Expanding and Factorising Quadratics

Expansion and factorisation are reverse processes

To expand brackets, multiply to remove the brackets. To factorise, reverse that process by rewriting an expression as a product. For example,

$(x+2)(x+5)=x^2+7x+10$

so the reverse statement is

$x^2+7x+10=(x+2)(x+5).$

This connection gives a reliable check: expand any proposed factorisation and see whether it returns the original expression.

Expanding two binomials

The distributive law means that every term in one bracket must multiply every term in the other. For $(x+3)(x-4)$:

$(x+3)(x-4)=x(x-4)+3(x-4)$

The middle terms are then collected because they are like terms. Take particular care with signs: in the example, $3\times(-4)=-12$.

More generally,

$(x+p)(x+q)=x^2+(p+q)x+pq.$

This explains the number-pair method used later for factorising: the two numbers inside the brackets add to make the coefficient of and multiply to make the constant term.

Expanding a product of two binomials is additional Foundation content. At Higher tier, the requirement extends to products of two or more binomials.

Expanding three or more binomials

Deal with two brackets at a time. The order may be chosen because multiplication is commutative and associative; a helpful choice can reduce the work. Keep the unexpanded brackets visible until they are needed.

For example,

$(x+2)(x-3)(x+4)$ $=(x^2-x-6)(x+4)$

An alternative is to expand $(x-3)(x+4)$ first. Either route must give the same final expression. When there are more brackets, continue multiplying by one bracket at a time and collect like terms after each stage.

Factorising $x^2+bx+c$

A quadratic has highest power 2. First arrange its terms in the standard order $x^2+bx+c$. To factorise it into $(x+p)(x+q)$, find two numbers and satisfying

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

For $x^2+6x+8$, the numbers 2 and 4 multiply to 8 and add to 6, so

$x^2+6x+8=(x+2)(x+4).$

Signs are part of the number pair. For $x^2-x-12$, the product is negative, so one number must be positive and the other negative. The pair 3 and $-4$ has product $-12$ and sum $-1$:

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

If $c$ is positive, the two numbers have the same sign; their sum determines whether both are positive or both are negative. If $c$ is negative, their signs differ.

Factorising $ax^2+bx+c$ when $a\ne1$

This is Higher-tier content. The brackets no longer both begin with $x$, because their first terms must multiply to $ax^2$. One systematic method is to split the middle term.

For $4x^2+8x+3$, calculate $ac=4\times3=12$. Find two numbers that multiply to 12 and add to 8: 6 and 2. Replace with and in pairs:

$4x^2+8x+3=4x^2+6x+2x+3$

The matching bracket $(2x+3)$ is what makes the grouping work. Expanding the answer gives $4x^2+6x+2x+3=4x^2+8x+3$, confirming the factorisation.

Where a common factor is present, take it out first. This simplifies the remaining quadratic and helps ensure that the final answer is fully factorised.

Difference of two squares

A subtraction of two complete squares has the special factorisation

$A^2-B^2=(A+B)(A-B).$

It works because expansion gives $A^2-AB+AB-B^2$ and the two middle terms cancel. For example,

$16a^2-25b^2=(4a)^2-(5b)^2=(4a+5b)(4a-5b).$

Both terms must be squares and they must be separated by subtraction. The letters $A$ and $B$ can represent whole algebraic terms, not merely single variables.

Look for a common factor first. For instance,

$3y^2-75=3(y^2-25)=3(y+5)(y-5).$

Stopping at $3(y^2-25)$ would leave a difference of two squares still available, so the expression would not yet be fully factorised.

Expand

Multiply every term in one bracket by every term in the other, then collect like terms:

$(x+p)(x+q)=x^2+(p+q)x+pq.$