Equations that need expanding first

Some equations have brackets on one or both sides, such as \((x+1)(x-4)=14\). Expand the brackets, collect all the terms on one side, then factorise again.

Why: \((x+1)(x-4)=14\) is a product equal to \(14\), not \(0\), so the factors \(x+1\) and \(x-4\) cannot be set to \(14\) separately. Expanding gives a new quadratic with \(0\) on one side, whose factors are different.

Method

1. Expand every bracket, taking care with a minus sign in front of a bracket.
2. Move all the terms to one side and collect like terms.
3. Factorise the new quadratic and solve.
4. If both sides share a factor such as \((x+3)\), do not divide by it; move everything to one side and take it out as a common factor.

Example

Solve \((x+1)(x-4)=14\).

\(\begin{aligned} x^2-3x-4 &= 14 \\ x^2-3x-18 &= 0 \\ (x-6)(x+3) &= 0 \\ x=6 \quad &\text{or}\quad x=-3 \end{aligned}\)

Solve \(x(x-5)=2(x-5)\).

\(\begin{aligned} x(x-5)-2(x-5) &= 0 \\ (x-5)(x-2) &= 0 \\ x=5 \quad &\text{or}\quad x=2 \end{aligned}\)

Check: \((6+1)(6-4)=14\) and \((-3+1)(-3-4)=14\). A common error is to write \(x+1=14\) or \(x-4=14\), which gives \(x=13\) or \(x=18\); neither value satisfies the equation. Dividing \(x(x-5)=2(x-5)\) by \((x-5)\) would lose the root \(x=5\).