A Level Maths: Remainder Factor Theorem
Introduction
When you did GCSE maths you were introduced to the factorisation of quadratics. As a part of A Level Maths you are taken beyond that in terms of being able to factorise cubics, quartics and so on.
A technique that we will introduce here is known as the remainder and factor theorem and they are very useful in helping us to deal with polynomials of high powers.
We will also see how the remainder theorem can help us to find a remainder very quickly which avoids the need to perform algebraic long division.
Notation
At GCSE level you would have met expressions such as y=x^2+1 and would have been required to determine the value of y when x=1.
Instead the expression can be written as f(x)=x^2+1 and to determine the value when x=1 you would be asked to evaluate f(1)
So f(1)=1^2+1=2
In either case the results are the same.
The factor theorem
The factor theorem states:
If (x-a) is a factor of f(x) then f(a)= 0 and x=a is a root of the equation f(x)= 0
Example
Given f(x)=x^3-6 x^2+11 x-6
Find f(0), f(1), f(2), f(3) \text { and } f(4)
Solution
\begin{aligned} & f(0)=0^3-6(0)^2+11(0)-6=-6 \\ & f(1)=1^3-6(1)^2+11(1)-6=0 \\ & f(2)=2^3-6(2)^2+11(2)-6=0 \\ & f(3)=3^3-6(3)^2+11(3)-6=0 \\ & f(4)=4^3-6(4)^2+11(4)-6=6 \end{aligned}
Now because f(1), f(2) and f(3) all equal zero then (x-1),(x-2) and (x-3) are all factors of f(x)
Example
Given f(x)=x^3-x^2-3 x+2
a) Show (x-2) is a factor
b) Solve f(x)=0
Solution
In order to answer part a), the output should equal 0. It is also important to substitute the correct value for x.
The factor theorem states:
If (x-a) is a factor of f(x) then f(a)= 0 and x=a is a root of the equation f(x)= 0; so the value of x will be 2.
a) f(2)=2^3-2^2-3(2)+2=8-4-6+2=0
This shows that (x-2)
b) In order to solve [latexlf(x)=0[/latex] you need to find perform long division (or compare coefficients or the grid method):
\begin{gathered} x^3-x^2-3 x+2=(x-2)\left(a x^2+b x+c\right) \\ x^3-x^2-3 x+2=a x^3+b x^2+c x-2 a x^2-2 b x-2 c \\ x^3-x^2-3 x+2 \equiv a x^3+x^2(b-2 a)+x(c-2 b)-2 c \end{gathered}Comparing coefficients of x^3: a=1
Comparing coefficients of x^2:
\begin{gathered} -1=b-2 a \\ b=1 \end{gathered}
Comparing coefficients of x:
\begin{gathered} -3=c-2 b \\ c=-1 \end{gathered}So a x^2+b x+c=x^2+x-1
Which means that \begin{gathered} f(x)=(x-2)\left(x^2+x-1\right) \\ \therefore(x-2)\left(x^2+x-1\right)=0 \\ \therefore x-2=0 \rightarrow x=2 \\ x^2+x-1=0 \end{gathered}
On the quadratic expression the quadratic formula needs to be used:
\begin{gathered} x=\frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2} \\ x=\frac{-1 \pm \sqrt{5}}{2} \end{gathered}So, x=2 \text { or } x=\frac{-1 \pm \sqrt{5}}{2}
Example
Given f(x)=x^3-3 x^2+2 x-6 find the root of f(x)
Solution
In this question no further information has been provided. What needs to be done is to list all the possible factors of -6 which are \pm 1, \pm 2, \pm 3, \pm 6 and to then use trial and error to find which of these possible factors when substituted into f(x) will equal 0
\begin{aligned} & f(1)=1^3-3(1)^2+2(1)-6=-6 \\ & f(2)=2^3-3(2)^2+2(2)-6=-6 \\ & f(3)=3^3-3(3)^2+2(3)-6=0 \end{aligned}You will see that f(3)=0 so (x-3) is a factor and so x=3 is one root of f(x)
If required you could then perform long division (or other method) to find the solutions to fx=0. It is also possible to continue with trial and error to find the other values that will result in 0, but this can be time consuming.
The remainder theorem
When dividing a polynomial by a linear expression there will sometimes be a remainder.
This remainder can be found by using the process of long division (or other methods which are the grid method or comparing coefficients).
However the remainder theorem is a much quicker alternative.
The remainder theorem states:
f(a) is the remainder when f(x) is divided by (x-a)
Example
Find the remainder when f(x)=2 x^3-3 x+5 is divided by (x+1)
Solution
Here you need to substitute x=-1 \text { into } f(x)
\begin{gathered} f(-1)=2(-1)^3-3(-1)+5 \\ f(-1)=-2+3+5=6 \end{gathered}The remainder is 6.
Example
When f(x)=x^2-6 x+a is divided by (x-3) the remainder is 2. Find the value of a.
Solution
Here you will substitute x=3 \text { into } f(x) to give:
f(3)=3^2-6(3)+aNow f(3) has a value of 2 so:
\begin{gathered} 2-9-18+a \\ a=11 \end{gathered}Example
When f(x)=x^3+a x^2+b x+8 is divided by (x-3) the remainder is 2 and when divided by (x+1) the remainder is -2. Find the values of a and b.
Solution
When x=3, f(3)=2:
\begin{gathered} f(3)=3^3+a(3)^2+b(3)+8 \\ 2=35+9 a+3 b \\ 0=33+9 a+3 b \\ 0=11+3 a+b \end{gathered}
When x=-1, f(-1)=-2:
\begin{gathered} f(-1)=(-1)^3+a(-1)^2+b(-1)+8 \\ -2=-1+a-b+8 \\ 0=9+a-b \end{gathered}Taking the information in stages, the result is a set of simultaneous equations which can be solved by adding:
\begin{gathered} 0=20+4 a \\ a=-5 \\ b=9+a=9-5=4 \end{gathered}So:
a=-5 \text { and } b=4You have seen a variety of questions here and even questions that require the creating and solving of simultaneous equations. Within A Level Maths the world of algebra is never ending. It is therefore why you were asked the question “do you enjoy algebra?” at any open event that you may have visited.
The main thing with these types of questions is to ensure that you are reading the question carefully and that you are doing what the question actually wants. For instance, a question may state to use the remainder theorem in order to find a remainder, but the mistake would be to use algebraic long division. Even though the same answer may be obtained, the question at hand has asked for a particular method to be used.
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What you should do now is to try all the examples in this article again without looking at the answers. If you have made a mistake check the solution but the chances are high that you have made mistakes within your arithmetic calculations.
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