Factor Theorem And Algebraic Division
Factor Theorem And Algebraic Division – Introduction
In this article we are going to look at questions that cover the year 12 topic of Factor Theorem and Algebraic Division.
It is important that you know what the factor theorem states which is:
\text { If } f(a)=0 \text { then } x-a \text { is a factor of } f(x)Questions involving the factor theorem can extend to solving equations as well as sketching curves and can also involve simultaneous equations.
We will consider some past exam questions and discuss how to approach these questions.
The factor theorem and algebraic division can also be extended to questions involving calculus so it is very important that you are able to master the skills needed to answer these questions.
Factor Theorem And Algebraic Division - Example 1
Consider the following factor theorem and algebraic division question:
Here the question is asking you to use specific methods to solve the question. If you use a different method, even if you obtain the same answer, you will not obtain any marks. So it is very important that you are reading the question carefully.
You will notice that no linear terms have been given so you need to start by consider the possible factors of 10 which are \pm 1, \pm 2, \pm 5, \pm 10
There will be some initial trial and error to determine a factor. Once a factor has been determined, you can then perform algebraic division.
\begin{aligned} & f(x)=2 x^3-7 x^2-17 x+10 \\ & f(1)=-12 \\ & f(2)=-36 \\ & f(-1)=18 \\ & f(-2)=0 \\ & f(-2)=0 \quad \therefore(x+2) \text { is a factor } \end{aligned}From the above you can see that after a few attempts, a factor has now been obtained. We can now perform long division as the question requests.This will then give us a quadratic factor as follows:
\begin{aligned} & 2 x^2-11 x+5 \\ & x + 2 \longdiv { 2 x ^ { 3 } – 7 x ^ { 2 } – 1 7 x + 1 0 } \\ & 2 x^3+4 x^2 \\ & -11 x^2-17 x \\ & -11 x^2-22 x \\ & 5 x+10 \\ & \frac{5 x+10}{0} \\ & \end{aligned}And this means f(x)=(x+2)\left(2 x^2-11 x+5\right)
Taking note of the question, it is asking us to factorise f(x) completely. The final step is to then factorise the quadratic.
The complete factorise answer is:
\begin{aligned} & (x+2)\left(2 x^2-11 x+5\right) \\ & (x+2)(2 x-1)(x-5) \end{aligned}Factor Theorem And Algebraic Division - Example 2
Consider the following factor theorem and algebraic division question;
Here the question is asking you to use a specific technique in order to answer the question. Take note of what the factor theorem is. So you are substituting “-3” into the expression as follows:
\begin{aligned} & g(x)=4 x^3-8 x^2-35 x+75 \\ & g(-3)=4(-3)^3-8(-3)^2-35(-3)+75 \\ & g(-3)=0 \quad \therefore \quad(x+3) \text { is a factor of } g(x) \end{aligned}So a factor has been confirmed.
Part b) is essentially asking you to perform a long division in order to obtain a quadratic and to then factorise that quadratic which can be done as follows:
\begin{aligned} & x + 3 \longdiv { 4 x ^ { 3 } – 8 x ^ { 2 } – 3 5 x + 7 5 } \\ & 4 x^3+12 x^2 \\ & -20 x^2-35 x \\ & -20 x^2-60 x \\ & 25 x+75 \\ & \frac{25 x+75}{0} \\ & \end{aligned}Make sure that you factorising g(x) completely:
\begin{aligned} & (x+3)\left(4 x^2-20 x+25\right) \\ & (x+3)(2 x-5)(2 x-5) \\ & (x+3)(2 x-5)^2 \end{aligned}
Factor Theorem And Algebraic Division – Example 3
Consider the following factor theorem and algebraic division question:
Part a) is now hopefully a standard question which you can easily and quickly do. This is simply a case of performing a substitution as shown:
\begin{aligned} & \quad f(x)=x^3+6 x^2+4 x-15 \\ & f(-3)=(-3)^3+6(-3)^2+4(-3)-15 \\ & =0 \\ & \therefore(x+3) \text { is a factor } \\ & x=-3 \text { is a solution } \end{aligned}Part b) is again asking you to perform an algebraic division and the other important wording of the question is to find the other solutions to 2 decimal places. This means that the quadratic to be obtained does not factorise and so the quadratic formula needs to be used.
\begin{aligned} & \frac{x^2+3 x-5}{x + 3 \longdiv { x ^ { 3 } + 6 x ^ { 2 } + 4 x – 1 5 }} \\ & x^3+3 x^2 \\ & 3 x^2+4 x \\ & 3 x^2+9 x \\ & -5 x-15 \\ & \frac{-5 x-15}{0} \\ & \end{aligned}This means that f(x) is equivalent to: (x+3)\left(x^2+3 x-5\right)
Solving the quadratic gives the following values:
\begin{aligned} & a=1 \quad b=3 \quad c=-5 \\ & x=\frac{-(3) \pm \sqrt{(3)^2-4(-5)}}{2} \\ & x=1.19 \quad x=-4.19 \end{aligned}The factor theorem relies generally on you being able to perform a substitution but it is important that the correct values are being used. Take note if a question is asking for the use of algebraic division. There are alternative methods such as comparing coefficients and also the grid method. The two most common being algebraic division and comparing coefficients.
We will consider one more example which involves the use of simultaneous equations.
Example:
For part a) you first need to perform the substitution to obtain two equations with two unknowns as follows:
\begin{aligned} f(4) & =0 \\ 0 & =(4)^3+6(4)^2+4 p+q \\ 0 & =160+4 p+q \end{aligned}\begin{aligned} f(-5) & =36 \\ 36 & =(-5)^3+6(-5)^2-5 p+q \\ 36 & =25-5 p+q \\ 11 & =-5 p+q \end{aligned}We have obtained the two equations and these now need to be solved simultaneously.
How you solve the questions is a matter of preference. You could make one variable the subject and then make a substitution as follows:
\begin{aligned} 0 & =160+4 p+11+5 p \\ 0 & =171+9 p \\ -171 & =9 p \\ p & =-19 \end{aligned}Following this we can then obtain a value for q as follows: \begin{aligned} & q=11+5(-19) \\ & q=-84 \end{aligned}
For part b) we are asked to perform a complete factorisation. One factor is known as so a algebraic division can be performed to obtain a quadratic:
\begin{aligned} & x – 4 \longdiv { x ^ { 3 } + 6 x ^ { 2 } – 1 9 x – 8 4 } \\ & x^3-4 x^2 \\ & 10 x^2-19 x \\ & 10 x^2-40 x \\ & 2 / x-84 \\ & \frac{21 x-84}{0} \\ & \end{aligned}Having obtained the quadratic, the function can now be factorised completely:
\frac{(x-4)\left(x^2+10 x+21\right)}{(x-4)(x+3)(x+7)}Questions surrounding the factor theorem and algebraic division require you to be knowledgeable in algebraic techniques such as that of substitution and also indices. These types of questions are very common in exams in both year 1 and year 2 maths exams. Try the questions above again without looking at the solutions.