Factorising quadratic expressions with other leading coefficients

When the coefficient of \(x^2\) is not \(1\), the brackets have the form \((px+q)(rx+s)\). The outside terms give \(pr=a\) and \(qs=c\), and the \(x\)-term comes from the two cross products, \(ps+qr=b\).

Why: expanding gives \((px+q)(rx+s)=prx^2+(ps+qr)x+qs\). Several pairs can give the right \(x^2\)-term and the right constant, but only one arrangement gives the right middle term, so the cross products must be checked.

Multiplication frame for \(3x^2+10x-8\)

Method

1. Write the factor pairs of the coefficient of \(x^2\) and of the constant term.
2. Choose one factor pair of each. Put one bracket down the side of the frame (an \(x\)-term, then a number) and the other bracket across the top (an \(x\)-term, then a number), then fill in the four products.
3. Check the cross products: they must add to the coefficient of \(x\). If not, swap the two constants or try another pair.
4. Product method (another way): find two numbers with product \(a\times c\) and sum \(b\), split the middle term with them, then factorise in pairs.

Example

Factorise \(3x^2+10x-8\).

Frame: put the bracket \(3x-2\) down the side (\(3x\), then \(-2\)) and the bracket \(x+4\) across the top (\(x\), then \(+4\)). The cross products are \(12x\) and \(-2x\), which add to \(10x\), so \(3x^2+10x-8=(3x-2)(x+4)\).

Product method: \(a\times c=3\times(-8)=-24\), and \(12+(-2)=10\).

\(\begin{aligned} 3x^2+10x-8 &= 3x^2+12x-2x-8 \\ &= 3x(x+4)-2(x+4) \\ &= (3x-2)(x+4) \end{aligned}\)

Check: expand \((3x-2)(x+4)=3x^2+12x-2x-8\). A common error is \((3x+4)(x-2)\): its outside terms are right, but its cross products give \(-6x+4x=-2x\).