Quadratic (Leaving Cert Mathematics): Revision Notes

Quadratic Factorisation

As the name implies, quadratic factorisation means factorising quadratic expressions. We'll revisit quadratic expressions more in detail.

Here are some examples of quadratics expressions :

• $3x^2+5x+2$
• $-81x^2-27x-9$
• $x^2-9$
• $x^2$ A quadratic in the form $ax^2+bx+c$ can be factorised in terms of it's roots, which are the two points where the curve crosses the $x$-axis.

The roots of a quadratic can be derived from the quadratic formula, commonly known as the $-b$-formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's go through the steps in factorising a quadratic using the example : $x^2-8x+15$.

1. Recall that every quadratic expression is in the form $ax^2+bx+c$.

• $a$ is the coefficient of the $x^2$ term.

• $b$ is the coefficient of the $x$ term.

• $c$ is the constant term. In our case :

• $1$ is the coefficient of the $x^2$ term.

• $-8$ is the coefficient of the $x$ term.

• $15$ is the constant term.

2. After finding out the values of the coefficients, substitute them into the quadratic formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

When subbed in :

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(15)}}{2(1)}$

Our results will be :

$x=3$ and $x=5$

3. We have our two roots solved for. Now bring the root onto the $x$'s side. For $x=3$, we subtract $3$ on both sides :

$x-3=0$

For $=5$, we subtract $5$ on both sides :

$x-5=0$

4. So our two factors of the original quadratic a $(x-3)$ and $(x-5)$ $(x-3)(x-5)$