**6.4 General Strategy for Factoring Polynomials**

Topics covered in this section are:

**6.4.1 Recognize and Use the Appropriate Method to Factor a Polynomial Completely**

**GENERAL STRATEGY FOR FACTORING POLYNOMIALS**

**HOW TO: 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 $ax^{2}+bx+c$?

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.

- Is it a sum?
- 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 1**

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

**Solution**

$7x^{3}-21x^{2}-70x$ | |

Is there a GCF? Yes, $7x$. | |

Factor out the GCF. | $7x(x^{2}-3x-10)$ |

In the parentheses, is it a binomial, trinomial, or are there more terms? | |

It is a trinomial with leading coefficient $1$. | |

“Undo” FOIL. | $7x(x \ \ \ \ \ )(x \ \ \ \ \ \ )$ |

$7x(x+2)(x-5)$ | |

Is the expressions completely factored? Yes, neither binomial can be factored. | |

Check your answer by multiplying. | $7x(x+2)(x-5)$ $7x(x^{2}-5x+2x-10)$ $7x(x^{2}-3x-10$ $7x^{3}-21x^{2}-70x \ \checkmark$ |

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

**Example 2**

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

**Solution**

$24y^{2}-150$ | |

Is there a GCF? Yes, $6$. | |

Factor out the GCF. | $6(4y^{2}-25)$ |

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, difference of squares. | $6((2y)^{2}-(5)^{2})$ |

Write as a product of conjugates. | $6(2y-5)(2y+5)$ |

Is the expression factored completely? Yes, neither binomial can be factored. | |

Check: | $6(2y-5)(2y+5)$ $6(4y^{2}-10y+10y-25)$ $24y^{2}-150 \ \checkmark$ |

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

**Example 3**

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

**Solution**

$4a^{2}-12ab+9b^{2}$ | |

Is there a GCF? No. | |

Is it a binomial, trinomial, or are there more terms? | |

Trinomial with $a≠1$. But the first term is a perfect square. | |

Is the last term a perfect square? Yes. | $(2a)^{2}-12ab+(3b)^{2}$ |

Does it fit the pattern, $a^{2}-2ab+b^{2}$? Yes. | $(2a)^{2} \searrow – 12ab + \swarrow (3b)^{2}$ $-2(2a)(3b)$ |

Write it as a square. | $(2a-3b)^{2}$ |

Is the expression factored completely? Yes, the binomial cannot be factored. | |

Check your answer by multiplying. | $(2a-3b)^{2}$ $(2a)^{2}-2 \cdot 2a \cdot 3b + (3b)^{2}$ $4a^{2}-12ab+9b^{2} \ \checkmark$ |

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

**Example 4**

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

**Solution**

$12x^{3}y^{2}+75xy^{2}$ | |

Is there a GCF? Yes, $3xy^{2}$. | |

Factor out the GCF. | $3xy^{2}(4x^{2}+25)$ |

In the parentheses, is it a binomial, trinomial, or are there more than three terms? It is a binomial. | |

Is it a sum? Of squares? Yes. | Sums of squares are prime. |

Is the expression factored completely? Yes. | |

Check: | $3xy^{2}(4x^{2}+25)$ $12x^{3}y^{2}+75xy^{2} \ \checkmark$ |

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

**Example 5**

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

**Solution**

$24x^{3}+81y^{3}$ | |

Is there a GCF? Yes, $3$. Factor it out. | $3(8x^{3}+27y^{3})$ |

In the parentheses, is it a binomial, trinomial, or are there more than three terms? Binomial. | |

Is it a sum or difference? Sum. | |

Sum of squares or cubes? Cubes. | |

Write it using the sum of cubes pattern. | |

Is the expression factored completely? Yes. | $3(2x+3y)(4x^{2}-6xy+9y^{2})$ |

Check by multiplying. | We leave this to you. |

**Example 6**

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

**Solution**

$3x^{5}y-48xy$ | |

Is there a GCF? Yes, $3xy$. Factor it out. | $3xy(x^{4}-16)$ |

Is the binomial a sum or difference? Of squares or cubes? Write it as a difference of squares. | $3xy \left( (x^{2})^{2}-(4)^{2} \right)$ |

Factor it as a product of conjugates. | $3xy(x^{2}-4)(x^{2}+4)$ |

The first binomial is again a difference of squares. | $3xy \left( (x)^{2}-(2)^{2} \right) (x^{2}+4)$ |

Factor it as a product of conjugates. | $3xy(x-2)(x+2)(x^{2}+4)$ |

Is the expression factored completely? Yes. | |

Check by multiplying. | $3xy(x-2)(x+2)(x^{2}+4)$ $3xy(x^{2}-4)(x^{2}+4)$ $3xy(x^{4}-16)$ $3x^{5}y-48xy \ \checkmark$ |

**Example 7**

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

**Solution**

$4x^{2}+8bx-4ax-8ab$ | |

Is there a GCF? Factor out the GCF, $4$. | $4(x^{2}+2bx-ax-2ab)$ |

There are four terms, use grouping. | $4[x(x+2b)-a(x+ab)]$ $4(x+2b)(x-a)$ |

Is the expression completely factored? Yes. | |

Check your answer by multiplying. | $4(x+2b)(x-a)$ $4(x^{2}-ax+2bx-2ab)$ $4x^{2}+8bx-4ax-8ab \ \checkmark$ |

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

**Example 8**

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

**Solution**

$40x^{2}y+44xy-24y$ | |

Is there a GCF? Factor out the GCF, $4y$. | $4y(10x^{2}+11x-6)$ |

Factor the trinomial with $a≠1$. | $4y(5x-2)(2x+3)$ |

Is the expression completely factored? Yes. | |

Check your answer by multiplying. | $4y(5x-2)(2x+3)$ $4y(10x^{2}+11x-6)$ $40x^{2}y+44xy-24y \ \checkmark$ |

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

**Solution**

$9x^{2}-12xy+4y^{2}-49$ | |

Is there a GCF? No. | |

With more than $3$ terms, use grouping. Last $2$ terms have no GCF. Try grouping first $3$ terms. | |

Factor the trinomial with $a≠1$. But the first term is a perfect square. | $(3x)^{2}-12xy+(2y)^{2}-49$ |

Is the last term of the trinomial a perfect square? Yes. | |

Does the trinomial fit the pattern, $a^{2}-2ab+b^{2}$? Yes. | $(3x)^{2} _\searrow -12xy+ _\swarrow (2y)^{2}$ $-2(3x)(2y)$ |

Write the trinomial as a square. | $(3x-2y)^{2}-49$ |

Is this binomial a sum or difference? Of squares or cubes? Write it as a difference of squares. | $(3x-2y)^{2}-7^{2}$ |

Write it as a product of conjugates. | $((3x-2y)-7)((3x-2y)+7)$ $(3x-2y-7)(3x-2y+7)$ |

Is the expression factored completely? Yes. |

Check your answer by multiplying.

$(3x-2y-7)(3x-2y+7)$

$9x^{2}-6xy+21x-6xy+4y^{2}-14y-21x+14y-49$

$9x^{2}-12xy+4y^{2}-49 \ \checkmark$

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*Marecek, L., & Mathis, A. H. (2020). General Strategy for Factoring Polynomials. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/6-4-general-strategy-for-factoring-polynomials*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction*