Polynomial Division

In algebra, the remainder theorem allows us to divide one polynomial by another polynomial of the same or lower degree, using a process similar to long division. You can do it without a calculator because it breaks a complex-looking problem up into smaller ones that you have learnt how to solve previously.

The remainder theorem

The Remainder Theorem is a useful mathematical theorem that can be used to factorize polynomials of any degree in a neat and fast manner.

 

The process using long division

Solve: $\left(3x^3+2x^2-6\right)\div\left(x+2\right)$(3x3+2x2−6)÷​(x+2) using long division

The dividend: $3x^3+2x^2-6$3x3+2x2−6

The divisor: $x+2$x+2

It's helpful to write the dividend like this: $3x^3+2x^2+0x-6$3x3+2x2+0x−6.

Then we can use long division to solve the problem.

1. Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of $x$x, which in this case is $x$x). Place the result above the bar ($3x^3\div x=3x^2$3x3÷​x=3x2). 

2. Multiply the divisor by the number you just wrote above the top line. Write the result under the first two terms of the dividend. ($3x^2\times\left(x+2\right)=3x^3+6x^2$3x2×(x+2)=3x3+6x2). 

3. Subtract these terms from the from those in the original dividend, making sure you pay attention to the positive and negative signs ($3x^3+2x^2-\left(3x^3+6x^2\right)=-4x^2+0x$3x3+2x2−(3x3+6x2)=−4x2+0x). 

4. Repeat the previous three steps, except this time use the two terms that have just been written as the dividend (ie. divide $-4x^2+0x$−4x2+0x by $x+2$x+2) 

. 

5. Repeat step 4. Keep going until there is nothing to "pull down".

The polynomial above the bar is the quotient, $Q(x)$Q(x), and the number left over, $-22$−22, is the remainder, $R(x)$R(x).

 

Speeding up the process

The remainder theorem also points out the relationship between division and multiplication. For example, since $20\div5=4$20÷​5=4, we can also say $4\times5=20$4×5=20. If you get a remainder, then you do the multiplication first, then add the remainder back in. For example, $17\div2=8$17÷​2=8 remainder $1$1, which we could also write as $2\times8+1=17$2×8+1=17. The process is the same for polynomials.

The remainder theorem states that when a polynomial ($P(x)$P(x)) is divided by a linear factor ($x-a$x−a), the result will be a quotient polynomial ($Q(x)$Q(x)) plus a remainder ($R(x)$R(x)). In other words:

$P(x)\div\left(x-a\right)=Q(x)+R(x)$P(x)÷​(x−a)=Q(x)+R(x)

 

This also means that: $P(x)=\left(x-a\right)\times Q(x)+R(x)$P(x)=(x−a)×Q(x)+R(x)

You can't always easily solve questions with this quick method but it's handy to know anyway.

 

Finding remainders

 If a polynomial $P(x)$P(x) is divided by $x-a$x−a, then: $P(a)=remainder$P(a)=remainder

 

Remember

If the quotient of a polynomial does not leave a remainder, then this value is a factor.

Worked examples

Question 1

Question 2

Question 3

Question 4