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)

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Categories: Algebra
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