# Factoring Polynomials Overview

Factoring polynomials is an important skill to master because it allows us to rewrite polynomials in a simpler form. The process of factoring helps us understand more about the equations we are working with and produces useful information. For example, solutions are often much easier to identify when a polynomial is in a factored form. Factoring polynomials also makes it easier to graph an equation.

Factoring Polynomials Sample Questions

There are many different factoring techniques. The most common strategy for factoring polynomials is to simply factor out the greatest common factor. If there is no clear factor in common, then another approach needs to be implemented. Another common approach is to split the polynomial into two sets of parentheses that are multiplied by each other.

Let’s take a closer look at the first strategy: factoring out the greatest common factor.

Consider the polynomial $$6x^2+9x$$. We can simplify this polynomial by pulling out the greatest common factor. The largest factor that is shared by both terms is $$3x$$. Factor out the $$3x$$ by dividing each term by $$3x$$. $$6x^2$$ becomes $$3x$$ and $$9x$$ becomes 3. So 6×2+9x becomes 3x(x+3). Notice that this factored form will multiply back to its original form. In this case, $$3x$$ times $$(x+3)$$ gives us $$6x^2+9x$$,  which proves that the factored form is equivalent to the original polynomial.

Let’s take a look at a tougher polynomial that has three terms.

Consider the following polynomial: $$4x^3-2x^2+6x$$. Look at each term and determine if there is a common factor shared by all terms. In this example, the greatest common factor is $$2x$$. Now “factor this out” by dividing each term by $$2x$$. $$4x^3-2x^2+6x$$ becomes $$2x(2x^2-x+3)$$. We can check our work to make sure that we have factored correctly by multiplying $$2x$$ by $$(2x-x+3)$$. If the product is the original polynomial, we have factored correctly. $$2x$$ multiplied by $$(2x^2-x+3)$$ equals $$4x^3-2x^2+6x$$, so we have factored correctly.

It is convenient when polynomials have a clear common factor. However, many times there will not be an obvious factor that is shared by all terms. For example, $$x^2+5x+6$$. There are no factors that are shared by all terms, so we need to approach this polynomial in a different way. This example can be factored using the approach of splitting the polynomial into two sets of parentheses that are multiplied by each other. Let’s use this approach with the polynomial $$x^2+5x+6$$.

$$x^2+5x+6$$ has no common factors, so set up two empty sets of parentheses that are multiplied by each other $$( )( )$$. Now, break apart the first term. $$x^2$$ can be split up into x times x, so place this in the parentheses $$(x\text{ }\text{ }) (x\text{ }\text{ })$$. Now take a close look at the next two terms $$5x+6$$. Ask yourself, “What multiplies to 6 and adds to 5?” In this case, 2 times 3 multiplies to 6, and 2 plus 3 adds to 5. Place 2 and 3 in the parentheses. $$(x\text{ }\text{ }) (x\text{ }\text{ })$$ becomes $$(x+2)(x+3)$$. This is the factored version of the original polynomial. Again, we can check that we have factored correctly by multiplying $$(x+2)$$ times $$(x+3)$$ to make sure that our product results in the original polynomial.

Let’s take a look at a polynomial that can be factored by pulling out the greatest common factor AND by splitting up a polynomial into two sets of parentheses that are multiplied.

For example, let’s look at the polynomial $$12x^2-27$$. When we look for a common factor, we notice that both terms are divisible by 3. 3 is the largest factor shared by both terms, so let’s factor this out. $$12x^2-27$$ becomes $$3(4x^2-9)$$. Many polynomials are successfully factored after just this single step. However, sometimes it takes multiple steps to completely factor a polynomial. $$3(4×62-9)$$ is close to complete, but consider what is inside the parentheses. $$(4x^2-9)$$ can be broken down even further. $$(4x^2-9)$$ can be split up into two sets of parentheses that are multiplied by each other. $$(x^2-9)$$ becomes $$(2x−3)(2x+3)$$. Now we are left with $$3x(2x−3)(2x+3)$$, which is factored completely.

Remember, when factoring a polynomial, you are simply rewriting it in a different way. The value has not changed at all, just the form. The objective of factoring is to break down a long and complicated polynomial into smaller pieces that are more manageable. This makes it easier to identify solutions, which inevitably makes the polynomial easier to graph.

## Factoring Polynomials Sample Questions

Here are a few sample questions going over factoring polynomials.

Question #1:

Factor the following polynomial: $$2x-4y+2$$

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

The correct answer is D. In order to factor the polynomial $$2x-4y+2$$, factor out the greatest common factor. In this case, the greatest common factor for all terms is 2. When 2 is factored out, we are left with $$(x-2y+1)$$. The product of 2 and $$(x-2y+1)$$ gives us the original polynomial, so $$2(x-2y+1)$$ is the factored form of $$2x-4y+2$$.

Question #2:

Factor the following polynomial: $$4x^3-8x^2$$

$$4x^2(x-2)$$
$$2x^2(x-2)$$
$$2x^2(x-4)$$
$$3x^3(x-4)$$

The correct answer is A. In order to factor the polynomial $$4x^3-8x^2$$, factor out the greatest common factor. In this case, the greatest common factor for all terms is $$4x^2$$. When this is factored out, we are left with $$(x-2)$$. The product of $$4x^2$$ and $$x-2$$ gives us the original polynomial, so $$4x^2(x-2)$$ is the factored form of $$4x^3-8x^2$$.

Question #3:

Factor the following polynomial: $$x^2-9$$

$$(x+3)(x-3)$$
$$(x-3)(x-3)$$
$$(x+3)(x+3)$$
$$x(x-9)$$

The correct answer is A. The polynomial $$x^2-9$$ does not have any clear common factors, so instead of factoring out the greatest common factor, it is necessary to try another approach. This polynomial can be split into two sets of parentheses that are multiplied by each other, like this:
( ) ( )

From here, look at the first term, which is $$x^2$$. This can be This can be split up into x times x, so we can place this in each parenthesis. Now we have $$(x )(x )$$. From here, think of two numbers that multiply to -9 but add to 0. $$3×-3=-9$$, and 3 and -3 add to 0, so place this in the parentheses. Now we have $$x+3x-3$$, which is the factored form of the polynomial $$x^2-9$$.

Question #4:

Factor the following polynomial: $$x^2+7x+10$$

$$(x+7)(x+10)$$
$$(x+5)(x+2)$$
$$x(x+5)$$
$$x(x+7)$$

The correct answer is B. The polynomial $$x^2+7x+10$$ does not have any clear common factors, so instead of factoring out the greatest common factor, it is necessary to try another approach. This polynomial can be split into two sets of parentheses that are multiplied by each other, like this:
( ) ( )

From here, look at the original polynomial $$x^2+7x+10$$ and focus on the first term, $$x^2$$. The term can be split by multiplying x times x, so place this inside each parenthesis: $$(x )(x )$$.

Now, focus on the remaining polynomial: $$7x+10$$. Think of two numbers that multiply to 10 but add to 7. In this case, those numbers would be 5 and 2: $$5×2=10$$ and $$5+2=7$$. Plug 5 and 2 into the parentheses.

$$x+5x+2$$

This can be checked by distributing each term to all other terms and simplifying.