Factorisation

NZ Level 6 (NZC) Level 1 (NCEA)

Simplify algebraic fractions I

Lesson

An interesting application of factorisation is helping to simplify fractions, as both processes involve looking at factors. Let's have a quick review of how to do exactly that with algebraic fractions.

Simplifying fractions already taught us how to simplify fractions that only involve numbers, and algebraic fractions work exactly the same way. Simply find common factors (highest common factor [HCF] is the fastest way) between the denominator and the numerator and cancel them out until you can't find any more.

For example, $\frac{49st^2}{42t}$49`s``t`242`t` involves numbers and the two variables $s$`s` and $t$`t`, so let's look at them all separately. Just looking at the numbers, we see that $7$7 is the $HCF$`H``C``F` between $49$49 and $42$42, so we can take it out from both to leave $7$7 and $6$6, respectively. As for $s$`s` terms, there are no common factors between the denominator and numerator except $1$1 so we leave them. Lastly the $t$`t` terms have a HCF of $t$`t`, so we are left with $t$`t` on the top and $1$1 on the bottom after cancelling out. So in the end we should have $\frac{7st}{6}$7`s``t`6 after simplifying.

As previously mentioned, factorisation can greatly help us simplify complicated algebraic fractions but we can use all the methods we've learnt so far to similarly simplify fractions. Have a look at the following examples to see what I'm talking about.

Factorise and simplify $\frac{x^2-6x+9}{x-3}$`x`2−6`x`+9`x`−3

**Think **about whether the numerator can be factorised using quadratic methods or perfect square methods

**Do**

$x^2-6x+9$`x`2−6`x`+9 can be factorised using the perfect square method as $9$9 is a square number, and $6$6 is double $\sqrt{9}=3$√9=3

Therefore it becomes $\left(x-3\right)^2$(`x`−3)2

So our fraction can be rewritten as:

$\frac{\left(x-3\right)^2}{x-3}=x-3$(`x`−3)2`x`−3=`x`−3 as $x-3$`x`−3 is a common factor of the numerator and denominator

Factorise and simplify $\frac{y+4}{y^2-3y-28}$`y`+4`y`2−3`y`−28

**Think **about which method to use for the denominator and how the negative $-28$−28 will affect it

**Do**

$y^2-3y-28$`y`2−3`y`−28 is a monic quadratic trinomial but not a perfect square as $-28$−28 is not a square number

Its negativity also means the two numbers $a$`a` and $b$`b` we need to find in $\left(y+a\right)\left(y+b\right)$(`y`+`a`)(`y`+`b`) have different signs.

Number pairs that give us $-28$−28 are:

$1$1 & $-28$−28, $-1$−1 & $28$28, $2$2 & $-14$−14, $-2$−2 & $14$14, $4$4 & $-7$−7, $-4$−4 & $7$7

The only pair to have a sum of $-3$−3 is $4$4 & $-7$−7, which must be our $a$`a` and $b$`b`

$\left(y+4\right)\left(y-7\right)$(`y`+4)(`y`−7) must then be the factorised form

Our fraction then becomes $\frac{y+4}{\left(y+4\right)\left(y-7\right)}=\frac{1}{y-7}$`y`+4(`y`+4)(`y`−7)=1`y`−7 as we take the $y+4$`y`+4 out

Factorise and simplify $\frac{jk-j+k^2-k}{k\left(j+k\right)}$`j``k`−`j`+`k`2−`k``k`(`j`+`k`)

**Think** about which method applied to the four termed numerator

**Do**

$jk-j+k^2-k$`j``k`−`j`+`k`2−`k` can be factorised by grouping in pairs, so it becomes

$j\left(k-1\right)+k\left(k-1\right)$`j`(`k`−1)+`k`(`k`−1) = $\left(k-1\right)\left(j+k\right)$(`k`−1)(`j`+`k`)

So the fraction becomes

$\frac{\left(k-1\right)\left(j+k\right)}{k\left(j+k\right)}=\frac{k-1}{k}$(`k`−1)(`j`+`k`)`k`(`j`+`k`)=`k`−1`k` if you take the $j+k$`j`+`k` out

Factorise $\frac{6x-16}{12}$6`x`−1612 and simplify.

Factorise and simplify $\frac{50m^2+70mn}{80m^2}$50`m`2+70`m``n`80`m`2.

Factorise and simplify $\frac{a^2-81}{9-a}$`a`2−819−`a`.

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

Generalise the properties of operations with rational numbers, including the properties of exponents

Apply algebraic procedures in solving problems