Highest common factor

Factorisation is finding common factors (usually the HCF) between all the terms in a long algebraic expression and using them to rewrite the expression as a product of many factors. In some ways it can be thought of as the opposite of expanding brackets as factorisation puts many terms into products of brackets. We've already come across some of the basics of factorisation in Adding in brackets and Letters as factors.

To factorise an expression, just follow the following format for a three-term expression:

$AB+AC+AD=A\left(B+C+D\right)$AB+AC+AD=A(B+C+D)

where $A$A is the HCF of all the terms, and can be extended to examples with more or less than three terms.

Where there are many different variables involved, just look at each one individually and see what the HCF is between the terms.

Example

Question 1

Question 2

Question 3

Factorise the following by taking out the highest negative common factor: $-5x^2+20x$−5x2+20x

Think: about how the signs will change after taking out a negative factor

Do:The negative HCF between the coefficients $-5$−5 and $20$20 is $-5$−5. The HCF between $x^2$x2 and $x$x is $x$x.

Therefore our overall negative HCF is $-5x$−5x.

The first term $-5x^2$−5x2 will be positive after dividing by $-5x$−5x, and $20x$20x will be negative after dividing by $-5x$−5x.

So $-5x^2+20x=-5x\left(x-4\right)$−5x2+20x=−5x(x−4)

Question 4

$4xy^2+2xy$4xy2+2xy	$=$=	$\left(2xy\right)\left(2y\right)+\left(2xy\right)\times\left(1\right)$(2xy)(2y)+(2xy)×(1)
Highest common factor is $2xy$2xy, so remove that from each component
	$=$=	$2xy\left(2y+1\right)$2xy(2y+1)	Factorising completely

Question 5