Factorising Harder Quadratics (Cambridge International IGCSE Maths)

Factorising harder quadratics

How do I factorise a quadratic expression where a ≠ 1 in ax² + bx + c?

Method 1: Factorising by grouping

• This is shown most easily through an example: factorising $4x^{2}-25x-21$

• We need a pair of numbers that, for $ax^{2}+bx+c$

  • both multiply to give ac

    • ac in this case is 4 × -21 = -84

  • and both add to give b

    • b in this case is -25

  • -28 and +3 satisfy these conditions

  • Rewrite the middle term using -28x and +3x

    • $4x^{2}-28x+3x-21$

  • Group and fully factorise the first two terms, using 4x as the common factor

  • and group and fully factorise the last two terms, using 3 as the common factor

    • 4x (x−7)+3 (x−7)

  • These terms now have a common factor of (x−7)

    • This whole bracket can be factorised out
    • This gives the answer (x−7)(4x+3)

Method 2: Factorising using a grid

• Use the same example: factorising $4x^{2}-25x-21$

• We need a pair of numbers that for $ax^{2}+bx+c$

  • multiply to give ac

    • ac in this case is 4 × -21 = -84

  • and add to give b

    • b in this case is -25

  • -28 and +3 satisfy these conditions

  • Write the quadratic expression in a grid

    • (as if you had used a grid to expand the brackets)
    • splitting the middle term up as -28x and +3*x *(either order)

  • The grid works by multiplying the row and column headings, to give a product in the boxes in the middle

• Write a heading for the first row, using 4x as the highest common factor of 4x² and -28x

• You can then use this to find the headings for the columns, e.g. “What does 4x need to be multiplied by to give 4x²?”

• We can then fill in the remaining row heading using the same idea, e.g. “What does x need to be multiplied by to give +3x?”

• We can now read off the brackets from the column and row headings:

  • (x−7)(4x+3)

Worked Example (a) Factorise $6x^{2}-7x-3$.

Answer:

We will factorise by grouping

We need two numbers that:

multiply to 6 × -3 = -18 and sum to -7

-9, and +2

Split the middle term up using these values

*6x²+ 2x - 9x *- 3

*Factorise 2**x *out of the first two terms

2x(3x + 1) - 9x - 3

Factorise -3 of out the last two terms

2x(3x + 1) - 3(3x + 1)

These have a common factor of (3x + 1) which can be factorised out

Final answer: (3x** + 1)(2x - 3)

(b) Factorise $10x^{2}+9x-7$.

Answer:

We will factorise using a grid

We need two numbers that:

multiply to 10 × -7 = -70 and sum to +9

-5, and +14

Use these values to split the 9x term and write in a grid

	10x2	*-5x
	*+14x	-7

Write a heading using a common factor of 5x from the first row

5x **	*10**x *2	*-5**x **
	*+14x	-7

Work out the headings for the rows, e.g. “What does 5x need to be multiplied by to make 10**x²?”

	2x**	-1
5x**	10x2	*-5x
	*+14x	-7

Repeat for the heading for the remaining row, e.g. “What does 2x need to be multiplied by to make +14x?”

	2x **	-1
5x**	10x2	*-5x
+7	*+14x	-7

Read off the brackets from the column and row headings

Final answer: (2x - 1)(5x + 7)