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E2.2 Algebraic Manipulation: Expanding, Factorising & Completing the Square

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How do I expand three brackets multiplied together?

Never try to expand all three at once. Multiply the first two brackets together first, simplify that result, and then multiply the whole thing by the third bracket.

(x-2)(x+3)(2x+1)

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

(x^2+x-6)(2x+1) = 2x^3+x^2+2x^2+x-12x-6 = 2x^3+3x^2-11x-6

How do I factorise an expression with four terms?

Group the terms into two pairs that share a common factor, factorise each pair separately, and then factor out the bracket the two pairs have in common:

ax+ay+bx+by = a(x+y)+b(x+y) = (a+b)(x+y)

How do I factorise a quadratic when the coefficient in front of $x^2$ isn't 1?

Find two numbers whose product is $a \times c$ and whose sum is $b$ , use them to split the middle term into two, and then factorise by grouping, exactly as above.

6x^2+x-12: \quad ac=-72,\ b=1 \implies \text{use } 9 \text{ and } -8

6x^2+9x-8x-12 = 3x(2x+3)-4(2x+3) = (3x-4)(2x+3)

Exam Tip

"Factorise fully" means checking your final pair of brackets for any common factor you may have missed — losing this mark is one of the most common ways to drop easy marks on an otherwise correct answer.

How can I rewrite the first two terms of a quadratic expression as the difference of two squares?

For $x^2+bx$ , halve the coefficient of $x$ and square it: this is the number you need to add and subtract to keep the expression unchanged.

x^2-10x = (x-5)^2-5^2 = (x-5)^2-25

The bracket $(x-5)^2$ absorbs the $x^2$ and the $-10x$ term exactly, but it also introduces an extra $+25$ that wasn't in the original expression — subtracting $25$ cancels that out.

How do I complete the square for a full quadratic expression?

Do the step above, then combine the leftover $-25$ with whatever constant term the original quadratic already had.

x^2-10x+3 = (x-5)^2-25+3 = (x-5)^2-22

How do I complete the square when there's a coefficient in front of the $x^2$ term?

Factor that coefficient out of the $x^2$ and $x$ terms only, complete the square inside the bracket, then multiply back out and simplify.

2x^2+8x-3 = 2(x^2+4x)-3 = 2\big[(x+2)^2-4\big]-3 = 2(x+2)^2-8-3 = 2(x+2)^2-11

How do I find the turning point of a quadratic graph by completing the square?

Once a quadratic is written as $a(x+p)^2+q$ , the turning point is simply $(-p, q)$ — because a squared term $(x+p)^2$ can never be negative, so it's smallest (zero) exactly when $x=-p$ , making $y=q$ the minimum (or maximum, if $a<0$ ) value.

y = (x-5)^2-22 \implies \text{turning point at } (5,-22)

Worked Example

Write $3x^2-12x+7$ in the form $a(x+p)^2+q$ , and hence state the coordinates of the minimum point of the curve.

Factor $3$ out of the first two terms:

3(x^2-4x)+7

Complete the square inside the bracket (half of $-4$ is $-2$ , squared is $4$ ):

3\big[(x-2)^2-4\big]+7

Multiply the $3$ back through and combine constants:

3(x-2)^2-12+7 = 3(x-2)^2-5

Read off the turning point: since $3>0$ , this is a minimum at $(2,-5)$ .