A18 - Solving quadratic equations by factorising | AQA GCSE Mathematics

A quadratic equation can have two solutions, one repeated solution, or no real solution. You will use factorised expressions to find exact solutions and graphs to read approximate ones. The key is to turn one quadratic equation into simpler linear equations.

Why a zero product gives solutions

Can one equation have two different correct answers? Try $x^2-x-12=0$ . Both $4$ and $-3$ work. Factorising will show why those two values belong together.

You already know

Factorising reverses expanding. For example, $(x-4)(x+3)=x^2-x-12$ . A root is an $x$ -value where a function is zero; a graph shows it at an $x$ -axis crossing.

Start with ordinary numbers. $0\times7=0$ , and $6\times0=0$ . If two non-zero real numbers are multiplied, the product is not zero. Therefore, if two factors have product zero, at least one factor must be zero. This is the zero-product rule.

Zero-product rule

For real numbers or expressions $A$ and $B$ ,

AB=0\quad\Longrightarrow\quad A=0\text{ or }B=0

The product must equal zero. For example, $(x-4)(x+3)=0$ gives $x-4=0$ or $x+3=0$ .

Not on the formulae sheet: learn it.

So the opening equation has solutions $x=4$ or $x=-3$ . These are alternatives: you substitute one value at a time. The same number does not have to make both brackets zero.

The number $1$ provides a neat surprise. It is equal to its own square, but so is $0$ : the equation $x^2=x$ has exactly these two real solutions. Writing it as $x(x-1)=0$ reveals both. Dividing by $x$ would hide the zero solution.

Key idea

Factorising reveals the possible zeros of a product. A bracket equal to zero gives one possible value of the unknown.

The right side matters. A product equal to $12$ does not mean either factor is zero.

Factorising a quadratic to solve it

When a quadratic is already equal to zero, factorise the expression and set each factor equal to zero.

For $x^2+5x+6$ , the two numbers in the brackets must multiply to $6$ and add to $5$ . They are $2$ and $3$ , so the expression is $(x+2)(x+3)$ . Use your earlier factorising skills, then finish by solving the linear equations. This solving method is not on the formulae sheet: learn it.

Worked example 1: two negative solutions

Question. Solve $x^2+5x+6=0$ . Non-calculator.

Plan. Find the factor pair, write the brackets, then make each bracket zero.

\begin{aligned} 2\times3&=6,\quad2+3=5&&\text{choose the pair}\\ (x+2)(x+3)&=0&&\text{factorise}\\ x+2=0&\quad\text{or}\quad x+3=0&&\text{zero-product rule}\\ x=-2&\quad\text{or}\quad x=-3&&\text{solve each equation} \end{aligned}

Why solve both brackets? Either zero factor makes the whole product zero.

Answer. $x=-2$ or $x=-3$ .

Check. At $-2$ , $4-10+6=0$ ; at $-3$ , $9-15+6=0$ .

Change only the constant term from $+6$ to $-6$ . The factors now have opposite signs, so the roots will too.

Worked example 2: opposite-sign solutions

Question. Solve $x^2+5x-6=0$ . Non-calculator.

Plan. Choose numbers with product $-6$ and sum $5$ , then use the zero-product rule.

\begin{aligned} 6\times(-1)&=-6,\quad6+(-1)=5&&\text{choose signed factors}\\ (x+6)(x-1)&=0&&\text{factorise}\\ x+6=0&\quad\text{or}\quad x-1=0&&\text{make each factor zero}\\ x=-6&\quad\text{or}\quad x=1&&\text{solve} \end{aligned}

Why use signed numbers? Both the product and sum must match the expression, including its signs.

Answer. $x=-6$ or $x=1$ .

Check. $36-30-6=0$ and $1+5-6=0$ .

A repeated factor gives a repeated root. For $(x-3)^2=0$ , both factors give $x=3$ , so there is only one distinct solution. Listing $3$ twice does not create another value.

Watch out

Wrong: $(x+6)(x-1)=0$ , so $x=6$ or $-1$ . Those numbers are in the brackets but do not make them zero. Right: solve $x+6=0$ and $x-1=0$ , giving $-6$ and $1$ .

Keep the negative constant, but change the middle coefficient.

Keeping the zero solution

A quadratic with no constant term has a common factor of $x$ . Taking it out reveals a root that division can accidentally remove.

For example, $x^2-4x=x(x-4)$ . At $x=0$ , the first factor is zero. At $x=4$ , the second factor is zero. You need both possibilities.

Worked example 3: a common factor of x

Question. Solve $x^2-4x=0$ . Non-calculator.

Plan. Take out $x$ , then solve the two zero-factor equations.

\begin{aligned} x(x-4)&=0&&\text{factor out }x\\ x=0&\quad\text{or}\quad x-4=0&&\text{zero-product rule}\\ x=0&\quad\text{or}\quad x=4&&\text{solve the second factor} \end{aligned}

Why factor instead of dividing by $x$ ? Division would be invalid when $x=0$ , which is one of the solutions.

Answer. $x=0$ or $x=4$ .

Check. $0^2-4(0)=0$ and $4^2-4(4)=0$ .

Change only the scale of the equation: multiply both terms by $3$ . A non-zero constant factor does not change where a product is zero.

Worked example 4: a numerical common factor too

Question. Solve $3x^2-12x=0$ . Non-calculator.

Plan. Take out the full common factor $3x$ , then identify the zero factors.

\begin{aligned} 3x(x-4)&=0&&\text{factor out }3x\\ x=0&\quad\text{or}\quad x-4=0&&3\ne0\\ x=0&\quad\text{or}\quad x=4&&\text{solve} \end{aligned}

Why does $3$ give no extra root? It is a fixed non-zero number, so it cannot make the product zero.

Answer. $x=0$ or $x=4$ .

Check. At $4$ , $3(16)-12(4)=48-48=0$ ; at $0$ , both terms are zero.

The same idea applies to a negative common factor. Factoring it out changes the brackets but does not remove either zero.

Watch out

Wrong: $3x^2=12x$ , divide by $x$ , so $x=4$ is the only answer. Right: $3x(x-4)=0$ , giving $0$ as well as $4$ .

Use a new middle coefficient while keeping the common factor visible.

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A18 - Solving quadratic equations by factorising

Why a zero product gives solutions

Factorising a quadratic to solve it

Keeping the zero solution

Continue this lesson.

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