# Factoring Polynomials

**Factoring a polynomial** is the process of decomposing a polynomial into a product of two or more polynomials. For example, \( f(x) = x^2 + 5x + 6 \) can be decomposed into \( f(x) = (x+5)(x+1) .\)

Another example:

Factor \(x^2 - x - 6 \).

We have

\[ x^2 - x - 6 = (x-3)(x+2).\ _\square\]

Note that \( x^2 - x - 6 \) can also be expressed as \( 1 \cdot (x^2 -x - 6) \). Thus, the factors of \( x^2 - x - 6 \) are \(1\), \(x^2 - x - 6\), \(x-3\) and \(x+2\).

In order to solve problems like this, you will need to understand prime factorization. Review this concept, if needed, before continuing. We will look at 3 common ways in which a polynomial can be factored: grouping, substitution, and using identities.

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## Factoring Polynomials by Grouping

We often see the grouping method applied to polynomials with 4 terms. The idea is to pair like terms together so that we can apply the distributive property in order to factorize them nicely.

## Factor \( x^3 - 3x^2 -x + 3 \).

We have

\[ \begin{align} x^3 - 3x^2 - x + 3 &= x^2(x - 3) - (x - 3) \\ &= (x^2 -1)(x - 3) \\ &= (x - 1)(x + 1)(x - 3). \ _\square \end{align} \]

## Factor \( x^2 + 4x - y^2 - 4y \).

We have

\[ \begin{align} x^2 + 4x - y^2 - 4y &= (x^2 - y^2) + (4x - 4y) \\ &= (x -y)(x + y) + 4(x-y) \\ &= (x - y)(x + y + 4). \ _\square \end{align} \]

## Factor \( x^2 - 16y^2 - 6x + 9 \).

We have

\[ \begin{align} x^2 - 16y^2 - 6x + 9 &= (x^2 - 6x + 9) - 16y^2 \\ &= (x-3)^2 - (4y)^2 \\ &= (x - 3 - 4y)(x - 3 + 4y). \ _\square \end{align} \]

## Factoring by Substitution

If a polynomial is complicated, you can try substituting one of the complicated terms with a simpler term to make it easier to factor.

## Factor \( (x - y)(x - y - 1) - 20 \).

Let \( S \) = \( x - y \). Substituting \( S \) in for \( x - y\) for ease of computation, we get

\[ \begin{align} (x - y)(x - y - 1) - 20 &= (S)(S - 1) - 20 \\ &= S^2 - S - 20 \\ &= (S - 5)(S + 4)\\ &= (x - y - 5)(x - y + 4). \ _\square \end{align} \]

## Factor \( (x + y)(x + y + 1) - 12 \).

Let \( S \) = \( x + y \). Substituting \( S \) in for \( x + y\) for ease of computation, we get

\[ \begin{align} (x + y)(x + y + 1) - 12 &= (S)(S + 1) - 12 \\ &= S^2 + S - 12\\ &= (S - 3)(S + 4)\\ &= (x + y - 3)(x + y + 4). \ _\square \end{align} \]

You can see more examples applying this method on the Factoring Polynomials by Substitution page.

## Factoring Polynomials using Identities

Factoring binomials of the form \( x^2 - y^2= (x-y)(x+y):\)

This approach applies the difference of two squares identity.

Factor \( x^2 - 16 \).

We have

\[ x^2 - 16 = x^2 - 4^2 = (x-4)(x+4).\ _\square\]

Factor \( x^4 - 25y^4 \).

We have

\[ x^4 - 25y^4 = \big(x^2\big)^2 - \big(5y^2\big)^2 = \big(x^2 - 5y^2\big)\big(x^2 + 5y^2\big).\ _\square\]

Factor \( 9x^3y - 36xy \).

We have

\[ 9x^3y - 36xy = 9xy\big(x^2 - 4\big) = 9xy\big(x^2 - 2^2\big) = 9xy(x-2)(x+2).\ _\square\]

Factor \( (x+1)^2 - 9(x-2)^2 \).

We have

\[ \begin{align} (x+1)^2 - 9(x-2)^2 &= (x+1)^2 - \big(3(x-2)\big)^2 \\ &= \big((x+1) - 3(x-2)\big) \big((x+1) + 3(x-2)\big) \\ &= (x + 1 -3x + 6)(x + 1 + 3x - 6) \\ &= (-2x + 7)(4x - 5). \ _\square \end{align} \]

Factoring binomials of the form \( x^3 + y^3 = (x+y)\big(x^2 - xy + y^2\big) \) and \(x^3 - y^3 = (x-y)\big(x^2 + xy+ y^2\big): \)

This approach applies the sum and difference of cubes identity.

Factor \( 27x^3 - y^3 \).

We have

\[ \begin{align} 27x^3 - y^3 &= (3x)^3 - y^3 \\ &= (3x - y)\big((3x)^2 + 3xy + y^2\big) \\ &= (3x-y)(9x^2 + 3xy + y^2). \ _\square \end{align} \]

Factor \( 64x^3 + 27y^3 \).

We have

\[ \begin{align} 64x^3 + 27y^3 &= (4x)^3 + (3y)^3 \\ &= (4x + 3y)\big((4x)^2 - 12xy + (3y)^2\big) \\ &= (4x + 3y)\big(16x^2 -12 xy + 9y^2\big). \ _\square \end{align} \]

Factoring trinomials of the form \( x^2 + 2xy + y^2 = (x+y)^2 \) and \(x^2 - 2xy + y^2 = (x-y)^2: \)

Factor \( x^2 + 6x + 9 \).

We have

\[ x^2 + 6x + 3 = x^2 + 2(x \cdot 3) + 3^2 = (x+3)^2. \ _\square \]

Factor \( x^2 - 4x + 4 \).

We have

\[ x^2 - 4x + 4 = x^2 - 2(x \cdot 2) + 2^2 = (x-2)^2. \ _\square \]

Factor \( 25x^2 - 20x + 4 \).

We have

\[ 25x^2 - 20x + 4 = (5x)^2 - 2(5x \cdot 2) + 2^2 = (5x-2)^2. \ _\square \]

Factor \( 16x^2 - 24xy + 9y^2 \).

We have

\[ 16x^2 - 24xy + 9y^2 = (4x)^2 - 2(4x \cdot 3y) + (3y)^2 = (4x-3y)^2. \ _\square \]

Factoring trinomials of the form \( x^2 + (a+b)x + ab = (x+a)(x+b): \)

This approach is also known as factorization by observation.

In cases where we don't have a perfect square of the form \( (x+y)^2 \) or \( (x-y)^2 \), but the leading coefficient of \( x \) is 1, we can try to find \(a \) and \( b\) such that \( a \cdot b\) is equal to the constant term, and \( a + b \) is equal to the coefficient of \( x \).

Factor \( x^2 + 5x + 6 \).

We have

\[ x^2 + 5x + 6 = x^2 + (2 + 3)x + (2 \times 3) = (x + 2)(x + 3). \ _\square \]

Factor \( x^2 - 3x - 4 \).

We have

\[ x^2 - 3x + 4 = x^2 + (1 - 4)x + \big(1 \times (-4)\big) = (x + 1)(x - 4). \ _\square \]

Factor \( x^2 - 12x + 35 \).

We have

\[ x^2 - 12x + 35 = x^2 + (-5 - 7)x + \big((-5) \times (-7)\big) = (x - 5)(x - 7). \ _\square \]

Factor \( 2x^2 + 4xy - 30y^2 \).

We have

\[ \begin{align} 2x^2 + 4xy - 30y^2 &= 2\big(x^2 + 2xy - 15y^2\big) \\ &= 2\big(x^2 + (5y - 3y)x + (5y) \times (-3y)\big) \\ &= 2(x + 5y)(x - 3y) . \ _\square \end{align} \]

Factoring trinomials of the form \( acx^2 + (ad+bc)x + bd = (ax+b)(cx+d): \)

When the leading coefficient of \(x\) is *not* 1, we must factor both the leading coefficient and the constant.

Factor \( 2x^2 + 4x + 2 \).

Factors of \( 2x^2: \) \( 2x \) and \( x \)

Factors of \( 2 :\) \( 1 \) and \( 2 \)Since we're looking at all-positive coefficients, we can ignore negative factors.

Given this, two simple possible factorizations are \( (2x + 1)(x + 2) \) and \( (2x + 2)(x + 1) \) .

In both of these factorizations, the leading term will be \(2x \cdot x = 2x^2 \) and the constant term will be \( 1 \cdot 2 = 2 \), So we don't have to worry about them. Let's turn our attention to getting the correct middle term \( 4x \).

In the factorization \( (2x + 1)(x + 2) \), the middle term would be \( (2x \cdot 2) + (1 \cdot x) = 5x \), which is incorrect.

In the factorization \( (2x + 2)(x + 1) \), the middle term would be \( (2x \cdot 1) + (2 \cdot x) = 4x \), which is what we want. So the factorization that gives us the correct expansion is \( (2x + 2)(x + 1) \). \(_\square\)

Factor \( 2xy^5 - 11xy^4 - 6xy^3 \).

We can factor out \( xy^3 \) to make our work simpler:

\[ 2xy^5 - 11xy^4 - 6xy^3 = xy^3\big(2y^2 - 11y - 6\big). \]

Now we can just concentrate on factoring \( 2y^2 - 11y - 6 \).

Factors of \( 2y^2: \) \( 2y \) and \( y \), \( -2y \) and \( -y \)

Factors of \( -6 :\) \( -2 \) and \( 3 \), \( 2 \) and \( -3 \), \( -1 \) and \( 6 \), \( 1 \) and \(-6 \)There are many different combinations we can try for the factorization. But one thing to notice is that since the middle term is \( -11y \), we need to use factors that can get us to the number \( -11 \): \( 2y \times (-6) \) and \(y \times 1 \) would give us the middle term we want.

Since we want \( 2y \) to be multiplied by \( -6 \) to give us the middle term, the \( -6 \) must exist in a different set of parentheses from \( 2y \) so that it can be distributed to and multiplied by \( 2y \).

So let's check the factorization \( (2y + 1 )( y - 6 ) \). The leading term is \( 2y \cdot y =2y^2 \), and the constant term is \( 1 \cdot -6 = -6 \) as desired. The middle term is \( \big(2y \cdot (-6)\big) + 1 \cdot y = -11y \), as we wanted.

So \( 2y^2 - 11y - 6 \) = \( (2y + 1 )( y - 6 ) \), implying that our final answer is

\[ 2xy^5 - 11xy^4 - 6xy^3 = xy^3\big(2y^2 - 11y - 6\big) = xy^3( 2y + 1 )( y - 6 ) .\ _\square\]

Factoring polynomials in this way involves some amount of guessing and checking. You can greatly improve your speed at this process by using your number sense to figure out which combinations of numbers will successfully get you the middle term that you want.

## Factoring Polynomials

Now that you're familiar with the different ways of factoring polynomials, let's work on some problems.

**Cite as:**Factoring Polynomials.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/factoring-polynomials/