Partial FractionsExercises on mastering the art of partial fraction decomposition. 
Express in partial fractions. Your answer must be in a particular format. Click the question number to see the template the answer must match.
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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:
$$ \frac{px+q}{(xa)(xb)} \equiv \frac{A}{xa} + \frac{B}{xb} $$Level 2  Linear or quadratic numerator and cubic denominator decomposing to three fractions with linear denominators
Example:
$$ \frac{px^2+qx+r}{(xa)(xb)(xc)} \equiv\frac{A}{xa} + \frac{B}{xb} + \frac{C}{xc} $$Level 3  Linear or quadratic numerator and a denominator with a repeated factor
Examples:
$$ \frac{px+q}{(xa)^2} \equiv \frac{A}{xa} + \frac{B}{(xa)^2} $$ $$ \frac{px^2+qx+r}{(xa)^2(xb)} \equiv \frac{A}{xa} + \frac{B}{(xa)^2} + \frac{C}{xb} $$Level 4  Linear or quadratic numerator and a denominator with a quadratic factor that cannot be factorised
Example:
$$ \frac{px^2+qx+r}{(xa)(x^2+bx+c)} \equiv \frac{A}{xa} + \frac{Bx+C}{x^2+bx+c} $$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/Alevel exam paper questions (worked solutions are available for Transum subscribers).
More on this topic including lesson Starters, visual aids, investigations and selfmarking exercises.
Express as partial fractions $$\frac{3x  2}{x^2 + x  12}$$
Solution:
We first factorise the denominator.
$$x^2 + x  12$$ factorises as $$ (x+4)(x3) $$
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 $$ (x+4)(x3) $$
$$3x  2 = A(x  3) + B(x + 4)$$
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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Ann R,
Saturday, January 27, 2024
"Do you ever check your solutions on a graphics calculator? I plot the question and the answer and make sure that one graph sits on top of the other."