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 x33x2+x+1 by x1.

Solution

Begin by writing the question out in long division form:

x1)x33x2+x+1¯

Starting with the first term in the polynomial, x3, calculate what x1 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×(x1)=x3x2, so we subtract this from the polynomial under the division sign. We write the x2 on top of the division line.

x2x1)x33x2+x+1¯x3x2

x2x1)x33x2+x+1¯x3x202x2+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 x1 by 2x to get 2x2+2x which we can then subtract to get rid of the x2 term.

x22xx1)x33x2+x+1¯x3x202x2+x+1¯2x2+2x

x22xx1)x33x2+x+1¯x3x202x2+x+1¯2x2+2x0x+1¯

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

x22x1x1)x33x2+x+1¯x3x202x2+x+1¯2x2+2x0x+1¯x+10¯

This happens to have eliminated the constant as well and we are left with zero. This means that x1 is a factor of the cubic. We have shown x33x2+x+1x1=x22x1

Example 2

Divide 3x35x2+10x3 by 3x+1.

Solution

Begin by writing the question out in long division form.

3x+1)3x35x2+10x3¯

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)3x35x2+10x3¯3x3+x2

x23x+1)3x35x2+10x3¯3x3+x206x2+10x3¯

We now have a new polynomial 6x2+10x3 with which to repeat the process.

This time we are looking to eliminate 6x2, so we multiply 3x+1 by 2x to get 6x22x which we can then subtract to get rid of the x2 term.

x22x3x+1)3x35x2+10x3¯3x3+x206x2+10x3¯6x22x

x22x3x+1)3x35x2+10x3¯3x3+x206x2+10x3¯6x22x0+12x3¯

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

x22x+43x+1)3x35x2+10x3¯3x3+x206x2+10x3¯6x22x0+12x3¯12x+47¯

We cannot divide this any more and are left with a remainder of 7. So 3x35x2+10x33x+1=x22x+473x+1

Video Examples

Prof. Robin Johnson uses the long division method to divide 2x3+x24x+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 anxn1(x+b)anxn1b. 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 xn1 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 x29x10 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

x29x10=x(x+1)x9x10=x(x+1)10x10

Now we consider the 10x term. As before, we take out a factor of x and replace it with (x+1). 10(x+1)=10x10, so we need to add 10 to preserve equality. We now have

x(x+1)10x10=x(x+1)10(x+1)+1010=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.

x29x10x+1=x(x+1)10(x+1)x+1=x10.

Note: There is no remainder, so we have shown that x+1 is a factor of x29x10.

Example 2

Divide x33x2+5x4 by x2.

Solution

First, consider the term x3. Take out a factor of x and replace it with x2. x2(x2)=x32x2, so we must add 2x2 to preserve equality.

x33x2+5x4=x2(x2)+2x23x2+5x4=x2(x2)x2+5x4

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

x2(x2)x2+5x4=x2(x2)x(x2)2x+5x4=x2(x2)x(x2)+3x4

Finally, consider the 3x term. This is replaced with 3(x2)+6.

x2(x2)x(x2)+3x4=x2(x2)x(x2)+3(x2)+64=x2(x2)x(x2)+3(x2)+2

Now as many terms as possible have a factor of x2, so it's straightforward to perform the division:

x33x2+5x4x2=x2(x2)x(x2)+3(x2)+2x2=x2x+3+2x2

Video Examples

Prof. Robin Johnson uses the factorisation method to divide x33x2+x+1 by x1, and 2x3+5x23x+2 by x+2.

In this example, Prof Johnson finds all the roots of the cubic equation x32x221x26=0 with one of the roots already known, by first finding a factor x+2 of the cubic x32x221x26 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

External Resources

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