Factorising Quadratics

Introduction: why factorising matters, students 👋

Quadratics appear everywhere in the IB Mathematics: Analysis and Approaches HL course, especially in the study of functions. A quadratic function has the form $f(x)=ax^2+bx+c$, where $a\neq 0$. One of the most useful ways to work with a quadratic is to factorise it, which means rewriting it as a product of simpler expressions such as $f(x)=(x-p)(x-q)$ or $f(x)=a(x-p)(x-q)$.

This skill is important because factorising helps you find the roots of a quadratic, solve equations, sketch graphs, and understand how a function behaves. It also connects directly to the broader function topic: when a quadratic is factorised, its zeros become visible, and those zeros are the $x$-intercepts of the graph 📈.

Learning objectives

By the end of this lesson, students, you should be able to:

explain the key ideas and vocabulary behind factorising quadratics,
factorise common quadratic expressions using IB-style reasoning,
connect factorising to graphs, roots, and function notation,
use factorised form to solve equations and inequalities,
see how quadratics fit into the wider study of functions.

What factorising a quadratic means

A quadratic expression is any polynomial expression of degree $2$, such as $x^2+5x+6$ or $2x^2-3x-2$. Factorising means rewriting the expression as a multiplication of simpler parts.

For example,

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

This works because expanding the brackets gives back the original expression:

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

The factorised form is useful because it reveals the values of $x$ that make the expression equal to zero. If

$$f(x)=(x+2)(x+3),$$

then $f(x)=0$ when $x=-2$ or $x=-3$.

These are called the roots, zeros, or solutions of the equation $f(x)=0$.

Key vocabulary

Quadratic expression: an expression of degree $2$.
Factorised form: writing an expression as a product.
Root/zero: a value of $x$ that makes the function equal to $0$.
Axis of symmetry: the vertical line halfway between the roots of a parabola.
Coefficient: a number multiplying a variable, such as $a$ in $ax^2+bx+c$.

Factoring monic quadratics

A monic quadratic has leading coefficient $1$, so it looks like $x^2+bx+c$.

The main idea is to find two numbers that:

multiply to $c$,
add to $b$.

Example 1

Factorise $x^2+7x+12$.

We need two numbers that multiply to $12$ and add to $7$. Those numbers are $3$ and $4$.

So,

$$x^2+7x+12=(x+3)(x+4).$$

Example 2

Factorise $x^2-x-12$.

We need two numbers that multiply to $-12$ and add to $-1$. Those numbers are $-4$ and $3$.

So,

$$x^2-x-12=(x-4)(x+3).$$

Checking your answer

A good habit in IB Mathematics is to verify by expanding. This reduces mistakes and strengthens understanding.

For instance,

$$ (x-4)(x+3)=x^2+3x-4x-12=x^2-x-12. $$

That confirms the factorisation is correct ✅.

Factoring quadratics with a leading coefficient other than 1

Now consider expressions such as $2x^2+7x+3$. These are still quadratics, but they are not monic because the coefficient of $x^2$ is not $1$.

A common method is to write the quadratic as a product of two brackets:

$$2x^2+7x+3=(2x+1)(x+3).$$

To see why, expand:

$$ (2x+1)(x+3)=2x^2+6x+x+3=2x^2+7x+3. $$

A useful structure

For a quadratic $ax^2+bx+c$, you can often look for factorisation in the form

$$ (px+q)(rx+s), $$

where $pr=a$ and $qs=c$.

Example 3

Factorise $3x^2-10x+8$.

We want two numbers that multiply to $3\cdot 8=24$ and add to $-10$. Those numbers are $-6$ and $-4$.

Split the middle term:

$$3x^2-10x+8=3x^2-6x-4x+8.$$

Now factor by grouping:

$$3x(x-2)-4(x-2).$$

So,

$$3x^2-10x+8=(3x-4)(x-2).$$

This method is especially useful in HL problems where the factorisation is not obvious at first glance.

Factorising and solving equations

Factorising is one of the fastest ways to solve quadratic equations.

$$f(x)=0,$$

then after factorising, use the zero-product property:

$$ab=0,$$

then $a=0$ or $b=0$.

Example 4

Solve

$$x^2+5x+6=0.$$

First factorise:

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

Then set each factor equal to zero:

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

So the solutions are

$$x=-2 \quad \text{or} \quad x=-3.$$

Why this matters in functions

If $f(x)=x^2+5x+6$, then the solutions to $f(x)=0$ are the $x$-intercepts of the graph of $y=f(x)$. The factorised form gives these intercepts immediately.

For example, if

$$f(x)=(x+2)(x+3),$$

then the graph crosses the $x$-axis at $x=-2$ and $x=-3$.

Factorising in graph interpretation

Quadratic functions graph as parabolas. Their factorised form provides direct information about the graph.

$$f(x)=a(x-r_1)(x-r_2),$$

then:

$r_1$ and $r_2$ are the roots,
the $x$-intercepts are $(r_1,0)$ and $(r_2,0)$,
the axis of symmetry is

$$x=\frac{r_1+r_2}{2}.$$

Example 5

Consider

$$f(x)=2(x-1)(x-5).$$

The roots are $x=1$ and $x=5$, so the parabola crosses the $x$-axis at $(1,0)$ and $(5,0)$.

The axis of symmetry is

$$x=\frac{1+5}{2}=3.$$

Because the coefficient $2$ is positive, the parabola opens upward. The factorised form also shows how the vertical stretch affects the graph.

Factorising and inequalities

Factorising is not only for equations. It also helps solve inequalities involving functions.

Example 6

Solve

$$x^2-4x-5>0.$$

First factorise:

$$x^2-4x-5=(x-5)(x+1).$$

Now determine where the product is positive. The critical values are $x=-1$ and $x=5$.

A product of two linear factors is positive when both factors are positive or both are negative. Testing intervals gives:

for $x<-1$, both factors are negative, so the product is positive,
for $-1<x<5$, one factor is positive and the other negative, so the product is negative,
for $x>5$, both factors are positive, so the product is positive.

So the solution is

$$x<-1 \quad \text{or} \quad x>5.$$

This interval method is a key IB technique for quadratic inequalities.

Common mistakes to avoid

When factorising quadratics, students, watch out for these errors:

forgetting the sign of the middle term,
choosing two numbers that multiply correctly but do not add correctly,
not checking by expanding,
mixing up factorised form with completed square form,
assuming every quadratic factors nicely over the integers.

Some quadratics cannot be factorised neatly using integers. For example,

$$x^2+x+1$$

does not factor over the real numbers because its discriminant is

$$b^2-4ac=1-4=-3,$$

which is negative. This means there are no real roots.

Connection to the wider functions topic

Factorising quadratics is part of the bigger study of functions because it links algebraic form to graphical behavior.

In the functions topic, you may be asked to:

interpret a quadratic in function notation such as $f(x)=ax^2+bx+c$,
find roots using factorisation,
compare the graph of $f(x)$ and transformations such as $f(x-h)$ or $af(x)$,
solve equations like $f(x)=0$ or inequalities like $f(x)>0$,
use the roots to understand intercepts and symmetry.

Factorising is also useful in composite and inverse function work when a quadratic expression appears inside a larger expression. Even when the main focus is not factorisation, it often provides the algebra needed to move forward.

Conclusion

Factorising quadratics is a core algebra skill with strong links to functions in IB Mathematics: Analysis and Approaches HL. It lets students rewrite a quadratic expression as a product, solve equations efficiently, identify roots, interpret graphs, and solve inequalities. In short, factorising turns a quadratic from a complicated-looking expression into a structure that is easier to understand and use.

When you practice, always check your factorisation by expanding, and always connect the algebra to the graph. That connection is what makes factorising quadratics such an important tool in the functions topic ✨.

Study Notes

A quadratic has the form $f(x)=ax^2+bx+c$ with $a\neq 0$.
Factorising means writing a quadratic as a product of simpler factors.
For $x^2+bx+c$, find two numbers that multiply to $c$ and add to $b$.
For $ax^2+bx+c$, factorisation often uses splitting the middle term or grouping.
The zero-product property says if $ab=0$, then $a=0$ or $b=0$.
The roots of a factorised quadratic are the values that make each factor equal to zero.
Roots correspond to the $x$-intercepts of the graph of $y=f(x)$.
The axis of symmetry of $f(x)=a(x-r_1)(x-r_2)$ is $x=\frac{r_1+r_2}{2}$.
Factorising helps solve quadratic equations and inequalities.
Always check a factorisation by expanding to confirm it is correct.