GCSE Factorising ax2 + bx + c Practice Questions

Set up the brackets

The $x^2$ term is $2x^2 = 2x \times x$, so each bracket starts with $2x$ and $x$.

Find the number terms

The two numbers must multiply to give $1$ and, after cross-multiplying, give the middle term $3x$. Those numbers are $1$ and $1$.

Write the factorised form

So $2x^2 + 3x + 1 = (2x + 1)(x + 1)$. Expanding $(2x + 1)(x + 1)$ returns $2x^2 + 3x + 1$, confirming the factorisation.

Set up the brackets

The $x^2$ term is $5x^2 = 5x \times x$, so each bracket starts with $5x$ and $x$.

Find the number terms

The two numbers must multiply to give $1$ and, after cross-multiplying, give the middle term $6x$. Those numbers are $1$ and $1$.

Write the factorised form

So $5x^2 + 6x + 1 = (5x + 1)(x + 1)$. Expanding $(5x + 1)(x + 1)$ returns $5x^2 + 6x + 1$, confirming the factorisation.

Identify the coefficients

Compare the quadratic with $ax^2 + bx + c$.

Multiply $a$ and $c$

The AC method starts by finding the product of the first and last coefficients.

Find a suitable pair

We need two numbers with product $-8$ and sum $2$: they are $-2$ and $4$.

Split the middle term

Rewrite the $2x$ term as -2x + 4x so the expression can be grouped.

Factorise in pairs

Take out the common factor of each pair; both leave the bracket $(4x - 1)$.

Write as a product

The shared bracket gives $8x^2 + 2x - 1 = (2x + 1)(4x - 1)$. Expanding confirms every term.

Identify the coefficients

Read off the coefficients of $ax^2 + bx + c$.

Find the product $ac$

The AC method needs the product of the outer coefficients.

List factor pairs

These are the factor pairs of $72$, and because $ac$ is negative one number in the chosen pair must be negative.

Choose the pair that adds to $b$

The pair $6$ and $-12$ multiplies to $-72$ and adds to $-6$.

Split the middle term

Replace $-6x$ with 6x - 12x.

Group into two pairs

Bracket the first two terms and the last two terms.

Factorise each pair

Each pair has a common factor, and both leave the identical bracket $(3x + 2)$.

Take out the common bracket

Factor out $(3x + 2)$ to get the product $(3x - 4)(3x + 2)$.

Expand to check

Multiplying the brackets back out returns the original quadratic.

State the answer

The fully factorised form is $(3x - 4)(3x + 2)$.

Check for a perfect square

The first and last terms are perfect squares, so test for a perfect-square trinomial.

Square root the first term

The first term of the bracket is $2x$.

Square root the last term

The last term of the bracket is $3$.

Propose the square

A perfect square would be $(2x + 3)^2$.

Check the middle term

The middle term matches, so the factorisation is exact.

State the factorisation

Part (a): the quadratic is a perfect square.

Set up the equation

Part (b): now solve the equation.

Use the factorised form

Replace the quadratic with its factorised form.

Take the factor to zero

A squared bracket is zero only when the bracket itself is zero.

Rearrange

Subtract $3$ from both sides.

Solve for $x$

Divide both sides by $2$.

Note the repeated root

Because the bracket is squared, there is a single repeated root.

Interpret the graph

A repeated root means the parabola just touches the $x$-axis at $x = -\tfrac{3}{2}$.

Verify by substitution

Substituting the root gives zero, as required.

State the answer

The factorisation and the repeated solution.