General Strategy for Factoring Polynomials

Lynn Marecek

Factoring

General Strategy for Factoring Polynomials

Learning Objectives

By the end of this section, you will be able to:

Recognize and use the appropriate method to factor a polynomial completely

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.

General Strategy for Factoring Polynomials

Use a general strategy for factoring polynomials.

Is there a greatest common factor?

Factor it out.
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 ${x}^{2}+bx+c?$ Undo FOIL.
- Is it of the form $a{x}^{2}+bx+c?$
  
  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.
Check.

Is it factored completely?

Do the factors multiply back to the original polynomial?

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

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

$\begin{array}{cccccc}& & & & & 7{x}^{3}-21{x}^{2}-70x\hfill \\ \text{Is there a GCF? Yes,}\phantom{\rule{0.2em}{0ex}}7x.\hfill & & & & & \\ \text{Factor out the GCF.}\hfill & & & & & 7x\left({x}^{2}-3x-10\right)\hfill \\ \begin{array}{c}\text{In the parentheses, is it a binomial, trinomial,}\hfill \\ \text{or are there more terms?}\hfill \\ \text{Trinomial with leading coefficient 1.}\hfill \end{array}\hfill & & & & & \\ \text{``Undo'' FOIL.}\hfill & & & & & 7x\left(x\phantom{\rule{1.6em}{0ex}}\right)\left(x\phantom{\rule{1.4em}{0ex}}\right)\hfill \\ & & & & & 7x\left(x+2\right)\left(x-5\right)\hfill \\ \text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Neither binomial can be factored.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \text{Multiply.}\hfill & & & & & \\ \\ \\ \hfill 7x\left(x+2\right)\left(x-5\right)\hfill & & & & & \\ \\ \hfill 7x\left({x}^{2}-5x+2x-10\right)\hfill & & & & & \\ \hfill 7x\left({x}^{2}-3x-10\right)\hfill & & & & & \\ \hfill 7{x}^{3}-21{x}^{2}-70x✓\hfill & & & & & \end{array}$

Factor completely: $8{y}^{3}+16{y}^{2}-24y.$

$8y\left(y-1\right)\left(y+3\right)$

Factor completely: $5{y}^{3}-15{y}^{2}-270y.$

$5y\left(y-9\right)\left(y+6\right)$

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

Factor completely: $24{y}^{2}-150.$

$\begin{array}{cccccc}& & & & & \hfill 24{y}^{2}-150\hfill \\ \text{Is there a GCF? Yes, 6.}\hfill & & & & & \\ \text{Factor out the GCF.}\hfill & & & & & \hfill 6\left(4{y}^{2}-25\right)\hfill \\ \text{In the parentheses, is it a binomial, trinomial}\hfill & & & & & \\ \text{or are there more than three terms? Binomial.}\hfill & & & & & \\ \text{Is it a sum? No.}\hfill & & & & & \\ \text{Is it a difference? Of squares or cubes? Yes, squares.}\hfill & & & & & \hfill 6\left({\left(2y\right)}^{2}-{\left(5\right)}^{2}\right)\hfill \\ \text{Write as a product of conjugates.}\hfill & & & & & \hfill 6\left(2y-5\right)\left(2y+5\right)\hfill \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Is the expression factored completely?}\hfill & & & & & \\ \phantom{\rule{2em}{0ex}}\text{Neither binomial can be factored.}\hfill & & & & & \\ \text{Check:}\hfill & & & & & \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Multiply.}\hfill & & & & & \\ \\ \hfill 6\left(2y-5\right)\left(2y+5\right)\hfill & & & & & \\ \\ \hfill 6\left(4{y}^{2}-25\right)\hfill & & & & & \\ \hfill 24{y}^{2}-150✓\hfill & & & & & \end{array}$

Factor completely: $16{x}^{3}-36x.$

$4x\left(2x-3\right)\left(2x+3\right)$

Factor completely: $27{y}^{2}-48.$

$3\left(3y-4\right)\left(3y+4\right)$

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

Factor completely: $4{a}^{2}-12ab+9{b}^{2}.$

$\begin{array}{cccccc}& & & & & \hfill 4{a}^{2}-12ab+9{b}^{2}\hfill \\ \text{Is there a GCF? No.}\hfill & & & & & \\ \text{Is it a binomial, trinomial, or are there more terms?}\hfill & & & & & \\ \text{Trinomial with}\phantom{\rule{0.2em}{0ex}}a\ne 1.\phantom{\rule{0.2em}{0ex}}\text{But the first term is a perfect square.}\hfill & & & & & \\ \text{Is the last term a perfect square? Yes.}\hfill & & & & & \hfill {\left(2a\right)}^{2}-12ab+{\left(3b\right)}^{2}\hfill \\ \text{Does it fit the pattern,}\phantom{\rule{0.2em}{0ex}}{a}^{2}-2ab+{b}^{2}?\phantom{\rule{0.2em}{0ex}}\text{Yes.}\hfill & & & & & \hfill {\left(2a\right)}^{2}{}_{\text{↘}}\underset{-2\left(2a\right)\left(3b\right)}{-12ab+}{}_{\text{↙}}{\left(3b\right)}^{2}\hfill \\ \text{Write it as a square.}\hfill & & & & & \hfill {\left(2a-3b\right)}^{2}\hfill \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \phantom{\rule{2em}{0ex}}\text{The binomial cannot be factored.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Multiply.}\hfill & & & & & \\ \hfill {\left(2a-3b\right)}^{2}\hfill & & & & & \\ \hfill {\left(2a\right)}^{2}-2·2a·3b+{\left(3b\right)}^{2}\hfill & & & & & \\ \hfill 4{a}^{2}-12ab+9{b}^{2}✓\hfill & & & & & \end{array}$

Factor completely: $4{x}^{2}+20xy+25{y}^{2}.$

${\left(2x+5y\right)}^{2}$

Factor completely: $9{x}^{2}-24xy+16{y}^{2}.$

${\left(3x-4y\right)}^{2}$

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

Factor completely $12{x}^{3}{y}^{2}+75x{y}^{2}.$

$\begin{array}{cccccc}& & & & & \hfill 12{x}^{3}{y}^{2}+75x{y}^{2}\hfill \\ \text{Is there a GCF? Yes,}\phantom{\rule{0.2em}{0ex}}3x{y}^{2}.\hfill & & & & & \\ \text{Factor out the GCF.}\hfill & & & & & \hfill 3x{y}^{2}\left(4{x}^{2}+25\right)\hfill \\ \text{In the parentheses, is it a binomial, trinomial, or are}\hfill & & & & & \\ \text{there more than three terms? Binomial.}\hfill & & & & & \\ \\ \text{Is it a sum? Of squares? Yes.}\hfill & & & & & \text{Sums of squares are prime.}\hfill \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Check:}\hfill & & & & & \\ \\ \\ \phantom{\rule{2em}{0ex}}\text{Multiply.}\hfill & & & & & \\ \hfill 3x{y}^{2}\left(4{x}^{2}+25\right)\hfill & & & & & \\ \hfill 12{x}^{3}{y}^{2}+75x{y}^{2}✓\hfill & & & & & \end{array}$

Factor completely: $50{x}^{3}y+72xy.$

$2xy\left(25{x}^{2}+36\right)$

Factor completely: $27x{y}^{3}+48xy.$

$3xy\left(9{y}^{2}+16\right)$

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

Factor completely: $24{x}^{3}+81{y}^{3}.$

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.

Factor completely: $250{m}^{3}+432{n}^{3}.$

$2\left(5m+6n\right)\left(25{m}^{2}-30mn+36{n}^{2}\right)$

Factor completely: $2{p}^{3}+54{q}^{3}.$

$2\left(p+3q\right)\left({p}^{2}-3pq+9{q}^{2}\right)$

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

$\begin{array}{cccccc}& & & & & \hfill 3{x}^{5}y-48xy\hfill \\ \text{Is there a GCF? Factor out}\phantom{\rule{0.2em}{0ex}}3xy\hfill & & & & & \hfill 3xy\left({x}^{4}-16\right)\hfill \\ \begin{array}{c}\text{Is the binomial a sum or difference? Of squares or cubes?}\hfill \\ \text{Write it as a difference of squares.}\hfill \end{array}\hfill & & & & & \hfill 3xy\left({\left({x}^{2}\right)}^{2}-{\left(4\right)}^{2}\right)\hfill \\ \text{Factor it as a product of conjugates}\hfill & & & & & \hfill 3xy\left({x}^{2}-4\right)\left({x}^{2}+4\right)\hfill \\ \text{The first binomial is again a difference of squares.}\hfill & & & & & \hfill 3xy\left({\left(x\right)}^{2}-{\left(2\right)}^{2}\right)\left({x}^{2}+4\right)\hfill \\ \text{Factor it as a product of conjugates.}\hfill & & & & & \hfill 3xy\left(x-2\right)\left(x+2\right)\left({x}^{2}+4\right)\hfill \\ \text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \text{Multiply.}\hfill & & & & & \\ \\ \\ \hfill 3xy\left(x-2\right)\left(x+2\right)\left({x}^{2}+4\right)\hfill & & & & & \\ \hfill 3xy\left({x}^{2}-4\right)\left({x}^{2}+4\right)\hfill & & & & & \\ \hfill 3xy\left({x}^{4}-16\right)\hfill & & & & & \\ \hfill 3{x}^{5}y-48xy✓\hfill & & & & & \end{array}$

Factor completely: $4{a}^{5}b-64ab.$

$4ab\left({a}^{2}+4\right)\left(a-2\right)\left(a+2\right)$

Factor completely: $7x{y}^{5}-7xy.$

$7xy\left({y}^{2}+1\right)\left(y-1\right)\left(y+1\right)$

Factor completely: $4{x}^{2}+8bx-4ax-8ab.$

$\begin{array}{cccccc}& & & & & \hfill 4{x}^{2}+8bx-4ax-8ab\hfill \\ \text{Is there a GCF? Factor out the GCF, 4.}\hfill & & & & & \hfill 4\left({x}^{2}+2bx-ax-2ab\right)\hfill \\ \text{There are four terms. Use grouping.}\hfill & & & & & \hfill 4\left[x\left(x+2b\right)-a\left(x+2b\right)\right]\hfill \\ & & & & & \hfill 4\left(x+2b\right)\left(x-a\right)\hfill \\ \text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \text{Multiply.}\hfill & & & & & \\ \\ \\ \hfill \phantom{\rule{4em}{0ex}}4\left(x+2b\right)\left(x-a\right)\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}4\left({x}^{2}-ax+2bx-2ab\right)\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}4{x}^{2}+8bx-4ax-8ab✓\hfill & & & & & \end{array}$

Factor completely: $6{x}^{2}-12xc+6bx-12bc.$

$6\left(x+b\right)\left(x-2c\right)$

Factor completely: $16{x}^{2}+24xy-4x-6y.$

$2\left(4x-1\right)\left(2x+3y\right)$

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

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

$\begin{array}{cccccc}& & & & & \hfill 40{x}^{2}y+44xy-24y\hfill \\ \text{Is there a GCF? Factor out the GCF,}\phantom{\rule{0.2em}{0ex}}4y.\hfill & & & & & \hfill 4y\left(10{x}^{2}+11x-6\right)\hfill \\ \text{Factor the trinomial with}\phantom{\rule{0.2em}{0ex}}a\ne 1.\hfill & & & & & \hfill 4y\left(10{x}^{2}+11x-6\right)\hfill \\ & & & & & \hfill 4y\left(5x-2\right)\left(2x+3\right)\hfill \\ \text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \text{Multiply.}\hfill & & & & & \\ \\ \\ \hfill \phantom{\rule{4em}{0ex}}4y\left(5x-2\right)\left(2x+3\right)\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}4y\left(10{x}^{2}+11x-6\right)\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}40{x}^{2}y+44xy-24y✓\hfill & & & & & \end{array}$

Factor completely: $4{p}^{2}q-16pq+12q.$

$4q\left(p-3\right)\left(p-1\right)$

Factor completely: $6p{q}^{2}-9pq-6p.$

$3p\left(2q+1\right)\left(q-2\right)$

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.

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

$\begin{array}{cccccc}& & & & & \hfill 9{x}^{2}-12xy+4{y}^{2}-49\hfill \\ \text{Is there a GCF? No.}\hfill & & & & & \\ \begin{array}{c}\text{With more than 3 terms, use grouping. Last 2 terms}\hfill \\ \text{have no GCF. Try grouping first 3 terms.}\hfill \end{array}\hfill & & & & & \hfill 9{x}^{2}-12xy+4{y}^{2}-49\hfill \\ \begin{array}{c}\text{Factor the trinomial with}\phantom{\rule{0.2em}{0ex}}a\ne 1.\phantom{\rule{0.2em}{0ex}}\text{But the first term is a}\hfill \\ \text{perfect square.}\hfill \end{array}\hfill & & & & & \\ \text{Is the last term of the trinomial a perfect square? Yes.}\hfill & & & & & \hfill {\left(3x\right)}^{2}-12xy+{\left(2y\right)}^{2}-49\hfill \\ \text{Does the trinomial fit the pattern,}\phantom{\rule{0.2em}{0ex}}{a}^{2}-2ab+{b}^{2}?\phantom{\rule{0.2em}{0ex}}\text{Yes.}\hfill & & & & & \hfill {\left(3x\right)}^{2}{}_{\text{↘}}\underset{-2\left(3x\right)\left(2y\right)}{-12xy+}{}_{\text{↙}}{\left(2y\right)}^{2}-49\hfill \\ \text{Write the trinomial as a square.}\hfill & & & & & \hfill {\left(3x-2y\right)}^{2}-49\hfill \\ \begin{array}{c}\text{Is this binomial a sum or difference? Of squares or}\hfill \\ \text{cubes? Write it as a difference of squares.}\hfill \end{array}\hfill & & & & & \hfill {\left(3x-2y\right)}^{2}-{7}^{2}\hfill \\ \text{Write it as a product of conjugates.}\hfill & & & & & \hfill \left(\left(3x-2y\right)-7\right)\left(\left(3x-2y\right)+7\right)\hfill \\ & & & & & \hfill \left(3x-2y-7\right)\left(3x-2y+7\right)\hfill \\ \text{Is the expression factored completely? Yes.}\hfill & & & & & \\ \text{Check your answer.}\hfill & & & & & \\ \text{Multiply.}\hfill & & & & & \\ \\ \\ \hfill \phantom{\rule{4em}{0ex}}\left(3x-2y-7\right)\left(3x-2y+7\right)\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}9{x}^{2}-6xy-21x-6xy+4{y}^{2}+14y+21x-14y-49\hfill & & & & & \\ \hfill \phantom{\rule{4em}{0ex}}9{x}^{2}-12xy+4{y}^{2}-49✓\hfill & & & & & \end{array}$

Factor completely: $4{x}^{2}-12xy+9{y}^{2}-25.$

$\left(2x-3y-5\right)\left(2x-3y+5\right)$

Factor completely: $16{x}^{2}-24xy+9{y}^{2}-64.$

$\left(4x-3y-8\right)\left(4x-3y+8\right)$

Key Concepts

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 ${x}^{2}+bx+c?$ Undo FOIL.
  
  Is it of the form $a{x}^{2}+bx+c?$
  
  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.

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

$\left(2n-1\right)\left(n+7\right)$

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

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

${a}^{3}\left({a}^{2}+9\right)$

$75{m}^{3}+12m$

$121{r}^{2}-{s}^{2}$

$\left(11r-s\right)\left(11r+s\right)$

$49{b}^{2}-36{a}^{2}$

$8{m}^{2}-32$

$8\left(m-2\right)\left(m+2\right)$

$36{q}^{2}-100$

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

${\left(5w-6\right)}^{2}$

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

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

${\left(m+7n\right)}^{2}$

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

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

$7\left(b+3\right)\left(b-2\right)$

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

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

$3\left(x-3\right)\left({x}^{2}+3x+9\right)$

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

${k}^{4}-16$

$\left(k-2\right)\left(k+2\right)\left({k}^{2}+4\right)$

${m}^{4}-81$

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

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

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

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

$3\left(5p+4\right)\left(q-1\right)$

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

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

$4\left(x+3\right)\left(x+7\right)$

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

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

${u}^{2}\left(u+1\right)\left({u}^{2}-u+1\right)$

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

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

prime

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

$10{m}^{4}-6250$

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

$3{v}^{4}-768$

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

$3y\left(3x+2\right)\left(4x-1\right)$

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

$8{x}^{3}-27{y}^{3}$

$\left(2x-3y\right)\left(4{x}^{2}+6xy+9{y}^{2}\right)$

$64{x}^{3}+125{y}^{3}$

${y}^{6}-1$

$\left(y+1\right)\left(y-1\right)\left({y}^{2}-y+1\right)\left({y}^{2}+y+1\right)$

${y}^{6}+1$

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

$\left(3x-y+7\right)\left(3x-y-7\right)$

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

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

$\left(3x-2{\right)}^{2}$

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

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 $\left({y}^{2}-25\right)\left({y}^{2}+25\right).$ 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?

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Learning Objectives

Recognize and Use the Appropriate Method to Factor a Polynomial Completely

Key Concepts

Practice Makes Perfect

Writing Exercises

Self Check

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