Factorising Quadratics

Take Note: The Factorising Trick

There is a quick trick to determine whether you should add a $+$ or $–$ sign to your brackets. There are three sub-types which we will go over here.

Sub-type (a): contains all positives

These quadratics contain all positive terms, e.g. $x^2 +5x + 6$. When factorised, both brackets will contain $bf{bf{large{+}}}$.

$x^2 + 5x + 6 = (x+3)(x+2)$

Sub-type (b): $b$ is negative and $c$ is positive

These quadratics contain a negative $b$ value and a positive $c$ value. When factorised, both brackets will contain $bf{large{-}}$.

$x^2 -10x +21 = (x-7)(x-3)$

Sub-type (c): $c$ is negative.

If $c$ is negative, when factorised, one bracket will contain a $bf{+}$ the other will contains a $bf{large{-}}$. The order or these will need to be determined. These are the hardest type and require the most thought.

$x^2 + 3x -18 = (x+6)(x-3)$

Type 1: Factorising quadratics ($a=1$)

When we say $a=1$, we mean the number before $x^2$ in  $x^2+bx+c$ will be $1$ (typically we don’t write the $1$). Any number that appears before an $x$ term is called a coefficient, so in this case, $a$ is the coefficient of $x^2$ which has a value of $1$.

Example: Factorise the following quadratic into two brackets, $x^2textcolor{blue}{-3}x+textcolor{red}{2}$

Step 1: First we can write two brackets with an $x$ placed in each bracket.

$(x,,,,,,,,,,,,,,,,,,,) (x,,,,,,,,,,,,,,,,,,,)$

Step 2: We can identify that this is a sub-type (b) quadratic, meaning both brackets will contain $large{-}$

$(x,,,,,,,,-,,,,,,,,) (x,,,,,,,-,,,,,,,,)$

Step 3: We have to find two numbers which multiply to make $textcolor{red}{2}$ and when added together make $textcolor{blue}{-3}$.

We know both numbers will be negative.

$-2 times -1 = textcolor{red}{2}$

$-2 + -1 = textcolor{blue}{-3}$

Finally add these numbers to the brackets.

$(x-2) (x-1)$

Type 2: Factorising quadratics ($a> 1$)

In this instance the general form of the equation is $ax^2+bx+c$ where $a>1$.

Example: Factorise the following quadratic $4x^2+textcolor{blue}{3x}textcolor{red}{-1}$

Step 1: When $a>1$ it makes things more complicated. It is not immediately obvious what the coefficient of each $x$ term should be. There are two possible options,

$(4x kern{1 cm} ) (x  kern{1 cm} )$ or $(2x kern{1 cm} ) (2x  kern{1 cm} )$

Step 2: We can identify that this quadratic is part of sub-type (c) meaning it can contain $+$ and $–$

This is most important for quadratic pairs which are non-symmetrically creating a third option, all three are shown below.

$begin{aligned}(4x kern{0.4 cm} +kern{0.4 cm} )&(x  kern{0.4 cm}-kern{0.4 cm} ) (4x kern{0.4 cm} -kern{0.4 cm} )&(x  kern{0.4 cm}+kern{0.4 cm} )(2x kern{0.4 cm}+kern{0.4 cm} )&(2x kern{0.4 cm}-kern{0.4 cm} )end{aligned}$

Step 3: We need to find two numbers which when multiplied make $textcolor{red}{-1}$

$textcolor{red}{-1}$ has only one factor.

$-1 times 1 = -1$

Step 4: We need to find a combination which gives $textcolor{blue}{3x}$

We can test our $3$ possibilities,

$begin{aligned}(4x+1)(x -1) &= 4x^2 -3x -1 (4x-1)(x+1) &= 4x^2 + 3x -1 (2x+1)(2x-1) &= 4x^2 -1end{aligned}$

As we can see $(4x-1) (x+1)$ gives the correct expansion and is therefore the answer.