How to Factorize Polynomials of Degree 3 and 4

When you factorize polynomials of degree 3 or higher, you want to use techniques that leave you with a quadratic expression. The most important technique for this is polynomial long division.

When you are left with a quadratic expression, you can factorize that as usual by finding

a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}),

where $x_{1}$ and $x_{2}$ are solutions to the equation

a x^{2} + b x + c = 0 .

You can find these solutions through inspection, the quadratic formula, or by using digital tools. But first, let’s take a closer look at polynomial long division:

Theory

Important Features of Polynomial Long Division

If $P (a) = 0$ , then $x = a$ is a solution to the equation $P (x) = 0$ . That means the division $P (x) \div (x - a)$ is solvable and has no remainders.
When $P (x) \div (x - a)$ is solvable without a remainder, you get $P (x) \div (x - a) = Q (x)$ , where $Q (x)$ is a new polynomial of a lower degree than $P (x)$ .
If $Q (x)$ is of degree 3 or higher, you have to find more solutions. Solutions to $Q (x)$ are also solutions to $P (x)$ . If $Q (x)$ is of degree 2, you can factorize the expression like you normally would.
$P (x) = (x - a) \cdot Q (x)$ is a factorization of $P (x)$ .

In exercises where you are factorizing polynomials of higher degrees, you are either given a value that you are supposed to check, or there is a clear hint about what is correct. If not, one of the solutions will often be

x = - 2, - 1, 0, 1,

2

. You check the solution by inserting the value of

x

into the polynomial to see if the answer is 0. If it is, you know that the value you inserted is a solution to the equation, and you can start factorizing.

Rule

Factorization of Polynomials Using Polynomial Long Division

1.: Check if there is a power of $x$ in all terms. If that’s the case, you pull out the highest degree of $x$ that is common between all terms. If that gives you a quadratic expression, you factorize it in one of the ways I talked about above. For example: $\begin{array}{l} 4 x^{4} + 2 x^{2} = 2 x^{2} (2 x^{2} + 1) . \end{array}$
If not, go to Item 2.
2.: If the starting polynomial doesn’t have $x$ in all of its terms, you have to guess at a solution. You can start with the values that were introduced above: $x = 1, - 1, 2, - 2 \dots$
3.: Call your expression $P (x)$ and check the values of $P (0), P (1), P (- 1), \dots$ until you find one that gives you $P (a) = 0$ .
4.: Next, you find $P (x) \div (x - a)$ through polynomial long division.
5.: Repeat this process until you have a polynomial of degree 2 or lower.
6.: Use any of the ways to factorize quadratic expressions to finish the factorization.

Here are some examples of factorization of polynomials of degree $n$ :

\begin{array}{l} a x^{2} + b x + c \\ = a (x - x_{1}) (x - x_{2}) \\ a x^{3} + b x^{2} + c x + d \\ = a (x - x_{1}) (x - x_{2}) (x - x_{3}) \\ a x^{4} + b x^{3} + c x^{2} + d x + f \\ = a (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4}) \end{array}

\begin{array}{l} a x^{2} + b x + c & = a (x - x_{1}) (x - x_{2}) \\ a x^{3} + b x^{2} + c x + d & = a (x - x_{1}) (x - x_{2}) (x - x_{3}) \\ a x^{4} + b x^{3} + c x^{2} + d x + f & = a (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4}) \\ a x^{5} + b x^{4} + c x^{3} + d x^{2} + f x + g & = a (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4}) (x - x_{5}) \end{array}

In general, you factorize polynomials like this:

\begin{array}{l} a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} \\ = a_{n} (x - x_{1}) \dots (x - x_{n}) \end{array}

a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = a_{n} (x - x_{1}) \dots (x - x_{n})

If a solution shows up several times, they have to be included as many times as they show up. For example, when you have a quadratic function that touches the

x

-axis in just one point

x = x_{1}

, you write the factorization like this:

a (x - x_{1}) (x - x_{1}) = a {(x - x_{1})}^{2} .

$n$ th-degree polynomials have a maximum of $n$ solutions, but they can also have fewer. For example, a cubic polynomial has either one, two, or three real solutions. In cases with only one real solution, the factorization looks like this:

x^{3} + x + 2 = (x + 1) (x^{2} - x + 2)

Here $x = - 1$ is the only real solution to the cubic polynomial. The entire graph of the function $x^{2} - x + 2$ lies above the $x$ -axis, which means it doesn’t produce any solutions.

Example 1

Factorize the polynomial $x^{3} + 2 x^{2} - 5 x - 6$

Because $x$ is not present in all of the terms, you have to guess at solutions. Let’s start with $x = 1$ : $\begin{array}{l} P (1) & = {(1)}^{3} + 2 {(1)}^{2} - 5 (1) - 6 \\ = 1 + 2 - 5 - 6 \\ = - 8 \\ \neq 0 . \end{array}$

You continue by trying $x = - 1$ : $\begin{array}{l} P (- 1) & = {(- 1)}^{3} + 2 {(- 1)}^{2} - 5 (- 1) - 6 \\ = - 1 + 2 + 5 - 6 \\ = 0 . \end{array}$

This tells you that $(x - (- 1)) = (x + 1)$ divides $P (x)$ . That means you can perform the polynomial long division

Polynomial long division of x^3+2x^2-5x-6 divided by x+1

You have found an expression of degree 2, just like you wanted. You can now factorize this like you always have, either with the quadratic formula, or through inspection.

Finding $x = x_{1}$ and $x = x_{2}$ : $\begin{array}{l} x & = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 6)}}{2 \cdot 1} \\ = \frac{- 1 \pm \sqrt{1 + 24}}{2} \\ = \frac{- 1 \pm \sqrt{25}}{2} \\ = \frac{- 1 \pm 5}{2}, \end{array}$

which gives you $x_{1} = 2$ and $x_{2} = - 3$ . That means the factorization of the quadratic expression is $(x - 2) (x + 3)$ . Now, you can set up the finished factorization of the cubic expression, which is $\begin{array}{l} P (x) & = x^{3} + 2 x^{2} - 5 x - 6 \\ = (x + 1) (x - 2) (x + 3) . \end{array}$

Example 2

Solve the equation $x^{3} + 2 x^{2} - 5 x - 6 = 0$

To solve equations like these, you begin by moving everything around to get zero on the right-hand side of the equation. In this case, that has already been done. Next, you factorize the expression on the left-hand side. As this is the same polynomial you had in Example 1, it is the following: