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Encyclopedia>Algebra>Equations and Inequalities>Equations>Polynomial Long Division>What Is Polynomial Long Division?

Polynomial division is division with polynomials. A polynomial is the sum of a collection of terms on the form $a x^{n}$ , where $a \in ℝ$ (real numbers) and $n \in ℕ$ (natural numbers). This technique is very similar to the one you learned for regular division in elementary school.

Polynomial division often appears when you are factorizing polynomials of higher degrees, such as

a x^{3} + b x^{2} + c x + d,

or when you’re solving higher degree polynomial equations.

One of the most important tricks is to guess one of the roots of the expression. That may sound a little crazy, but you’re not just pulling arbitrary numbers out of a hat, you’re making a qualified guess. More specifically, guessing a solution means trying to insert numbers like

x = 1, - 1, 2, - 2, 3, - 3, \dots,

into the expression and seeing whether it becomes $0$ . It sounds odd to just guess at solutions, but don’t worry, because it’s pretty simple.

Theory

Important Information About Polynomial Division

The division $P (x) \div (x - a)$ has no remainder if $P (a) = 0$ . This also means that if $P (a) = 0$ , $x = a$ is a solution to the equation $P (x) = 0$ .
The division $P (x) \div (x - a)$ is not without a remainder when $P (a) \neq 0$ . In this case, $x = a$ is not a solution to the equation $P (x) = 0$ , and you will get a remainder when you perform the division.

Polynomial division is easier to learn through examples than through cumbersome recipes. Here’s an example to check whether $x = a$ is a root, and some examples of polynomial division with explanations.

Note! Dividend is the same as numerator, and divisor is the same as denominator.

Example 1

Check if $x^{3} + 2 x^{2} - 3 x - 2$ is divisible by $x - 1$ .

You know that $P (x) = x^{3} + 2 x^{2} - 3 x - 2$ is divisible by $x - 1$ if $P (1) = 0$ . That means you can just insert $x = 1$ into $P (x)$ to check.

\begin{array}{l} P (1) & = {(1)}^{3} + 2 {(1)}^{2} - 3 (1) - 2 \\ = 1 + 2 - 3 - 2 \\ = - 2 \\ \neq 0 \end{array}

That means you can conclude that

x^{3} + 2 x^{2} - 3 x - 2

is not divisible by $x - 1$ .

Example 2

Polynomial Division Without a Remainder

Follow the steps below and make sure you understand the procedure. Try doing it yourself as well, and see whether you get the same answer!

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

1.: First, you ask yourself “what do I multiply $x$ by to get $x^{3}$ ?” The answer is $x^{2}$ .
2.: Write $x^{2}$ on top of the division, and $- x^{3}$ below $x^{3}$ on the left hand side.
3.: Multiply $x^{2}$ by $- 2$ . That gives you $- 2 x^{2}$ . Put $- (- 2 x^{2}) = 2 x^{2}$ to the right of $- x^{3}$ on line $2$ in the calculation.
4.: Add the terms on line $2$ to the similar terms above them on line $1$ , and pull down the first term in the dividend that doesn’t have a term underneath it yet.
5.: Repeat this process until you you run out of terms in the dividend.
6.: If the remainder is zero, you are done.
7.: If the remainder is not zero, make a fraction with the remainder as the numerator, and the divisor as the denominator. Add that fraction to the expression on top of the division, which is your answer.

Example 3

Polynomial Division with a Remainder

Follow the steps below and make sure you understand the procedure. Try doing it yourself as well, and see whether you get the same answer!

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

1.: First, ask yourself “What do I multiply $x$ by to get $2 x^{2}$ ?” The answer is $2 x$ .
2.: Write $2 x$ on top of the division, and $- 2 x^{2}$ below $2 x^{2}$ on the left side of the dividend.
3.: Multiply $2 x$ by $- 1$ . That’s $- 2 x$ . Put $- (- 2 x) = 2 x$ to the right of $- x^{3}$ on line $2$ of the division.
4.: Add the terms on line $2$ to the similar terms above them on line $1$ , and pull down the first term in the dividend that doesn’t have a term underneath it.
5.: Repeat this process until you run out of terms in the dividend.
6.: If the remainder is zero, you are finished.
7.: If the remainder is not zero, make a fraction with the remainder as the numerator, and the divisor as the denominator. Add that fraction to the expression on top of the division, which is your answer.

Example 4

If you have the polynomial division

(a x^{2} - x + 4) \div (x - 1),

what values can $a$ have so that you’re left with no remainder?

Alternative 1 (harder)

In this method, you have to perform the polynomial division and set the remainder equal to zero to find $a$ .

Polynomial long division of ax^2-x+4 divided by x-1

You set the remainder

3 + a

equal to

0

and solve for

a

, which gives you

a = - 3

. That means there is no remainder when

a = - 3

Alternative 2 (easier)

If the division

(a x^{2} - x + 4) \div (x - 1)

has no remainder, you know that

P (x) = (a x^{2} - x + 4) = 0

when $x = 1$ . That means you can insert $x = 1$ to find $a$ . $\begin{array}{l} P (1) = a \cdot 1^{2} - 1 + 4 & = 0 \\ a + 3 & = 0 \\ a & = - 3 \end{array}$

If $a = - 3$ , the division $(a x^{2} - x + 4) \div (x - 1)$ does not have a remainder.

This is considerably easier and faster than Alternative 1!

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