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Remainder and Factor Theorem


The Remainder Theorem:

To divide two polynomials, say $P\left(x\right)$P(x) divided by $D\left(x\right)$D(x), we need the degree of the divisor polynomial $D\left(x\right)$D(x) to be less than or equal to the degree of the dividend polynomial $P\left(x\right)$P(x). An earlier entry on polynomial division can be found here.

As an example suppose we divide $P\left(x\right)=x^2-5x+6$P(x)=x25x+6 by the monic polynomial $D\left(x\right)=x-1$D(x)=x1. We proceed in a manner similar to long division given as follows:

Thus we state:


We can express the result slightly differently by multiplying both sides by the divisor so that:


Stating the result like this is known as the division transformation.

In general, dividing $P\left(x\right)$P(x) by $\left(x-a\right)$(xa) will always produce a result that looks like:


Here, $Q\left(x\right)$Q(x) is the quotient polynomial of one less degree than $P\left(x\right)$P(x) and $R$R is the remainder.

This last equation holds the key to the remainder theorem. Because the polynomial holds true for all values of $x$x, by putting $x=a$x=a into this general result we see that:


This means that substituting $x=a$x=a into $P\left(x\right)$P(x) before dividing will reveal the remainder. It's a little mathematical magic! We can actually know the remainder even before the division by $\left(x-a\right)$(xa) is done. This nice result is known as the remainder theorem.

Let's try it with our example. With $P\left(x\right)=x^2-5x+6$P(x)=x25x+6 and $D\left(x\right)=x-1$D(x)=x1, before dividing note that $P\left(1\right)=\left(1\right)^2-5\left(1\right)+6=2$P(1)=(1)25(1)+6=2 and this is indeed the remainder!


The Factor Theorem

The factor theorem is an extension of the remainder theorem.

If a polynomial equation $P(x)=0$P(x)=0 has a root $x=a$x=a, meaning $P(a)=0$P(a)=0, then $x-a$xa must be a factor of $P(x)$P(x). We could write $P(x)=(x-a)Q(x)$P(x)=(xa)Q(x) where $Q$Q is a polynomial of degree one less than the degree of $P$P.

This means that if we can find by any means a number $a$a such that $P(a)=0$P(a)=0, then we know immediately that $x-a$xa is a factor of $P$P.


Factorise $p(x)=x^3-x^2-x-2$p(x)=x3x2x2.

By experiment, we find that $p(2)=0$p(2)=0. Therefore, $p(x)=(x-2)q(x)$p(x)=(x2)q(x). We can use the division algorithm to discover $q(x)$q(x). That is, we divide $p(x)$p(x) by $x-2$x2. In this way, we find that $q(x)=x^2+x+1$q(x)=x2+x+1 which is an irreducible quadratic. So, $p(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$p(x)=x3x2x2=(x2)(x2+x+1) is the complete factorisation.



Worked Examples

Question 1

Christa wants to test whether various linear expressions divide exactly into $P\left(x\right)$P(x), or whether they leave a remainder. For each linear expression below, state the value of $x$x that needs to be substituted into $P\left(x\right)$P(x) to find the remainder.

  1. $x+3$x+3

  2. $8-x$8x

  3. $5+4x$5+4x

  4. $6-x$6x

Question 2

Using the remainder theorem, find the remainder when $P\left(x\right)=-4x^4+6x^3+4x^2-7x+7$P(x)=4x4+6x3+4x27x+7 is divided by $A\left(x\right)=3x-1$A(x)=3x1.

Question 3

When $3x^3-2x^2-4x+k$3x32x24x+k is divided by $x-3$x3, the remainder is $47$47. Find the value of $k$k.


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