Polynomials

Hong Kong

Stage 4 - Stage 5

Lesson

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`)=`x`2−5`x`+6 by the monic polynomial $D\left(x\right)=x-1$`D`(`x`)=`x`−1. We proceed in a manner similar to long division given as follows:

Thus we state:

$\frac{P\left(x\right)}{D\left(x\right)}=\frac{x^2-5x+6}{x-1}=\left(x-4\right)+\frac{2}{x-1}$`P`(`x`)`D`(`x`)=`x`2−5`x`+6`x`−1=(`x`−4)+2`x`−1

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

$P\left(x\right)=x^2-5x+6=\left(x-1\right)\left(x-4\right)+2$`P`(`x`)=`x`2−5`x`+6=(`x`−1)(`x`−4)+2

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

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

$P\left(x\right)=\left(x-a\right)Q\left(x\right)+R$`P`(`x`)=(`x`−`a`)`Q`(`x`)+`R`

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:

$P\left(a\right)=\left(a-a\right)Q\left(a\right)+R=R$`P`(`a`)=(`a`−`a`)`Q`(`a`)+`R`=`R`

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 $\left(x-a\right)$(`x`−`a`) 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`)=`x`2−5`x`+6 and $D\left(x\right)=x-1$`D`(`x`)=`x`−1, before dividing note that $P\left(1\right)=\left(1\right)^2-5\left(1\right)+6=2$`P`(1)=(1)2−5(1)+6=2 and this is indeed the remainder!

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 $x-a$`x`−`a` must be a factor of $P(x)$`P`(`x`). We could write $P(x)=(x-a)Q(x)$`P`(`x`)=(`x`−`a`)`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 $x-a$`x`−`a` is a factor of $P$`P`.

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

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

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.

$x+3$

`x`+3$8-x$8−

`x`$5+4x$5+4

`x`$6-x$6−

`x`

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

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