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

ax2 + bx + c = a (x x 1) (x x2) ,

where x1 and x2 are solutions to the equation

ax2 + bx + 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) ÷(x a) is solvable and has no remainders.

  • When P (x) ÷(x a) is solvable without a remainder, you get P (x) ÷(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) 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, or 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: 4x4 + 2x2 = 2x2 (2x2 + 1) .

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
3.
Call your expression P (x) and check the values of P (0) ,P (1) ,P (1) , until you find one that gives you P (a) = 0.
4.
Next, you find P (x) ÷(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:

= ax2 + bx + c = a (x x1) (x x2) = ax3 + bx2 + cx + d = a (x x1) (x x2) (x x3) = ax4 + bx3 + cx2 + dx + f = a (x x1) (x x2) (x x3) (x x4)

ax2 + bx + c = a (x x 1) (x x2) ax3 + bx2 + cx + d = a (x x 1) (x x2) (x x3) ax4 + bx3 + cx2 + dx + f = a (x x 1) (x x2) (x x3) (x x4) ax5 + bx4 + cx3 + dx2 + fx + g = a (x x 1) (x x2) (x x3) (x x4) (x x5)

In general, you factorize polynomials like this:

= anxn + a n1xn1 + + a 1x + a0 = an (x x1) (x xn)

anxn + a n1xn1 + + a 1x + a0 = an (x x1) (x xn)

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 = x1, you write the factorization like this:

a (x x1) (x x1) = a (x x1) 2.

nth-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:

x3 + x + 2 = (x + 1) (x2 x + 2)

Here x = 1 is the only real solution to the cubic polynomial. The entire graph of the function x2 x + 2 lies above the x-axis, which means it doesn’t produce any solutions.

Example 1

Factorize the polynomial x3 + 2x2 5x 6

Because x is not present in all of the terms, you have to guess at solutions. Let’s start with x = 1:

P (1) = (1) 3 + 2 (1) 2 5 (1) 6 = 1 + 2 5 6 = 8 0.

You continue by trying x = 1:

P (1) = (1) 3 + 2 (1) 2 5 (1) 6 = 1 + 2 + 5 6 = 0.

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 = x1 and x = x2:

x = 1 ±12 4 1 (6) 2 1 = 1 ±1 + 24 2 = 1 ±25 2 = 1 ± 5 2 ,

which gives you x1 = 2 and x2 = 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

P (x) = x3 + 2x2 5x 6 = (x + 1) (x 2) (x + 3) .

Example 2

Solve the equation x3 + 2x2 5x 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:

= x3 + 2x2 5x 6 = (x + 1) (x 2) (x + 3) = 0

x3 + 2x2 5x 6 = (x + 1) (x 2) (x + 3) = 0

The zero product property says that if a b = 0, then a = 0 or b = 0. If you use this on the factors in the equation, you get that

x + 1 = 0 x = 1 x + 3 = 0 x = 3 x 2 = 0 x = 2,

x + 1 = 0 x = [1] x + 3 = 0 x = 3 x 2 = 0 x = 2,

which means the solutions to the equation are x1 = 1, x2 = 2 and x3 = 3.

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