Solving Quadratic Equations: Factorising

From a quadratic expression to an equation

A quadratic equation contains a squared variable as its highest power. It can be written in the standard form

$ax^2+bx+c=0,$

where $a$, $b$ and $c$ are numbers and $a\ne0$. For example, $x^2+11x+24=0$ is quadratic.

An expression such as $x^2+11x+24$ can be factorised, but it cannot be solved until it forms part of an equation. Factorising rewrites the expression as a product. Solving finds the value or values of the variable that make the equation true.

Why the equation must equal zero

Factorising is useful because of the zero-product property:

$A\times B=0\quad\Longrightarrow\quad A=0\text{ or }B=0.$

A product can equal zero only if at least one of its factors is zero. This conclusion cannot be made when the product equals a non-zero number. For instance, if $A\times B=12$, there are many possible values for both factors.

That is why the first step is always to rearrange a quadratic equation into the form

$ax^2+bx+c=0.$

Solving a monic quadratic

A monic quadratic has a coefficient of $1$ in front of $x^2$. To factorise

$x^2+bx+c,$

find two numbers $p$ and $q$ such that

$pq=c\qquad\text{and}\qquad p+q=b.$

Then

$x^2+bx+c=(x+p)(x+q).$

Consider

$x^2+11x+24=0.$

The numbers $3$ and $8$ multiply to give $24$ and add to give $11$, so

$x^2+11x+24=(x+3)(x+8).$

The equation becomes

$(x+3)(x+8)=0.$

Therefore,

$x+3=0\quad\text{or}\quad x+8=0,$

so

$x=-3\quad\text{or}\quad x=-8.$

The signs in the factors reverse when the equations are solved: $x+3=0$ gives $x=-3$, not $x=3$.

Rearranging before factorising

Some quadratic equations do not initially equal zero. Apply the same operation to both sides so that every term is on one side.

For example,

$x^2=14x-49$

rearranges to

$x^2-14x+49=0.$

This factorises as

$(x-7)(x-7)=0,$

or $(x-7)^2=0$. Both factors give $x=7$, so the equation has one .

When there is a common factor

If every term has a common factor, take it outside a bracket. For example,

$y^2=-12y$

first becomes

$y^2+12y=0.$

Factorising gives

$y(y+12)=0.$

Now use the zero-product property:

$y=0\quad\text{or}\quad y+12=0,$

so

$y=0\quad\text{or}\quad y=-12.$

Dividing the original equation by $y$ would lose the valid solution $y=0$, because division by zero is not permitted. Factorising preserves both possibilities.

When the coefficient of $x^2$ is not 1

For a quadratic $ax^2+bx+c$, the factors may begin with expressions such as $2x$ or $$, rather than both beginning with .

Consider

$2x^2-3x=9.$

First rearrange:

$2x^2-3x-9=0.$

To factorise, find two numbers whose product is $a\times c=2\times(-9)=-18$ and whose sum is $b=-3$. These numbers are and . Split the middle term and factorise by grouping:

$2x^2-6x+3x-9=0,$

$2x(x-3)+3(x-3)=0,$

$(2x+3)(x-3)=0.$

Therefore,

$2x+3=0\quad\text{or}\quad x-3=0,$

which gives

$x=-\frac32\quad\text{or}\quad x=3.$

A solution does not have to be an integer even when all the coefficients are integers.

Useful factor patterns

Some quadratics factorise using familiar patterns. A difference of two squares has the form

$a^2-b^2=(a-b)(a+b).$

For example,

$x^2-25=0$

becomes

$(x-5)(x+5)=0,$

so $x=5$ or $x=-5$.

A perfect-square quadratic produces identical factors, such as

$x^2-10x+25=(x-5)^2.$

If $(x-5)^2=0$, the repeated root is .

Checking the solutions

There are two useful checks:

1. Expand the factors. Expanding $(2x+3)(x-3)$ gives $2x^2-3x-9$, so the factorisation is correct.

Checking against the original equation is especially helpful when rearrangement has been required.