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.


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