Factor completely: $9 x^{2} - 12 x y + 4 y^{2} - 49 .$

	$9 x^{2} - 12 x y + 4 y^{2} - 49$
Is there a GCF? No.
With more than 3 terms, use grouping. Last 2 terms have no GCF. Try grouping first 3 terms.	$9 x^{2} - 12 x y + 4 y^{2} - 49$
Factor the trinomial with $a not equal to 1$ . But the first term is a perfect square.
Is the last term of the trinomial a perfect square? Yes.	${(3 x)}^{2} - 12 x y + {(2 y)}^{2} - 49$
Does the trinomial fit the pattern, $a^{2} - 2 a b + b^{2}$ ? Yes.	${(3 x)}^{2}_{↘} \underset{−2 (3 x) (2 y)}{−12 x y +}_{↙} {(2 y)}^{2} - 49$
Write the trinomial as a square.	${(3 x - 2 y)}^{2} - 49$
Is this binomial a sum or difference? Of squares or cubes? Write it as a difference of squares.	${(3 x - 2 y)}^{2} - 7^{2}$
Write it as a product of conjugates.	$((3 x - 2 y) - 7) ((3 x - 2 y) + 7)$
	$(3 x - 2 y - 7) (3 x - 2 y + 7)$
Is the expression factored completely? Yes.
Check your answer.
Multiply.
$(3 x - 2 y - 7) (3 x - 2 y + 7)$
$9 x^{2} - 6 x y - 21 x - 6 x y + 4 y^{2} + 14 y + 21 x - 14 y - 49$
$9 x^{2} - 12 x y + 4 y^{2} - 49 ✓$

Key Concepts

This chart shows the general strategies for factoring polynomials. It shows ways to find GCF of binomials, trinomials and polynomials with more than 3 terms. For binomials, we have difference of squares: a squared minus b squared equals a minus b, a plus b; sum of squares do not factor; sub of cubes: a cubed plus b cubed equals open parentheses a plus b close parentheses open parentheses a squared minus ab plus b squared close parentheses; difference of cubes: a cubed minus b cubed equals open parentheses a minus b close parentheses open parentheses a squared plus ab plus b squared close parentheses. For trinomials, we have x squared plus bx plus c where we put x as a term in each factor and we have a squared plus bx plus c. Here, if a and c are squares, we have a plus b whole squared equals a squared plus 2 ab plus b squared and a minus b whole squared equals a squared minus 2 ab plus b squared. If a and c are not squares, we use the ac method. For polynomials with more than 3 terms, we use grouping.

How to use a general strategy for factoring polynomials.
1. Is there a greatest common factor?
  Factor it out.
2. Is the polynomial a binomial, trinomial, or are there more than three terms?
  If it is a binomial:
  Is it a sum?
  Of squares? Sums of squares do not factor.
  Of cubes? Use the sum of cubes pattern.
  Is it a difference?
  Of squares? Factor as the product of conjugates.
  Of cubes? Use the difference of cubes pattern.
  If it is a trinomial:
  Is it of the form $x^{2} + b x + c ?$ Undo FOIL.
  Is it of the form $a x^{2} + b x + c ?$
  If a and c are squares, check if it fits the trinomial square pattern.
  Use the trial and error or “ac” method.
  If it has more than three terms:
  Use the grouping method.
3. Check.
  Is it factored completely?
  Do the factors multiply back to the original polynomial?

Practice Makes Perfect

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

In the following exercises, factor completely.

$2 n^{2} + 13 n - 7$

$(2 n - 1) (n + 7)$

$8 x^{2} - 9 x - 3$

$a^{5} + 9 a^{3}$

$a cubed open parenthesis a squared plus 9 close parenthesis$

$75 m cubed plus 12 m$

$121 r squared minus s squared$

$(11 r - s) (11 r + s)$

$49 b squared minus 36 a squared$

$8 m squared minus 32$

$8 (m - 2) (m + 2)$

$36 q squared minus 100$

$25 w^{2} - 60 w + 36$

$open parenthesis 5 w minus 6 close parenthesis squared$

$49 b^{2} - 112 b + 64$

$m^{2} + 14 m n + 49 n^{2}$

$open parenthesis m plus 7 n close parenthesis squared$

$64 x^{2} + 16 x y + y^{2}$

$7 b^{2} + 7 b - 42$

$7 (b + 3) (b - 2)$

$30 n^{2} + 30 n + 72$

$3 x^{4} y - 81 x y$

$3 x y (x - 3) (x^{2} + 3 x + 9)$

$4 x^{5} y - 32 x^{2} y$

$k^{4} - 16$

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

$m^{4} - 81$

$5 x^{5} y^{2} - 80 x y^{2}$

$5 x y^{2} (x^{2} + 4) (x + 2) (x - 2)$

$48 x^{5} y^{2} - 243 x y^{2}$

$15 p q - 15 p + 12 q - 12$

$3 (5 p + 4) (q - 1)$

$12 a b - 6 a + 10 b - 5$

$4 x^{2} + 40 x + 84$

$4 (x + 3) (x + 7)$

$5 q^{2} - 15 q - 90$

$4 u^{5} + 4 u^{2} v^{3}$

$4 u^{2} (u + v) (u^{2} - u v + v^{2})$

$5 m^{4} n + 320 m n^{4}$

$4 c^{2} + 20 c d + 81 d^{2}$

prime

$25 x^{2} + 35 x y + 49 y^{2}$

$10 m^{4} - 6250$

$10 (m - 5) (m + 5) (m^{2} + 25)$

$3 v^{4} - 768$

$36 x^{2} y + 15 x y - 6 y$

$3 y (3 x + 2) (4 x - 1)$

$60 x^{2} y - 75 x y + 30 y$

$8 x cubed minus 27 y cubed$

$(2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2})$

$64 x cubed plus 125 y cubed$

$y^{6} - 1$

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

$y^{6} + 1$

$9 x^{2} - 6 x y + y^{2} - 49$

$(3 x - y + 7) (3 x - y - 7)$

$16 x^{2} - 24 x y + 9 y^{2} - 64$

${(3 x + 1)}^{2} - 6 (3 x + 1) + 9$

$open parenthesis 3 x minus 2 close parenthesis squared$

${(4 x - 5)}^{2} - 7 (4 x - 5) + 12$

Writing Exercises

Explain what it mean to factor a polynomial completely.

Answers will vary.

The difference of squares $y^{4} - 625$ can be factored as $(y^{2} - 25) (y^{2} + 25) .$ But it is not completely factored. What more must be done to completely factor.

Of all the factoring methods covered in this chapter (GCF, grouping, undo FOIL, ‘ac’ method, special products) which is the easiest for you? Which is the hardest? Explain your answers.

Answers will vary.

Create three factoring problems that would be good test questions to measure your knowledge of factoring. Show the solutions.

Self Check

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

This table has 4 columns, 1 row and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statement: recognize and use the appropriate method to factor a polynomial completely. The remaining columns are blank.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

	$4 x^{2} + 8 b x - 4 a x - 8 a b$
Is there a GCF? Factor out the GCF, 4.	$4 (x^{2} + 2 b x - a x - 2 a b)$
There are four terms. Use grouping.	$\begin{matrix} 4 [x (x + 2 b) - a (x + 2 b)] \\ 4 (x + 2 b) (x - a) \end{matrix}$
Is the expression factored completely? Yes.
Check your answer. Multiply. $\begin{matrix} 4 (x + 2 b) (x - a) \\ 4 (x^{2} - a x + 2 b x - 2 a b) \\ 4 x^{2} + 8 b x - 4 a x - 8 a b ✓ \end{matrix}$