Home > Algebra > Factorising quadratics with rational roots

## Factorising quadratics with rational roots

We are going to consider quadratics of the form $ax^2+bx+c$ where we can put them in the form $(mx+n)(px+q)$

A prerequisite of this post is to have read Factorising into a single bracket
Also it should be noted that it is not possible to factorise all quadratics this way, since they do not all have rational roots and some can not be factorised at all.

Step by step instructions

1. Multiply a by c (the $x^2$ coefficient and the constant)
2. Find all the factor pairs to make this number
3. Total the pairs and see which pair equals the x coefficient
4. Split the x term according to the chosen pair
5. Factorise into single brackets the first and second pair of terms.  Make sure that you use the sign of the third term
6. Finally factorise the two resulting terms, where the common factor is the bracket
7. If the brackets are different then the expression does not have rational roots and another method, like completing the square must be used

Examples

Factorise $6x^2+11x+4$

$6 \times 4 = 24$

Factor pairs of 24

1 & 24, -1 & -24, 2& 12, -2 & -12, 3 & 8, -3 & -8, 4 & 6, -4 & -6

The pair that have a total of 11 are 3 & 8

$\therefore$ we change the $11x \rightarrow 3x + 8x$

So the expression becomes $6x^2+3x+8x+4$

Factorising each pair gives $3x(2x+1)+4(2x+1)$

Now factorise each term (common factor $(2x+1)$

We get $(2x+1)(3x+4)$

(Note: the brackets could be in either order)

Factorise $3x^2+10x-8$

$3 \times -8 = -24$

Factor pairs of -24

1 & -24, -1 & 24, 2& -12, -2 & 12, 3 & -8, -3 & 8, 4 & -6, -4 & 6

The pair that have a total of 10 are -2 & 12

$\therefore$ we change the $10x \rightarrow 12x - 2x$

So the expression becomes $3x^2+12x-2x-8$

Factorising each pair gives $3x(x+4)-2(x+4)$.  (Note we factor out -2, we take the sign of the third term)

Now factorise each term (common factor $(x+4)$

We get $(x+4)(3x-2)$