General Strategy for Factoring Polynomials

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.

Form after factoring out the GCF	Method
Binomial: difference of squares	$a^2-b^2=(a-b)(a+b)$
Binomial: sum of squares	Sums of squares do not factor.
Binomial: sum of cubes	$a^3+b^3=(a+b)(a^2-ab+b^2)$
Binomial: difference of cubes	$a^3-b^3=(a-b)(a^2+ab+b^2)$
Trinomial: $x^2+bx+c$	$(x\phantom{{}+{}})(x\phantom{{}+{}})$
Trinomial: $ax^2+bx+c$	If $a$ and $c$ are squares, check the trinomial-square pattern; otherwise use trial and error or the “ $ac$ ” method.
More than three terms	Use grouping.

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?
- For a binomial, decide whether it is a sum or difference and whether it involves squares or cubes. Sums of squares do not factor. Use the sum or difference of cubes pattern for cubes, and factor a difference of squares as a product of conjugates.
- For $x^2+bx+c$ , “undo” FOIL. For $ax^2+bx+c$ , check the trinomial-square pattern when $a$ and $c$ are squares; otherwise use trial and error or the “ $ac$ ” method.
- For more than three terms, use grouping.
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!

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

First factor out the GCF, $7x$ . The expression in parentheses is a trinomial with leading coefficient $1$ , so “undo” FOIL. Neither resulting binomial can be factored further.

\begin{array}{lrcl} \text{Factor out the GCF.} & 7x^3-21x^2-70x &=& 7x(x^2-3x-10) \\[4pt] \text{“Undo” FOIL.} &&=& 7x(x+2)(x-5) \end{array}

Check.

\begin{array}{rcl} 7x(x+2)(x-5) &=& 7x(x^2-5x+2x-10) \\[4pt] &=& 7x(x^2-3x-10) \\[4pt] &=& 7x^3-21x^2-70x \checkmark \end{array}

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

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

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

Example. Factor completely: $24y^2-150$ .

First factor out the GCF, $6$ . The remaining binomial is a difference of squares, so write it as a product of conjugates. Neither binomial can be factored further.

\begin{array}{lrcl} \text{Factor out the GCF.} & 24y^2-150 &=& 6(4y^2-25) \\[4pt] \text{Write as a product of conjugates.} &&=& 6(2y-5)(2y+5) \end{array}

Check: $6(2y-5)(2y+5)=6(4y^2-25)=24y^2-150\checkmark$ .

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

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

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

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

There is no GCF. The trinomial has $a\ne1$ , but the first and last terms are perfect squares. It fits the pattern $a^2-2ab+b^2$ .

4a^2-12ab+9b^2=(2a)^2-2(2a)(3b)+(3b)^2=(2a-3b)^2

The binomial cannot be factored. Check by multiplying:

(2a-3b)^2=(2a)^2-2(2a)(3b)+(3b)^2=4a^2-12ab+9b^2\checkmark

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

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

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

Example. Factor completely: $12x^3y^2+75xy^2$ .

Factor out the GCF, $3xy^2$ . The remaining binomial is a sum of squares, which is prime.

12x^3y^2+75xy^2=3xy^2(4x^2+25)

Check: $3xy^2(4x^2+25)=12x^3y^2+75xy^2\checkmark$ .

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

Factor completely: $27xy^3+48xy$ .

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

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

Factor out the GCF, $3$ . The remaining binomial is a sum of cubes. Apply the sum of cubes pattern.

\begin{array}{lrcl} \text{Factor out the GCF.} & 24x^3+81y^3 &=& 3(8x^3+27y^3) \\[4pt] \text{Recognize a sum of cubes.} &&=& 3\big((2x)^3+(3y)^3\big) \\[4pt] \text{Use the sum of cubes pattern.} &&=& 3(2x+3y)(4x^2-6xy+9y^2) \end{array}

The expression is factored completely. Check by multiplying.

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

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

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

Factor out the GCF, $3xy$ . The remaining binomial is a difference of squares. After factoring it as a product of conjugates, the first binomial is again a difference of squares.

\begin{array}{lrcl} \text{Factor out the GCF.} & 3x^5y-48xy &=& 3xy(x^4-16) \\[4pt] \text{Factor the difference of squares.} &&=& 3xy(x^2-4)(x^2+4) \\[4pt] \text{Factor the difference of squares again.} &&=& 3xy(x-2)(x+2)(x^2+4) \end{array}

The expression is factored completely. Check by multiplying.

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

Factor completely: $7xy^5-7xy$ .

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

Factor out the GCF, $4$ . There are four terms, so use grouping.

\begin{array}{lrcl} \text{Factor out the GCF.} & 4x^2+8bx-4ax-8ab &=& 4(x^2+2bx-ax-2ab) \\[4pt] \text{Group and factor.} &&=& 4\big[x(x+2b)-a(x+2b)\big] \\[4pt] \text{Factor the common binomial.} &&=& 4(x+2b)(x-a) \end{array}

Check: $4(x+2b)(x-a)=4(x^2-ax+2bx-2ab)=4x^2+8bx-4ax-8ab\checkmark$ .

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

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

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

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

Factor out the GCF, $4y$ , then factor the trinomial with $a\ne1$ .

\begin{array}{lrcl} \text{Factor out the GCF.} & 40x^2y+44xy-24y &=& 4y(10x^2+11x-6) \\[4pt] \text{Factor the trinomial.} &&=& 4y(5x-2)(2x+3) \end{array}

Check: $4y(5x-2)(2x+3)=4y(10x^2+11x-6)=40x^2y+44xy-24y\checkmark$ .

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

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

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

There is no GCF. With more than three terms, use grouping. The last two terms have no GCF, so group the first three terms. They form a perfect-square trinomial, and the resulting binomial is a difference of squares.

\begin{array}{lrcl} \text{Write the trinomial as a square.} & 9x^2-12xy+4y^2-49 &=& (3x-2y)^2-49 \\[4pt] \text{Write as a difference of squares.} &&=& (3x-2y)^2-7^2 \\[4pt] \text{Write as a product of conjugates.} &&=& (3x-2y-7)(3x-2y+7) \end{array}

Check.

\begin{array}{rcl} (3x-2y-7)(3x-2y+7) &=& 9x^2-6xy-21x-6xy+4y^2+14y+21x-14y-49 \\[4pt] &=& 9x^2-12xy+4y^2-49 \checkmark \end{array}

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

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

Key terms

completely factored — written as a product whose nonmonomial factors are prime. general factoring strategy — first factor out the GCF, then choose a method according to the number of terms and applicable special patterns, and finally check the factorization. product of conjugates — the factorization $(a-b)(a+b)$ of a difference of squares.

This section is adapted from Intermediate Algebra 2e, Section 6.4: General Strategy for Factoring Polynomials by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the factoring-method chart as an accessible table, reformatted the worked-example tables as aligned math, omitted the end-of-section exercises, and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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