Partial Fractions Practice Problems With Answers

Partial Fractions Practice Problems

The partial fractions lesson covers the decomposition rules and when each applies. This set drills the step after that: clearing denominators and solving for the constants.

How the difficulty levels work

Difficulty	Denominator	Unknowns
Easy	two distinct linear factors	and
Medium	distinct or repeated linear factors	and
Hard	a linear factor and an irreducible quadratic	, and

Easy is the case substitution handles cleanly — each root isolates one constant.

Medium mixes in the repeated factor , where the denominator has only one root. One substitution gets you ; needs coefficient comparison.

Hard adds an irreducible quadratic, so one numerator is linear. Three unknowns, one useful substitution, and the rest from matching powers of .

The method

Multiply both sides by the whole denominator to clear it, then find the constants by either route:

Substitute a root. Each value that zeroes a denominator kills all but one term.

Compare coefficients. Expand both sides and match the coefficient of each power of . This is what you need whenever substitution cannot isolate a constant.

The two mix freely — use substitution for whatever it reaches, then coefficients for the rest.

A worked problem — two distinct linear factors

Decompose . Find .

Step 1 — multiply through by .

Step 2 — substitute , which kills the term.

Step 3 — solve.

For , substitute : , so . Checking by recombining, .

A worked problem — an irreducible quadratic

Decompose . Find .

Step 1 — clear denominators.

Step 2 — substitute , the only real root available.

Step 3 — expand the right side and collect powers.

Step 4 — match the coefficients.

So , and matching the terms gives .

Where people lose marks

Dropping a sign when clearing denominators. Every term gets multiplied, including the ones you are not currently interested in.
Substituting into the original fraction. Do it after clearing, into the polynomial identity — the original is undefined at exactly those values.
Putting a single constant over an irreducible quadratic. It takes . One constant cannot match both a linear and a constant term.
Giving a repeated factor only its highest power. needs both and .
Not checking. Recombining the pieces over the common denominator takes one line and confirms every constant at once.

Frequently Asked Questions

Why is the decomposition form given to me?+

So the answer can be one number. Writing the correct form is a separate skill drilled in the lesson's own practice problems; here the focus is on clearing denominators and solving for the constants, which is where the arithmetic errors actually happen.

What is the fastest way to find A and B?+

Substitute the value that makes one denominator zero. Setting x equal to a root kills every term except one, giving that constant in a single line. It works for distinct linear factors; quadratic factors need coefficient comparison instead.

Why do some problems have an irreducible quadratic?+

Because a quadratic that cannot be factored over the reals takes a linear numerator, Bx + C, not a single constant. Those problems have three unknowns and no substitution isolates B or C, so you have to compare coefficients.

How do I know a quadratic is irreducible?+

Check the discriminant b² − 4ac. If it is negative there are no real roots, so it stays whole. Every quadratic in this set is of the form x² + c with c positive, which is always irreducible.

What if I get a fraction instead of a whole number?+

Then something has gone wrong — every problem here is built backwards from integer constants, so A, B and C are always integers. A fractional answer usually means a sign slipped when clearing denominators.

Is there a limit on how many sets I can generate?+

No. Problems are produced on demand, free, with no account needed. Reload for a fresh set whenever you like.

A/BPartial Fractions Practice Problems