Factorising a Quadratic Trinomial

Factorising a Quadratic Trinomial

Factorising is the reverse of calculating the product of factors. In order to factorise a quadratic, we need to find the factors which, when multiplied together, equal the original quadratic.

Consider a quadratic expression of the form \(a{x}^{2} + bx\). We see here that \(x\) is a common factor in both terms. Therefore \(a{x}^{2} + bx\) factorises as \(x(ax + b)\). For example, \(8{y}^{2} + 4y\) factorises as \(4y(2y + 1)\).

Another type of quadratic is made up of the difference of squares. We know that:

\[(a + b)(a - b) = {a}^{2} - {b}^{2}\]

So \({a}^{2} - {b}^{2}\) can be written in factorised form as \((a + b)(a - b)\).

This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down the factors. These types of quadratics are very simple to factorise. However, many quadratics do not fall into these categories and we need a more general method to factorise quadratics.

We can learn about factorising quadratics by looking at the opposite process, where two binomials are multiplied to get a quadratic. For example:

\begin{align*} (x + 2)(x + 3) & = {x}^{2} + 3x + 2x + 6 \\ & = {x}^{2} + 5x + 6 \end{align*}

We see that the \({x}^{2}\) term in the quadratic is the product of the \(x\)-terms in each bracket. Similarly, the \(\text{6}\) in the quadratic is the product of the \(\text{2}\) and \(\text{3}\) in the brackets. Finally, the middle term is the sum of two terms.

So, how do we use this information to factorise the quadratic?

Let us start with factorising \({x}^{2} + 5x + 6\) and see if we can decide upon some general rules. Firstly, write down the two brackets with an \(x\) in each bracket and space for the remaining terms.

\[(x \qquad )(x \qquad)\]

Next, decide upon the factors of \(\text{6}\). Since the \(\text{6}\) is positive, possible combinations are: 1 and 6, 2 and 3, \(-\text{1}\) and \(-\text{6}\) or \(-\text{2}\) and \(-\text{3}\).

Therefore, we have four possibilities:

Next, we expand each set of brackets to see which option gives us the correct middle term.

We see that Option 3, \((x+2)(x+3)\), is the correct solution.

The process of factorising a quadratic is mostly trial and error but there are some strategies that can be used to ease the process.