Simplifying Algebraic Fractions
Introduction
When it comes to algebra, one of the things that you need to be able to do is to be able to factorise. Having this skill will then allow you to simplifying algebraic fractions. The main aim with algebraic fractions is to simplify any expressions that you have as much as possible and one of the main skills that you will be using is that of factorising.
In this article we are going to look at a number of questions around simplifying algebraic fractions and to then discuss how to approach them.
But as mentioned the key skill that you need is that of factorising.
Simplifying Algebraic Fractions - Example 1
Consider the following Simplifying Algebraic Fractions question:
\frac{7 x^4-2 x^3+6 x}{x}There are two ways in which you can attempt this question. First you can factorise the numerator where x is common and this will then cancel with the denominator. Or you can write each term in the numerator over the denominator and then simplify each term as follows:
\begin{aligned} & =\frac{7 x^4}{x}-\frac{2 x^3}{x}+\frac{6 x}{x} \\ & =7 x^3-2 x^2+6 \end{aligned}Consider the following Simplifying Algebraic Fractions question:
\frac{(x+7)(2 x-1)}{(2 x-1)}In a situation such as this, do not expand the brackets. You have been given an algebraic fraction that already is in factorised form so you just need to look for any terms that can be cancelled. You can see that 2x – 1 is common to both the numerator and the denominator and these terms can then cancel out.
\frac{(x+7)(2 x-1)}{(2 x-1)}=x+7Consider the following question:
\frac{2 x^2+11 x+12}{(x+3)(x+4)}What needs to be done here is that the numerator needs to be factorised. Do not make the mistake of expanding the denominator and this is not going to help. Factorising allows you to then obtain common terms in this case it is (x + 4) and this can then be cancelled out.
\begin{aligned} & \frac{2 x^2+11 x+12}{(x+3)(x+4)} \\ & =\frac{(2 x+3)(x+4)}{(x+3)(x+4)} \\ & =\frac{2 x+3}{x+3} \end{aligned}Simplifying Algebraic Fractions - Exam Style Question
Consider the following Simplifying Algebraic Fractions question:
\frac{6 x^3+3 x^2-84 x}{6 x^2-33 x+42}=\frac{a x(x+b)}{x+c}.Find the values of a, b and c.
The numerator contains a quadratic and each number in the numerator is a multiple of 3x and so this can be taken out as a factor.
The quadratic has 3 as a common factor and this can be taken out as a factor. The process of factorising both the numerator and denominator can then be performed and any common terms can then be cancelled as shown in the solution below:
\begin{aligned} &\begin{aligned} 6 x^3+3 x^2-84 x & =3 x\left(2 x^2+x-28\right) \\ & =3 x(2 x-7)(x+4) \end{aligned}\\ &\begin{aligned} 6 x^2-33 x+42 & =3\left(2 x^2-11 x+14\right) \\ & =3(x-2)(2 x-7) \end{aligned}\\ &\begin{aligned} \frac{6 x^3+3 x^2-84 x}{6 x^2-33 x+42} & =\frac{3 x(2 x-7)(x+4)}{3(x-2)(2 x-7)} \\ & =\frac{x(x+4)}{(x-2)} \\ a=1, b=4, c= & -2 \end{aligned} \end{aligned}
Simplifying Algebraic Fractions – Further Exam Style Question
Part a) \frac{3 x^4-21 x}{3 x}
Part b) \frac{x^2-2 x-24}{x^2-7 x+6}
Part c) \frac{2 x^2+7 x-4}{2 x^2+9 x+4}
For these questions it is a case of factorising as much as possible before you cancel out any terms. Some expressions are not quadratics and are to a higher power such as 3, 4, or higher. Be looking for a common term as not everything will have a double bracket.
It is important that terms are not cancelled until you have terms expressed in terms of a product.
For part a) you can either factorise the numerator or you can write each term over the denominator and cancel any terms as follows:
\begin{aligned} \frac{3 x^4-21 x}{3 x} & =\frac{3 x^4}{3 x}-\frac{21 x}{3 x} \\ & =x^3-7 \end{aligned}For part b) you need to factorise both the numerator and the denominator before you can cancel any terms which is shown below:
\begin{aligned} & \frac{x^2-2 x-24}{x^2-7 x+6} \\ & =\frac{(x-6)(x+4)}{(x-6)(x-1)} \\ & =\frac{x+4}{x-1} \end{aligned}It is important that you do not cancel the x terms at the final stage. This is a very common mistake. If you were to put some numbers into the final answer, you will see that this step is not mathematically correct.
For part c) Again you need to factorise both the numerator and denominator. The quadratics are a little more challenging to factorise and the coefficient of x2 is not 1. The working is shown below:
\begin{aligned} & \frac{2 x^2+7 x-4}{2 x^2+9 x+4} \\ & =\frac{(2 x-1)(x+4)}{(2 x+1)(x+4)} \\ & =\frac{2 x-1}{2 x+1} \end{aligned}Again do not make the mistake by thinking that the 2x terms cancel as they don’t.
It is important that you are able to deal with algebraic fractions as you can be asked to perform further tasks on them such as differentiation or integration as well. Make sure you are able to simplify algebraically fractions quickly and proficiently.