Cover Up Method

The Cover Up Method is a fast technique for finding partial fraction coefficients when all factors are linear.

How it works:

1. Identify the linear factor you want to find the coefficient for
2. Find the root of that factor (set it equal to zero and solve for \(x\))
3. Substitute this root into the original fraction, but "cover up" (ignore) the factor whose coefficient you are finding
4. The result is your coefficient

Example 1:

\[\frac{x}{(1-x)(2x-1)} = \frac{A}{1-x} + \frac{B}{2x-1}\]

Finding A: Root of \((1-x)\) is \(x=1\). Cover up \((1-x)\) and substitute \(x=1\):

\[A = \frac{1}{2(1)-1} = \frac{1}{1} = 1\]

Finding B: Root of \((2x-1)\) is \(x=\frac{1}{2}\). Cover up \((2x-1)\) and substitute \(x=\frac{1}{2}\):

\[B = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1\]

Example 2:

\[\frac{2x+5}{(3x-1)(x+1)} = \frac{A}{3x-1} + \frac{B}{x+1}\]

Finding A: Root of \((3x-1)\) is \(x=\frac{1}{3}\). Cover up \((3x-1)\) and substitute \(x=\frac{1}{3}\):

\[A = \frac{2\left(\frac{1}{3}\right)+5}{\frac{1}{3}+1} = \frac{\frac{2}{3}+5}{\frac{4}{3}} = \frac{\frac{17}{3}}{\frac{4}{3}} = \frac{17}{4}\]

Finding B: Root of \((x+1)\) is \(x=-1\). Cover up \((x+1)\) and substitute \(x=-1\):

\[B = \frac{2(-1)+5}{3(-1)-1} = \frac{-2+5}{-3-1} = \frac{3}{-4} = -\frac{3}{4}\]

Important Limitation:

The Cover Up Method only works for linear factors. If you have a quadratic factor like \((x^2+1)\), you cannot use cover up for those terms. Instead, use the comparing coefficients method (equate coefficients of like powers of \(x\)) for the quadratic part.

Tip: In A-Level exams, you typically deal with linear factors, so the Cover Up Method is very efficient. Once you solve for the linear coefficients, any remaining quadratic terms can be handled using the secondary school method.