Simplify algebraic fractions III

If you need to, take some time to revise how to factorise algebraic expressions, binomial expansions, and evaluating rational expressions.

We can simplify the fraction $\frac{84}{270}$84270​ by factorising the numerator and the denominator and cancelling common factors, like so.

$\frac{84}{270}$84270​	$=$=	$\frac{2\times2\times3\times7}{2\times3\times3\times3\times5}$2×2×3×72×3×3×3×5​
	$=$=	$\frac{2\times7}{3\times3\times5}$2×73×3×5​
	$=$=	$\frac{14}{45}$1445​

In a similar way, we can simplify algebraic fractions by factorising the numerator and the denominator and cancelling common factors.

For instance, say we wanted to simplify the algebraic fraction $\frac{3x-12y}{x-4y}$3x−12yx−4y​, what could we do? Well, first we want to make sure the numerator and denominator are fully factorised so we can then move to cancel any common factors.

• Let's start with the numerator. How would we factorise $3x-12y$3x−12y? Well, we want to find the highest common factor of both terms. $3x$3x is $3\times x$3×x and $12y$12y can be broken down into $3\times4\times y$3×4×y, so the highest common factor of the two terms is $3$3! We divide this $3$3 out of both terms and put it outside of brackets to get $3\left(x-4y\right)$3(x−4y).
• The denominator $x-4y$x−4y on the other hand actually can't be factorised further. There is no common factor between the terms $x$x and $4y$4y.

We can therefore write the expression like this.

$\frac{3x-12y}{x-4y}$3x−12yx−4y​

$=$=

$\frac{3\left(x-4y\right)}{x-4y}$3(x−4y)x−4y​

Notice that the numerator and denominator actually have a common factor of $\left(x-4y\right)$(x−4y). We can therefore simplify the expression by cancelling this out.

$\frac{3x-12y}{x-4y}$3x−12yx−4y​	$=$=	$\frac{3\left(x-4y\right)}{x-4y}$3(x−4y)x−4y​
	$=$=	$3$3

Remember when factorising algebraic expressions to first try taking out the highest common factor (HCF) from all terms. After that, look for binomial factorisations of quadratic expressions, including differences of two squares.

Worked Examples

Question 1

Simplify $\frac{8x-20}{2x^3-5x^2}$8x−202x3−5x2​.

$\frac{8x-20}{2x^3-5x^2}$8x−202x3−5x2​	$=$=	$\frac{4\left(2x-5\right)}{2x^3-5x^2}$4(2x−5)2x3−5x2​	Take out the HCF of $4$4 in the numerator
	$=$=	$\frac{4\left(2x-5\right)}{x^2\left(2x-5\right)}$4(2x−5)x2(2x−5)​	Take out the HCF of $x^2$x2 in the denominator
	$=$=	$\frac{4}{x^2}$4x2​	Cancel out the common factor $\left(2x-5\right)$(2x−5)

Question 2

Simplify $\frac{16x^2+40x+25}{25-16x^2}$16x2+40x+2525−16x2​.

Firstly, how do we factorise the numerator $16x^2+40x+25$16x2+40x+25? We are looking for a binomial factorisation $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) for some integers $a$a, $b$b, $c$c, $d$d. 

The product of the constant terms $bd$bd will have to equal $25$25. Since $25$25 factorises to $5\times5$5×5, we should try $b=5$b=5 and $d=5$d=5 first, which would give $\left(ax+5\right)\left(cx+5\right)$(ax+5)(cx+5).

Now, can we find the $x$x coefficients $a$a and $c$c such that this expands to $16x^2+40x+25$16x2+40x+25? The product of these will be $16$16, so let's test $a=4$a=4 and $c=4$c=4 first.

$\left(4x+5\right)\left(4x+5\right)=16x^2+20x+20x+25$(4x+5)(4x+5)=16x2+20x+20x+25 so does indeed equal $16x^2+40x+25$16x2+40x+25 when expanded. Hence our factorisation is correct. Note that you could have also figured out this factorisation by using the quadratic formula to find the zeros of the quadratic.

What about the denominator? Well, notice that 25-16*x^2 is a difference of two squares so can be factorised to $\left(5-4x\right)\left(5+4x\right)$(5−4x)(5+4x).

Let's now use these factorisations to simplify the fraction.

$\frac{16x^2+40x+25}{25-16x^2}$16x2+40x+2525−16x2​	$=$=	$\frac{\left(4x+5\right)\left(4x+5\right)}{\left(5-4x\right)\left(5+4x\right)}$(4x+5)(4x+5)(5−4x)(5+4x)​	Using our factorisations
	$=$=	$\frac{4x+5}{5-4x}$4x+55−4x​	Cancel the common factors $4x+5$4x+5 and $5+4x$5+4x, which are equivalent

Question 3

Simplify $\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k2−5)4−10k(2k2−5)58(2k2−5)8​.

$\frac{6\left(2k^2-5\right)^4-10k\left(2k^2-5\right)^5}{8\left(2k^2-5\right)^8}$6(2k2−5)4−10k(2k2−5)58(2k2−5)8​	$=$=	$\frac{2\left(2k^2-5\right)^4\left(3-5k\left(2k^2-5\right)\right)}{8\left(2k^2-5\right)^8}$2(2k2−5)4(3−5k(2k2−5))8(2k2−5)8​	Take out a HCF of $2\left(2k^2-5\right)^4$2(2k2−5)4 from both terms.
	$=$=	$\frac{\left(2k^2-5\right)^4\left(3-5k\left(2k^2-5\right)\right)}{4\left(2k^2-5\right)^8}$(2k2−5)4(3−5k(2k2−5))4(2k2−5)8​	Cancel a common factor of $2$2 from $2$2 and $8$8.
	$=$=	$\frac{3-5k\left(2k^2-5\right)}{4\left(2k^2-5\right)^4}$3−5k(2k2−5)4(2k2−5)4​	Cancel a common factor of $\left(2k^2-5\right)^4$(2k2−5)4 from $\left(2k^2-5\right)^4$(2k2−5)4 and $\left(2k^2-5\right)^8$(2k2−5)8

Recall our index law that states that $b^m\div b^n=b^{m-n}$bm÷​bn=bm−n to notice that dividing out $\left(2k^2-5\right)^4$(2k2−5)4 from $\left(2k^2-5\right)^8$(2k2−5)8 will leave $\left(2k^2-5\right)^4$(2k2−5)4.

Finally, we expand what remains in the numerator to get our final answer of $\frac{3+25k-10k^3}{4\left(2k^2-5\right)^4}$3+25k−10k34(2k2−5)4​.

 

Further Simplifications

We can use these new techniques to finish off what we know about factorisation and simplifying complicated expressions.

Example

question 4

Factorise and simplify completely: $\frac{1}{9x-15}+\frac{4x+6}{6x^2-x-15}$19x−15​+4x+66x2−x−15​

Think about which of the three methods would be the easiest, and where you can cancel out to simplify

Do

$\frac{1}{9x-15}+\frac{4x+6}{6x^2-x-15}$19x−15​+4x+66x2−x−15​

$=$=

$\frac{1}{3\left(3x-5\right)}+\frac{2\left(2x+3\right)}{6x^2-x-15}$13(3x−5)​+2(2x+3)6x2−x−15​

Using the cross method (both PSF methods involve figuring out factors of a large number $6\times15=90$6×15=90):

Therefore:

$\frac{1}{3\left(3x-5\right)}+\frac{2\left(2x+3\right)}{6x^2-x-15}$13(3x−5)​+2(2x+3)6x2−x−15​

 

Further Examples

Question 5

 

Question 6

 

Question 7