Factorising trinomials

A trinomial expression takes the form:

\(a{x^2} + bx + c\)

To factorise a trinomial expression, put it back into a pair of brackets.

To find the terms that go in each bracket, look for a pair of numbers which multiply to give the last number and add together to give the middle number.

Example

Factorise the following trinomial expression.

\({a^2} + 7a + 12\)

Firstly find the numbers that go into each bracket.

Look for a pair of numbers which multiply to give the last number and add to give the middle number.

For this expression, we're looking for two numbers which multiply to give 12 and add to give 7.

Factors of 12	Adding the factors
\(1 \times 12\)	\(13\)
\(-1\times-12\)	\(-13\)
\(2 \times 6\)	\(8\)
\(-2 \times -6\)	\(- 8\)
\(3 \times 4\)	\(7\)
\(- 3 \times - 4\)	\(- 7\)

Factors of 12	\(1 \times 12\)
Adding the factors	\(13\)

Factors of 12	\(-1\times-12\)
Adding the factors	\(-13\)

Factors of 12	\(2 \times 6\)
Adding the factors	\(8\)

Factors of 12	\(-2 \times -6\)
Adding the factors	\(- 8\)

Factors of 12	\(3 \times 4\)
Adding the factors	\(7\)

Factors of 12	\(- 3 \times - 4\)
Adding the factors	\(- 7\)

Therefore the answer is: \((a + 3)(a + 4)\)

When factorising expressions, always check by multiplying out the brackets again.

Now try the example questions below.

Question

Factorise the following:

\({b^2} + 9b + 20\)

We want factors of 20 that will also add to give 9.

Factors of 20	Adding the factors
\(1x20\)	\(21\)
\(-1x-20\)	\(- 21\)
\(2x10\)	\(12\)
\(-2x-10\)	\(- 12\)
\(4 x 5\)	\(9\)
\(-4x - 5\)	\(- 9\)

Factors of 20	\(1x20\)
Adding the factors	\(21\)

Factors of 20	\(-1x-20\)
Adding the factors	\(- 21\)

Factors of 20	\(2x10\)
Adding the factors	\(12\)

Factors of 20	\(-2x-10\)
Adding the factors	\(- 12\)

Factors of 20	\(4 x 5\)
Adding the factors	\(9\)

Factors of 20	\(-4x - 5\)
Adding the factors	\(- 9\)

The required factors are 4 and 5.

Therefore the answer is: \((b + 4)(b + 5)\)

Question

Factorise the following:

\({c^2} - 3c - 10\)

Look for two numbers which multiply together to give -10 and add together to give -3.

Factors of -10	Adding the factors
\(1x - 10\)	\(- 9\)
\(-1 x 10\)	\(9\)
\(2 x -5\)	\(-3\)
\(-2 x 5\)	\(3\)

Factors of -10	\(1x - 10\)
Adding the factors	\(- 9\)

Factors of -10	\(-1 x 10\)
Adding the factors	\(9\)

Factors of -10	\(2 x -5\)
Adding the factors	\(-3\)

Factors of -10	\(-2 x 5\)
Adding the factors	\(3\)

The required factors are 2 and -5

Therefore the answer is \((c - 5)(c + 2)\)

Question

Factorise the following:

\({y^2} + y - 30\)

Firstly find factors of -30 that also add to give 1.

Factors of -30	Adding the factors
\(1x - 30\)	\(- 29\)
\(-1 x 30\)	\(29\)
\(2x - 15\)	\(-13\)
\(-2 x 15\)	\(13\)
\(3x - 10\)	\(- 7\)
\(-3 x 10\)	\(7\)
\(5x - 6\)	\(-1\)
\(- 5 x 6\)	\(1\)

Factors of -30	\(1x - 30\)
Adding the factors	\(- 29\)

Factors of -30	\(-1 x 30\)
Adding the factors	\(29\)

Factors of -30	\(2x - 15\)
Adding the factors	\(-13\)

Factors of -30	\(-2 x 15\)
Adding the factors	\(13\)

Factors of -30	\(3x - 10\)
Adding the factors	\(- 7\)

Factors of -30	\(-3 x 10\)
Adding the factors	\(7\)

Factors of -30	\(5x - 6\)
Adding the factors	\(-1\)

Factors of -30	\(- 5 x 6\)
Adding the factors	\(1\)

The required factors are -5 and 6.

Therefore the answer is \((y - 5)(y + 6)\)

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Factorising an algebraic expressionFactorising trinomials