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 strategy below summarizes all the factoring methods we have covered and outlines the order in which to try them when factoring a polynomial.

General strategy for factoring polynomials.

Always begin at the top and work down.

Is there a greatest common factor? Factor it out first.
Now look at what is left in the parentheses. Is it a binomial, a trinomial, or does it have more than three terms?
- If it is a binomial:
  - Is it a sum? A sum of squares does not factor. A sum of cubes factors with the sum-of-cubes pattern $a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right)$ .
  - Is it a difference? A difference of squares factors as a product of conjugates, $a^2 - b^2 = (a - b)(a + b)$ . A difference of cubes uses the difference-of-cubes pattern $a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right)$ .
- If it is a trinomial:
  - If it is of the form $x^2 + bx + c$ , undo FOIL to write it as $(x\ \ )(x\ \ )$ .
  - If it is of the form $ax^2 + bx + c$ and both $a$ and $c$ are perfect squares, check whether it fits the perfect-square-trinomial pattern $a^2 \pm 2ab + b^2 = (a \pm b)^2$ . Otherwise use trial and error or the “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!

Example. Factor completely: $4x^5 + 12x^4$ .

There is a GCF of $4x^4$ . Factor it out. What remains in the parentheses is a binomial. It is a sum, but neither a sum of squares nor a sum of cubes, so it does not factor further.

\begin{array}{lrcl} \text{Factor out the GCF } 4x^4. & 4x^5 + 12x^4 &=& 4x^4(x + 3) \end{array}

Check by multiplying: $4x^4 \cdot x + 4x^4 \cdot 3 = 4x^5 + 12x^4$ ✓, and the binomial $x + 3$ is prime, so the polynomial is factored completely.

Factor completely: $3a^4 + 18a^3$ .

Factor completely: $x^3 + 36x$ .

Example. Factor completely: $12x^2 - 11x + 2$ .

There is no GCF. What is left is a trinomial of the form $ax^2 + bx + c$ . Since $a = 12$ is not a perfect square, it is not a perfect-square trinomial, so we use trial and error (the “ac” method also works).

\begin{array}{lrcl} \text{Undo FOIL.} & 12x^2 - 11x + 2 &=& (3x - 2)(4x - 1) \end{array}

Check by multiplying: $(3x - 2)(4x - 1) = 12x^2 - 3x - 8x + 2 = 12x^2 - 11x + 2$ ✓.

Factor completely: $10a^2 - 17a + 6$ .

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

Example. Factor completely: $g^3 + 25g$ .

There is a GCF of $g$ . Factor it out. What remains is the binomial $g^2 + 25$ , a sum of squares — and sums of squares are prime, so it does not factor further.

\begin{array}{lrcl} \text{Factor out the GCF } g. & g^3 + 25g &=& g\left(g^2 + 25\right) \end{array}

Check: $g \cdot g^2 + g \cdot 25 = g^3 + 25g$ ✓.

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

Example. Factor completely: $12y^2 - 75$ .

There is a GCF of $3$ . Factor it out. What remains, $4y^2 - 25$ , is a difference of squares — $(2y)^2 - (5)^2$ — so we write it as a product of conjugates.

\begin{array}{lrcl} \text{Factor out the GCF } 3. & 12y^2 - 75 &=& 3\left(4y^2 - 25\right) \\[4pt] \text{Write as conjugates.} & &=& 3(2y - 5)(2y + 5) \end{array}

Neither remaining binomial is a difference of squares, so the polynomial is factored completely.

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

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

There is no GCF. It is a trinomial in which the first term $4a^2 = (2a)^2$ and the last term $9b^2 = (3b)^2$ are both perfect squares, and the middle term is $-2 \cdot 2a \cdot 3b = -12ab$ , so it fits the perfect-square-trinomial pattern.

\begin{array}{lrcl} \text{Write it as a square.} & 4a^2 - 12ab + 9b^2 &=& (2a - 3b)^2 \end{array}

Check: $(2a - 3b)^2 = (2a)^2 - 2 \cdot 2a \cdot 3b + (3b)^2 = 4a^2 - 12ab + 9b^2$ ✓.

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

Example. Factor completely: $6y^2 - 18y - 60$ .

There is a GCF of $6$ . Factor it out. What remains, $y^2 - 3y - 10$ , is a trinomial with leading coefficient $1$ , so we undo FOIL.

\begin{array}{lrcl} \text{Factor out the GCF } 6. & 6y^2 - 18y - 60 &=& 6\left(y^2 - 3y - 10\right) \\[4pt] \text{Undo FOIL.} & &=& 6(y + 2)(y - 5) \end{array}

Check by multiplying: $6(y + 2)(y - 5) = 6\left(y^2 - 3y - 10\right) = 6y^2 - 18y - 60$ ✓.

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

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

There is a GCF of $3$ . Factor it out. What remains, $8x^3 + 27$ , is a binomial that is a sum of cubes: $(2x)^3 + (3)^3$ . Apply the sum-of-cubes pattern $a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right)$ with $a = 2x$ and $b = 3$ .

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

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

Example. Factor completely: $2x^4 - 32$ .

There is a GCF of $2$ . Factor it out. What remains, $x^4 - 16$ , is a difference of squares — $\left(x^2\right)^2 - (4)^2$ . The first resulting binomial, $x^2 - 4$ , is again a difference of squares, so factor it once more. The binomial $x^2 + 4$ is a sum of squares and does not factor.

\begin{array}{lrcl} \text{Factor out the GCF } 2. & 2x^4 - 32 &=& 2\left(x^4 - 16\right) \\[4pt] \text{Product of conjugates.} & &=& 2\left(x^2 - 4\right)\left(x^2 + 4\right) \\[4pt] \text{Factor again.} & &=& 2(x - 2)(x + 2)\left(x^2 + 4\right) \end{array}

None of the remaining binomials is a difference of squares, so the polynomial is factored completely.

Factor completely: $4a^4 - 64$ .

Example. Factor completely: $3x^2 + 6bx - 3ax - 6ab$ .

There is a GCF of $3$ . Factor it out. What remains, $x^2 + 2bx - ax - 2ab$ , has more than three terms, so use grouping.

\begin{array}{lrcl} \text{Factor out the GCF } 3. & 3x^2 + 6bx - 3ax - 6ab &=& 3\left(x^2 + 2bx - ax - 2ab\right) \\[4pt] \text{Group and factor each pair.} & &=& 3[\,x(x + 2b) - a(x + 2b)\,] \\[4pt] \text{Factor out } (x + 2b). & &=& 3(x + 2b)(x - a) \end{array}

Check by multiplying: $3(x + 2b)(x - a) = 3\left(x^2 - ax + 2bx - 2ab\right) = 3x^2 - 3ax + 6bx - 6ab$ ✓.

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

Key terms

greatest common factor (GCF) — the largest monomial that divides every term of a polynomial; always factor it out first. difference of squares — a binomial $a^2 - b^2$ , which factors as the product of conjugates $(a - b)(a + b)$ . sum of squares — a binomial $a^2 + b^2$ , which is prime (does not factor). sum and difference of cubes — the patterns $a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right)$ and $a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right)$ . perfect-square trinomial — a trinomial $a^2 \pm 2ab + b^2$ , which factors as $(a \pm b)^2$ . grouping — a method for factoring a polynomial with more than three terms by pairing terms that share a common factor. factored completely — a polynomial written so that, other than monomials, all of its factors are prime.

This section is adapted from Elementary Algebra 2e, Section 7.5: General Strategy for Factoring Polynomials by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the “General Strategy for Factoring Polynomials” figure (Figure 7.3) as an accessible nested decision list and merged it with the How-To steps; recast the worked examples as prose with {lrcl} step tables and inline checks; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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