Polynomial Division

Definition

Polynomial division is the process whereby a polynomial, p(x), is divided by a polynomial, d(x), usually of degree less than or equal to that of p(x). The result is of the form p(x)d(x)=q(x)+r(x)d(x)

where q(x) is the quotient polynomial and r(x) the remainder polynomial of degree less than the degree of d(x).

If r(x)=0 then d(x) is said to divide p(x) exactly.

There are two common methods of polynomial division: long division and factorisation.

Long Division

Definition

The process is very like long division of numbers, except with algebra. This method is best explained using an example.

Worked Examples
Example 1

Divide x3−3x2+x+1 by x−1.

Solution

Begin by writing the question out in long division form:

x−1)x3−3x2+x+1¯

Starting with the first term in the polynomial, x3, calculate what x−1 needs to be multiplied by so that when we subtract it from the polynomial, it will eliminate the x3 term. In this case it is x2, as x2×(x−1)=x3−x2, so we subtract this from the polynomial under the division sign. We write the x2 on top of the division line.

x2x−1)x3−3x2+x+1¯x3−x2

x2x−1)x3−3x2+x+1¯x3−x20−2x2+x+1¯

We now have a new polynomial −2x2+x+1 with which to repeat the process.

This time we are looking to eliminate −2x2, so we multiply x−1 by −2x to get −2x2+2x which we can then subtract to get rid of the x2 term.

x2−2xx−1)x3−3x2+x+1¯x3−x20−2x2+x+1¯−2x2+2x

x2−2xx−1)x3−3x2+x+1¯x3−x20−2x2+x+1¯−2x2+2x0−x+1¯

Finally, we look to eliminate −x so we multiply our divisor by −1 to get −x+1, and subtract.

x2−2x−1x−1)x3−3x2+x+1¯x3−x20−2x2+x+1¯−2x2+2x0−x+1¯−x+10¯

This happens to have eliminated the constant as well and we are left with zero. This means that x−1 is a factor of the cubic. We have shown x3−3x2+x+1x−1=x2−2x−1

Example 2

Divide 3x3−5x2+10x−3 by 3x+1.

Solution

Begin by writing the question out in long division form.

3x+1)3x3−5x2+10x−3¯

Starting with the first term in the polynomial, 3x3, calculate what 3x+1 needs to be multiplied by so that when we subtract it from the polynomial, it will eliminate the x3 term. In this case it is x2, as x2×(3x+1)=3x3+x2, so we subtract this from the polynomial under the division sign. We write the x2 on top of the division line.

x23x+1)3x3−5x2+10x−3¯3x3+x2

x23x+1)3x3−5x2+10x−3¯3x3+x20−6x2+10x−3¯

We now have a new polynomial −6x2+10x−3 with which to repeat the process.

This time we are looking to eliminate −6x2, so we multiply 3x+1 by −2x to get −6x2−2x which we can then subtract to get rid of the x2 term.

x2−2x3x+1)3x3−5x2+10x−3¯3x3+x20−6x2+10x−3¯−6x2−2x

x2−2x3x+1)3x3−5x2+10x−3¯3x3+x20−6x2+10x−3¯−6x2−2x0+12x−3¯

Now, we look to eliminate 12x so we multiply our divisor by 4 to get 12x+4, and subtract.

x2−2x+43x+1)3x3−5x2+10x−3¯3x3+x20−6x2+10x−3¯−6x2−2x0+12x−3¯12x+4−7¯

We cannot divide this any more and are left with a remainder of −7. So 3x3−5x2+10x−33x+1=x2−2x+4−73x+1

Video Examples

Prof. Robin Johnson uses the long division method to divide 2x3+x2−4x+4 by x+2.

In this example, Prof Johnson finds all the roots of a cubic equation, with one of the roots already known, by first finding a factor of the cubic given by the known root and then dividing by that factor to obtain a quadratic and then find its roots to find the remaining two roots of the cubic.

Factorisation

Definition

The factorisation method involves forcing each term in p(x) to have a factor d(x).

Consider the case when d(x) is a polynomial of degree one, i.e. of the form x+b. For each term anxn in p(x), replace it with anxn−1(x+b)−anxn−1b. This gives us one term with x+b as a factor, and a second term of lower degree which can be combined with the other xn−1 term in the original expression. After applying this method to each non-constant term, we end up with an expression whose terms all have a factor of d(x), apart from the constant term (if there is one).

Worked Examples
Example 1

Divide x2−9x−10 by x+1.

Solution

First, we want to take a factor of x out of the x2 term and replace it with (x+1). x(x+1)=x2+x, so we need to subtract an x term in order to have the same quantity as we started with. We now have

x2−9x−10=x(x+1)−x−9x−10=x(x+1)−10x−10

Now we consider the −10x term. As before, we take out a factor of x and replace it with (x+1). −10(x+1)=−10x−10, so we need to add 10 to preserve equality. We now have

x(x+1)−10x−10=x(x+1)−10(x+1)+10−10=x(x+1)−10(x+1)

Now every term in the expression has a factor of (x+1) so it's straightforward to divide by x+1.

x2−9x−10x+1=x(x+1)−10(x+1)x+1=x−10.

Note: There is no remainder, so we have shown that x+1 is a factor of x2−9x−10.

Example 2

Divide x3−3x2+5x−4 by x−2.

Solution

First, consider the term x3. Take out a factor of x and replace it with x−2. x2(x−2)=x3−2x2, so we must add 2x2 to preserve equality.

x3−3x2+5x−4=x2(x−2)+2x2−3x2+5x−4=x2(x−2)−x2+5x−4

Now consider the −x2 term. This is replaced with −x(x−2)−2x.

x2(x−2)−x2+5x−4=x2(x−2)−x(x−2)−2x+5x−4=x2(x−2)−x(x−2)+3x−4

Finally, consider the 3x term. This is replaced with 3(x−2)+6.

x2(x−2)−x(x−2)+3x−4=x2(x−2)−x(x−2)+3(x−2)+6−4=x2(x−2)−x(x−2)+3(x−2)+2

Now as many terms as possible have a factor of x−2, so it's straightforward to perform the division:

x3−3x2+5x−4x−2=x2(x−2)−x(x−2)+3(x−2)+2x−2=x2−x+3+2x−2

Video Examples

Prof. Robin Johnson uses the factorisation method to divide x3−3x2+x+1 by x−1, and 2x3+5x2−3x+2 by x+2.

In this example, Prof Johnson finds all the roots of the cubic equation x3−2x2−21x−26=0 with one of the roots already known, by first finding a factor x+2 of the cubic x3−2x2−21x−26 given by the known root x=−2 and then dividing by that factor to obtain a quadratic. Finally, he solves the quadratic to give the remaining two roots of the cubic.

Test Yourself

Test yourself: Numbas test on polynomial division

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