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Partial Fractions

Exercises on mastering the art of partial fraction decomposition.

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Express in partial fractions. Your answer must be in a particular format. Click the question number to see the template the answer must match.

1 $$\dfrac{7x-2}{(x+1)(x-2)}$$

\(\equiv\)

\(\frac{\square }{x+\square }+\frac{\square }{x-\square }\)
2 $$\dfrac{7x+13}{(x+3)(x-1)}$$

\(\equiv\)

\(\frac{\square }{x+\square }+\frac{\square }{x-\square }\)
3 $$\dfrac{4x-41}{x^2-3x-10}$$

\(\equiv\)

\(\frac{\square }{x+\square }-\frac{\square }{x-\square }\)
4 $$\dfrac{x}{2x^2+5x+3}$$

\(\equiv\)

\(\frac{\square }{\square x+\square }-\frac{\square }{x+\square }\)
5 $$\dfrac{x}{x^2-5x+4}$$

\(\equiv\)

\(\frac{\square }{\square (x-\square )}-\frac{\square }{\square (x-\square )}\)
6 $$\dfrac{3x+2}{x^2-4}$$

\(\equiv\)

\(\frac{\square }{x+\square }+\frac{\square }{x-\square }\)
7 $$\dfrac{2x-1}{x^2-2x-3}$$

\(\equiv\)

\(\frac{\square }{\square (x+\square )}+\frac{\square }{\square (x-\square )}\)
8 $$\dfrac{x+9}{3x^2-5x-2}$$

\(\equiv\)

\(\frac{\square \square }{\square (x-\square )}-\frac{\square \square }{\square (\square x+\square )}\)
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This is Partial Fractions level 1. You can also try:
Level 2 Level 3 Level 4

Instructions

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Description of Levels

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Before starting this exercise you may want to check out Algebraic Fractions.

Refer to the syllabus for the course you are following to establish which levels are required.

Level 1 - Linear numerator and quadratic denominator decomposing to two fractions with linear denominators

Example:

$$ \frac{px+q}{(x-a)(x-b)} \equiv \frac{A}{x-a} + \frac{B}{x-b} $$

Level 2 - Linear or quadratic numerator and cubic denominator decomposing to three fractions with linear denominators

Example:

$$ \frac{px^2+qx+r}{(x-a)(x-b)(x-c)} \equiv\frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c} $$

Level 3 - Linear or quadratic numerator and a denominator with a repeated factor

Examples:

$$ \frac{px+q}{(x-a)^2} \equiv \frac{A}{x-a} + \frac{B}{(x-a)^2} $$ $$ \frac{px^2+qx+r}{(x-a)^2(x-b)} \equiv \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b} $$

Level 4 - Linear or quadratic numerator and a denominator with a quadratic factor that cannot be factorised

Example:

$$ \frac{px^2+qx+r}{(x-a)(x^2+bx+c)} \equiv \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c} $$

If the degree of the numerator is greater than or equal to the degree of the denominator you will need to divide first. See the exercise on Polynonial Division.

Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).

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Example - Level 1

Express as partial fractions $$\frac{3x - 2}{x^2 + x - 12}$$

Solution:

We first factorise the denominator.

$$x^2 + x - 12$$ factorises as $$ (x+4)(x-3) $$

Thus, we rewrite the expression in the form:

$$\frac{3x - 2}{(x + 4)(x - 3)} = \frac{A}{x + 4} + \frac{B}{x - 3}$$

Where A and B are constants to be determined.

By multiplying through by the common denominator $$ (x+4)(x-3) $$

$$3x - 2 = A(x - 3) + B(x + 4)$$

Expanding and equating coefficients, we obtain:

$$3x - 2 = Ax - 3A + Bx + 4B$$

Combining like terms, this becomes:

$$3x - 2 = (A + B)x + (-3A + 4B)$$

Equating the coefficients of x and the constant terms, we have:

For the coefficients of x: $$3 = A + B$$

For the constant terms: $$-2 = -3A + 4B$$

Solving these equations simultaneously gives:

Let's substitute $$B = 3 - A$$ into the second equation:

$$-2 = -3A + 4(3 - A)$$

This simplifies to:

$$-2 = -3A + 12 - 4A$$

$$-2 = -7A + 12$$

$$-7A = -14$$

$$A = 2$$

Substituting $$A = 2$$ into $$B = 3 - A$$ gives:

$$B = 3 - 2 = 1$$

Therefore, the expression in partial fractions is:

$$\frac{3x - 2}{x^2 + x - 12} = \frac{2}{x + 4} + \frac{1}{x - 3}$$

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