Home > Algebra > Factorising quadratics with integer roots

## Factorising quadratics with integer roots

The general form of a quadratic is $ax^2+bx+c$

We are going to consider cases where $a=1$

Consider $(x+\alpha)(x+\beta)$

If we expand the brackets we get $x^2+\alpha x + \beta x + \alpha \beta$

This simplifies to $x^2+(\alpha + \beta)x + \alpha \beta$

It can be seen from above that the sum of $\alpha$ and $\beta$ is equal to the coefficient of x

and that the product of $\alpha$ and $\beta$ is equal to the value of the constant.

So to factorise a quadratic first:

• Find all the pairs of integers, positive and negative, that multiply together to make the constant value
• See if one of the pairs if added instead can equal the coefficient of x
• If such a pair can be found use them to replace $\alpha$ and $\beta$ in the formula $(x+\alpha)(x+\beta)$
• If a pair can not be found then a different method is needed

Eamples

Factorise $x^2+8x+15$

First lets find the pairs of numbers that when multiplied together make +15.

 1 15 -1 -15 3 5 -3 -5

Now see if any pair when added together we get the value +8

 1 15 16 -1 -15 -16 3 5 8 -3 -5 -8

Since 3 & 5 multiply together to make 15 and add together to make 8 these are out $\alpha$ and $\beta$

So $x^2+8x+15=(x+3)(x+5)$

(Note that I tried to be systematic when finding the factor pairs which equalled 15.  Also note that it does not matter if you give the factorised expression as  $(x+3)(x+5)$ or $(x+5)(x+3)$

Factorise $x^2-7x+12$

The factor pairs

 1 12 -1 -12 2 6 -2 -6 3 4 -3 -4

Now total each pair

 1 12 13 -1 -12 -13 2 6 8 -2 -6 -8 3 4 7 -3 -4 -7

Since -3 & -4 multiply together to make 12 and add together to make -7 these are out $\alpha$ and $\beta$

So $x^2-7x+12=(x-3)(x-4)$

Factorise $x^2+3x-10$

The factor pairs

 1 -10 -1 10 2 -5 -2 5

Now total each pair

 1 -10 -9 -1 10 9 2 -5 -3 -2 5 3

Since -2 & 5 multiply together to make -10 and add together to make 3 these are out $\alpha$ and $\beta$

So $x^2+3x-10=(x-2)(x+5)$

Factorise $x^2-4x-21$

The factor pairs

 1 -21 -1 21 3 -7 -3 7

Now total each pair

 1 -21 -20 -1 21 21 3 -7 -4 -3 7 4

Since 3 & -7 multiply together to make -21 and add together to make -4 these are out $\alpha$ and $\beta$

So $x^2-7x-21=(x+3)(x-7)$

Factorise $x^2-2x-7$

The factor pairs

 1 -7 -1 7

Now total each pair

 1 -7 -6 -1 7 6

Since no pair has a total of -2 that we required and because we have been systematic we know we have considered all possible pairs, we can assume that either this expression does not factorise or that a different method is required.