Factorising quadratics

Sometimes we wish to take an expression, such as \({x^2}\) + 4\({x}\) + 3, and express it in the form (\({x}\) + a)(\({x}\) + \({b}\)).

To do this, we note that if we were to expand (\({x}\) + \({a}\))(\({x}\) + \({b}\)) we would obtain

(\({x}\) + \({a}\))(\({x}\) + \({b}\)) = \({x^2}\) + \({ax}\) + \({bx}\) + \({ab}\)

= \({x^2}\) + (\({a}\) + \({b}\))\({x}\) + \({ab}\)

Looking closely at this expression and comparing it to \({x^2}\) + 4\({x}\) + 3, we can see (\({a}\) + \({b}\)) = 4

ab = 3

We can solve the above by trial and error. We are looking for two numbers which, when multiplied give us 3, and when added give us 4. Clearly if \({a}\) = 1 and \({b}\) = 3, we have a solution that satisfies both criteria.

This means that \({x^2}\) + 4\({x}\) + 3 = (\({x}\) + 3)(\({x}\) + 1)

Example one

Factorise \({x^2}\) - 8\({x}\) + 7

As we learned above we are looking for two numbers which when multiplied give the result +7 and when added give -8.

The obvious choice for numbers that multiply to give 7 are:

(1, 7) and (-1, -7)

-1 + -7 gives the result -8 so clearly this is the correct choice. Therefore:

x² -8\({x}\) + 7 = (\({x}\) - 1)(\({x}\) - 7)

Example two

Factorise \({x^2}\) -4\({x}\) - 5

We are looking for two numbers that when multiplied give the result -5 and when added give -4. The factor pairs of -5 are:

(-5, 1)

(1, -5)

By adding 1 and -5 we obtain the result -4, which is exactly what we need. Therefore:

\({x^2}\) - 4\({x}\) - 5 = (\({x}\) + 1)(\({x}\) - 5)

Question

Factorise \({x^2}\) - 4\({x}\) - 12

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Quadratic expressions - Intermediate & Higher tier – WJECFactorising quadratics