2A - Factorising quadratic expressions

Can you turn a sum of three terms into a product of two brackets? You will use the connections between a quadratic's coefficients and its factors, then decide when integer factors are possible and when exact surds are needed.

Quadratics with leading coefficient one

Which two brackets multiply to give $x^2+7x+12$ ?

Think of a rectangle with side lengths $x+3$ and $x+4$ , for $x>0$ . Multiplying gives four area pieces:

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

The $x$ -terms add because they are like terms. Factorising reverses this expansion: we write the sum as a product. The algebraic identity remains true for every real $x$ , even when the rectangle is no longer a useful length model.

Quadratic expression

A quadratic expression has the form $ax^2+bx+c$ , where $a\ne0$ . The coefficient $a$ is its leading coefficient. A quadratic with leading coefficient $1$ is called monic.

For a monic quadratic, the constant terms in the brackets must have the right product and the right sum.

Product and sum

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

To factorise $x^2+bx+c$ , seek $p,q$ with $p+q=b$ and $pq=c$ .

Not on the formulae sheet: learn it.

If $c>0$ , the two numbers have the same sign; their sum tells you which sign. If $c<0$ , their signs are opposite. Testing the product alone is insufficient.

Worked example 1: positive constant

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

Plan. Find two integers with product $12$ and sum $7$ .

\begin{aligned} 12&=1\times12=2\times6=3\times4 &&\text{list factor pairs}\\ 3+4&=7 &&\text{choose the required sum}\\ x^2+7x+12&=(x+3)(x+4) &&\text{put the numbers in brackets} \end{aligned}

Answer. $(x+3)(x+4)$ .

Check. Expanding gives $x^2+4x+3x+12=x^2+7x+12$ , including the correct middle coefficient.

Keep the middle coefficient $7$ , but change the constant from $12$ to $-18$ . The two bracket numbers must now have opposite signs.

Worked example 2: negative constant

Question. Factorise $x^2+7x-18$ .

Plan. Find integers whose product is $-18$ and whose sum is $7$ .

\begin{aligned} 9\times(-2)&=-18 &&\text{opposite signs give a negative product}\\ 9+(-2)&=7 &&\text{the larger magnitude is positive}\\ x^2+7x-18&=(x+9)(x-2) &&\text{use both signed numbers} \end{aligned}

Answer. $(x+9)(x-2)$ .

Check. $(x+9)(x-2)=x^2-2x+9x-18=x^2+7x-18$ .

A negative middle coefficient also matters. For $x^2-7x+12$ , the numbers are $-3$ and $-4$ : their product is positive but their sum is negative.

Watch out

$(x+2)(x+6)$ is not a factorisation of $x^2+7x+12$ : it expands to $x^2+8x+12$ . Check both the product and the sum.

Use the sign of the constant to narrow the next search.

Other leading coefficients

When the leading coefficient is not one, the coefficients of $x$ in the two brackets must multiply to it.

For example, expanding $(3x+2)(2x-1)$ produces $6x^2$ , two middle terms $-3x+4x$ , and the constant $-2$ . The middle terms have coefficients whose product is $6\times(-2)=-12$ . This suggests a reliable way to reverse the expansion.

First take out any common factor. For the remaining $ax^2+bx+c$ , find integers with product $ac$ and sum $b$ . Split the middle term using those integers, then group the terms so a common bracket appears. This method finds integer factors when such factors exist; an unsuccessful search is not a reason to invent a pair.

Worked example 3: splitting the middle term

Question. Factorise $6x^2+x-2$ .

Plan. Use $ac=-12$ , split $x$ into $4x-3x$ , then factor by grouping.

\begin{aligned} 6x^2+x-2&=6x^2+4x-3x-2 &&\text{since }4+(-3)=1\\ &=2x(3x+2)-(3x+2) &&\text{factor each pair}\\ &=(3x+2)(2x-1) &&\text{take out the common bracket} \end{aligned}

Why this split? The numbers $4$ and $-3$ have product $-12$ ; grouping therefore exposes matching brackets.

Answer. $(3x+2)(2x-1)$ .

Check. Expansion gives $6x^2-3x+4x-2=6x^2+x-2$ .

Now multiply every coefficient by $3$ . The quadratic inside the common factor stays the same, so there is no need to repeat a harder factor search.

Worked example 4: keeping a common factor

Question. Factorise fully $18x^2+3x-6$ .

Plan. Take out $3$ , then factorise the simpler quadratic.

\begin{aligned} 18x^2+3x-6&=3(6x^2+x-2) &&\text{divide every term by }3\\ &=3(6x^2+4x-3x-2) &&\text{split the middle term}\\ &=3(3x+2)(2x-1) &&\text{group as before} \end{aligned}

Answer. $3(3x+2)(2x-1)$ .

Check. The two brackets give $6x^2+x-2$ ; multiplying all three terms by $3$ recovers $18x^2+3x-6$ .

A common factor can also include the variable. For instance, $5x^2-10x=5x(x-2)$ ; the missing constant term is zero. Keep the factor $x$ , since cancelling it would change the expression.

Watch out

When grouping $-3x-2$ , taking out $-1$ gives $-(3x+2)$ , not $-(3x-2)$ . The minus sign multiplies every term in the bracket.

Take out the common factor before splitting the next middle term.

Choosing an exact factorisation

The form of an expression can reveal its factors before any product-and-sum search.

For example, $(x-2)(x+2)=x^2-4$ : the middle terms cancel. The same cancellation works for any two expressions in place of $x$ and $2$ .

Difference of two squares

U^2-V^2=(U-V)(U+V).

This is an identity: expanding the right side cancels $UV$ and $-UV$ .

Not on the formulae sheet: learn it.

Use a difference, not a sum, of squares. The word “factorise” also needs a number system: factors with integer coefficients are more restrictive than factors with real coefficients, which may contain surds.

Worked example 5: recognising two squares

Question. Factorise $9x^2-16$ .

Plan. Recognise $(3x)^2-4^2$ , then use the difference of two squares.

\begin{aligned} 9x^2-16&=(3x)^2-4^2 &&\text{identify each square}\\ &=(3x-4)(3x+4) &&\text{use opposite signs} \end{aligned}

Answer. $(3x-4)(3x+4)$ .

Check. Multiplication gives $9x^2+12x-12x-16=9x^2-16$ .

Change only $16$ to $5$ . The constant is no longer an integer square, but $5=(\sqrt{5})^2$ , so an exact real factorisation still exists.

Worked example 6: exact surd factors

Question. Factorise $9x^2-5$ into real linear factors. Can those factors have integer coefficients?

Plan. Use $3x$ and $\sqrt{5}$ as the two quantities being squared.

\begin{aligned} 9x^2-5&=(3x)^2-(\sqrt{5})^2 &&\text{keep the root exact}\\ &=(3x-\sqrt{5})(3x+\sqrt{5}) &&\text{apply the identity} \end{aligned}

For integer factors $(mx+n)(px+q)$ , we would need $mp=9$ , $nq=-5$ and $mq+np=0$ . The possible magnitudes are $1,9$ or $3,3$ for $m,p$ , and $1,5$ for $n,q$ ; none makes the two middle terms cancel.

Answer. $(3x-\sqrt{5})(3x+\sqrt{5})$ . There is no factorisation into two integer-coefficient linear factors.

Check. The cross terms cancel and the constant is $-(\sqrt{5})^2=-5$ , giving the original expression exactly.

Some quadratics have no real linear factors at all. Since $x^2+4\ge4$ for every real $x$ , it is never zero. A real linear factor would become zero at some real value of $x$ , making the whole product zero. That contradiction rules out real linear factors of $x^2+4$ . Writing $(x-2)(x+2)$ would instead give $x^2-4$ .

Key idea

Failure to find integer factors does not prove that real factors are impossible. $x^2-2$ has exact real factors $(x-\sqrt{2})(x+\sqrt{2})$ , while $x^2+2$ has no real linear factors.

You can now answer the opening question and also recognise when its integer-factor method is unsuitable.

Quick Recap

For a monic quadratic, find two numbers with the required product and sum.
Take out common factors first; split the middle term when the leading coefficient is not one.
A difference of squares has two brackets with opposite signs.
Check the required number system: integer factors, exact real factors or no real linear factors are different outcomes.

The opening expression is $x^2+7x+12=(x+3)(x+4)$ , as expansion confirms.

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2A - Factorising quadratic expressions

Quadratics with leading coefficient one

Other leading coefficients

Choosing an exact factorisation

Quick Recap

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