2.2 - Factorising quadratic expressions

2.2 - Factorising quadratic expressions

You will reverse bracket expansion by choosing factors with the right coefficients and signs. Both tiers factorise quadratics with leading coefficient 1, including missing terms. Higher learners also factorise quadratics with other leading coefficients and extract common factors first.

Quadratics with leading coefficient one

Can every quadratic be written as two linear factors with integer constants? Keep x2+1x^2+1 in mind as we build a reliable way to find factors and decide when a proposed pair cannot work.

You already know

Expanding (x+p)(x+q)(x+p)(x+q) gives four products. The middle products collect, and expanding an answer checks a factorisation.

A quadratic expression has highest power 2. Its general form is ax2+bx+cax^2+bx+c, with a≠0a\ne0. Either bb or cc may be zero. A quadratic with a=1a=1 is called monic. For example, x2−3xx^2-3x is quadratic; 3x+23x+2 and x3+2x^3+2 are not.

Sum and product for a monic quadratic

(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x^2+(p+q)x+pq

To factorise x2+bx+cx^2+bx+c, find pp and qq such that p+q=bp+q=b and pq=cpq=c. Both conditions must hold.

Not in either tier's formulae sheet: learn it.

List the integer factor pairs of cc, then check their sums. If c>0c>0, the signs match; the sign of bb decides whether both are positive or both negative. If c<0c<0, the signs are opposite. The pair with greater magnitude decides the sign of the sum.

Worked example 1: positive sum and product

Question. Factorise x2+7x+12x^2+7x+12.

Plan. List factor pairs of 12 and choose the pair with sum 7.

1+12=13,2+6=8,3+4=7check factor pairsx2+7x+12=(x+3)(x+4)sum 7, product 12\begin{aligned} 1+12&=13,\quad2+6=8,\quad3+4=7 &&\text{check factor pairs}\\ x^2+7x+12&=(x+3)(x+4) &&\text{sum }7,\text{ product }12 \end{aligned}

Why this step? The constant checks the product, while the middle coefficient checks the sum; matching just one is insufficient.

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

Check. Expanding gives x2+4x+3x+12=x2+7x+12x^2+4x+3x+12=x^2+7x+12.

Change only the sign of the middle coefficient. The product is still positive, but the required sum is negative.

Worked example 2: changing the middle sign

Question. Factorise x2−7x+12x^2-7x+12.

Plan. Use a pair with product 12 and sum −7-7.

(−3)(−4)=12check the product−3+(−4)=−7check the sumx2−7x+12=(x−3)(x−4)write the factors\begin{aligned} (-3)(-4)&=12 &&\text{check the product}\\ -3+(-4)&=-7 &&\text{check the sum}\\ x^2-7x+12&=(x-3)(x-4) &&\text{write the factors} \end{aligned}

Why this step? Two negative constants give a positive constant term and a negative middle term.

Answer. (x−3)(x−4)(x-3)(x-4).

Check. Expanding gives x2−4x−3x+12=x2−7x+12x^2-4x-3x+12=x^2-7x+12.

For opposite signs, the same rule still works. In x2−x−20x^2-x-20, the pair −5-5 and 4 has product −20-20 and sum −1-1, so the factors are (x−5)(x+4)(x-5)(x+4). Similarly, x2+2x−15=(x+5)(x−3)x^2+2x-15=(x+5)(x-3).

Watch out

The line x2+7x+12=(x+2)(x+6)x^2+7x+12=(x+2)(x+6) is wrong. The product of the constants is 12, but their sum is 8. The correct pair is 3 and 4.

“Factorise” asks for a product, not values of xx. There is no equation here to solve. Factor order does not matter.

Missing terms and repeated factors

A missing term has coefficient zero. The same sum-and-product rule handles that case: you do not need to invent a new expression.

If the linear term is missing, the pair must add to zero, so use opposite numbers. If the constant is zero, take out a common factor xx. If the two factors are identical, write them as a square.

Special monic factorisations

x2−d2=(x−d)(x+d)x2+2dx+d2=(x+d)2x2+bx=x(x+b)\begin{aligned} x^2-d^2&=(x-d)(x+d)\\ x^2+2dx+d^2&=(x+d)^2\\ x^2+bx&=x(x+b) \end{aligned}

These identities hold for real values. The first is the difference of two squares; the second has a repeated factor.

Not in either tier's formulae sheet: learn them.

Each identity comes from expansion. In the first, the middle terms cancel. In the second, they add; this is why the middle coefficient must be checked even when the constant is a square.

Worked example 3: three useful patterns

Question. Factorise (a) x2−25x^2-25, (b) x2−10x+25x^2-10x+25, (c) x2+5xx^2+5x.

Plan. Use a zero middle coefficient, a repeated pair, and a zero constant in turn.

x2−25=(x−5)(x+5)opposites: sum 0x2−10x+25=(x−5)2sum −10, product 25x2+5x=x(x+5)take out x\begin{aligned} x^2-25&=(x-5)(x+5) &&\text{opposites: sum }0\\ x^2-10x+25&=(x-5)^2 &&\text{sum }-10,\text{ product }25\\ x^2+5x&=x(x+5) &&\text{take out }x \end{aligned}

Why this step? Missing terms still have coefficients: zero is information that guides the factor pair.

Answer. (a) (x−5)(x+5)(x-5)(x+5), (b) (x−5)2(x-5)^2, (c) x(x+5)x(x+5).

Check. Expanding gives x2+5x−5x−25x^2+5x-5x-25, x2−5x−5x+25x^2-5x-5x+25, and x2+5xx^2+5x, respectively.

Change only the shared pattern number from 5 to 3. The same three decisions still apply.

Worked example 4: the same patterns with a new number

Question. Factorise (a) x2−9x^2-9, (b) x2−6x+9x^2-6x+9, (c) x2+3xx^2+3x.

Plan. Choose opposite constants, repeated constants, or a common letter factor.

x2−9=(x−3)(x+3)opposites: product −9x2−6x+9=(x−3)2two constants of −3x2+3x=x(x+3)common factor x\begin{aligned} x^2-9&=(x-3)(x+3) &&\text{opposites: product }-9\\ x^2-6x+9&=(x-3)^2 &&\text{two constants of }-3\\ x^2+3x&=x(x+3) &&\text{common factor }x \end{aligned}

Why this step? A squared factor needs both the square constant 9 and the middle coefficient −6-6.

Answer. (a) (x−3)(x+3)(x-3)(x+3), (b) (x−3)2(x-3)^2, (c) x(x+3)x(x+3).

Check. At x=4x=4, the original values are 7, 1 and 28; the factorised forms give 1(7)1(7), 121^2 and 4(7)4(7).

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