Algebra · Eqora guide
Partial fractions homework: keep the factors
Match each constant to its denominator, justify the polynomial identity and check a fresh problem independently.
Partial fractions homework becomes confusing when the two unknown numerators look interchangeable. You may find plausible constants, copy a neat answer and still attach each constant to the wrong denominator. The useful question is not whether the final expression looks simpler. It is whether adding the proposed pieces recovers the original rational expression on its original domain. That gives you a way to judge every step, including an explanation supplied by an app.
Consider (5x + 1)/((x − 1)(x + 2)). We will decompose it, explain why evaluating a polynomial identity at an excluded input is legitimate, and test a changed problem independently. Eqora can help you capture the exercise and discuss a specific step, but you must check the transcription, assumptions and result yourself. Follow your course's rules about assistance; this workflow is for learning rather than completing an assessment that requires unaided work.
Partial fractions homework starts with the domain
Start by marking the numerator, the denominator and every parenthesis. Here the numerator has degree one and the denominator has degree two, so the fraction is proper. The denominator already factors into two distinct linear factors. Those two facts determine the simple decomposition we will use. A small minus sign matters: x − 1 and x + 1 lead to different excluded values and different coefficient equations. Do not begin by guessing constants from a blurry photograph.
The original expression is defined only when x ≠ 1 and x ≠ −2. Write those restrictions before changing the form. A decomposition describes the same rational function at allowed inputs; it does not give the original function values at its poles. The RMIT explanation of partial fractions distinguishes distinct factors from repeated factors and quadratic factors. That distinction is the reason to inspect the denominator before applying a remembered recipe, rather than treating every rational expression as the same exercise.
Give each denominator its own unknown numerator
For distinct linear factors, set (5x + 1)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2). The constants A and B belong to particular denominators. Keep that pairing visible throughout your work. A is not automatically the coefficient of x in the original numerator, and B is not automatically its constant term. The numerators change when the two fractions are put over their common denominator.
Adding the right side gives [A(x + 2) + B(x − 1)]/[(x − 1)(x + 2)]. The A term acquires x + 2 because that is the missing factor in its denominator. The B term acquires x − 1 for the same reason. If you multiply A by its own denominator factor instead, you are no longer carrying out ordinary fraction addition. This is the most useful line to write explicitly when an otherwise tidy solution goes wrong.
Compare polynomial coefficients in a justified identity
At allowed inputs, multiplying by the nonzero common denominator gives 5x + 1 = A(x + 2) + B(x − 1). Expanding produces 5x + 1 = (A + B)x + (2A − B). For this to hold as a polynomial identity, the coefficients must agree: A + B = 5 and 2A − B = 1. Add these equations to get 3A = 6, so A = 2. Substituting into the first equation gives B = 3.
This reasoning differs from solving the original fraction equation at one chosen input. We need equality for every allowed x, so matching both coefficients is necessary. It is also sufficient here: with A = 2 and B = 3, the two numerator polynomials have the same coefficients. You can now write the decomposition as 2/(x − 1) + 3/(x + 2), with the original restrictions attached. Keeping the short system next to the result makes the justification easy to inspect later.
Understand why the excluded roots can still find constants
An alternative uses 5x + 1 = A(x + 2) + B(x − 1). Substituting x = 1 into this polynomial identity gives 6 = 3A, and substituting x = −2 gives −9 = −3B. These inputs are useful because one term vanishes. They recover A = 2 and B = 3 with less arithmetic. They do not assert that the original fractions are defined at those inputs.
The distinction is the object being evaluated. After finding constants, the numerator equality is an identity between polynomials, which are defined at every real input. Two polynomials agreeing at all allowed real inputs agree identically; the finite excluded set does not change that conclusion. The rational expression still excludes 1 and −2. If your course has not introduced polynomial identities, coefficient matching supplies a direct route that avoids any apparent substitution into an undefined fraction. Either method needs the domain statement in the final answer.
Capture a legible exercise and locate the first uncertain line
If you use Eqora, first make your own attempt through the denominator restrictions and proposed decomposition. Photograph the entire expression squarely, with even light and enough margin to retain the fraction bar and both factors. Check the recognized task against the source before studying any explanation. A mistaken exponent or a missing bracket changes the mathematical problem, even if the later algebra is internally consistent. Re-enter unclear notation rather than accepting an attractive result for the wrong exercise.
Read the steps one at a time and identify the first line you cannot justify. For this example it may be the common-denominator numerator A(x + 2) + B(x − 1), rather than solving the small linear system. The guide to checking answers without an answer key explains why an independent recombination is stronger evidence than a matching-looking final line. An app explanation can prompt that check, but the check itself should remain something you can reproduce without the app open.
Ask one question that exposes the missing connection
A focused follow-up would be: “Why does the numerator over x − 1 get multiplied by x + 2 when the fractions are combined?” Another useful question is: “Which expression is being evaluated at x = 1, and why does that not define the original fraction there?” These requests name a precise gap. They invite a reason you can test, instead of merely asking for another complete solution that repeats the same unexplained step.
Then close or cover the explanation and reconstruct that step in your own words. Draw the missing factor beside each denominator, or expand both sides of the polynomial identity. If the explanation supplied a coefficient pair, verify the pair rather than assuming it is correct because the method sounds familiar. Eqora is a learning aid, not a guarantee of mathematical accuracy or grades. Check the task, notation, domain and final recombination, and use a teacher's clarification when a course convention remains uncertain.
Recombine the pieces before trusting numerical checks
With A = 2 and B = 3, the combined numerator is 2(x + 2) + 3(x − 1) = 2x + 4 + 3x − 3 = 5x + 1. The denominator is unchanged, so this symbolic recombination proves equality at every allowed input. It also detects a common swap: 3/(x − 1) + 2/(x + 2) produces 5x + 4, not 5x + 1. The constants are individually plausible but incorrectly attached.
A numerical check is still useful as a quick alarm. At x = 0, the original value is −1/2, while the decomposition gives −2 + 3/2 = −1/2. Avoid 1 and −2 because they are excluded. Passing one or two numerical checks does not prove an identity; two different expressions can agree at selected points. Use the symbolic numerator comparison for proof, and substitution to catch arithmetic mistakes while you work. If they disagree, revisit the earliest changed sign or denominator pairing.
Notice when this short template is no longer sufficient
If the numerator degree is at least the denominator degree, carry out polynomial division first. For example, (x² + 5x + 1)/(x² + x − 2) = 1 + (4x + 3)/(x² + x − 2). The remainder fraction is proper and can then be decomposed. Applying only A/(x − 1) + B/(x + 2) to the original improper fraction would miss its polynomial part. The guide to polynomial division with missing terms is useful when arranging that division.
A repeated factor also changes the form. For a denominator (x − 1)², include A/(x − 1) + B/(x − 1)²; a single constant over the squared denominator cannot represent every possible proper numerator. An irreducible quadratic over the real numbers generally needs a linear numerator, such as (Cx + D)/(x² + 1). These are boundaries of today's example, not invitations to memorize more symbols without checking the structure. Factor, compare degrees and choose the template before solving for coefficients.
Change the numerator and solve without the explanation
Try (7x − 1)/((x − 1)(x + 2)) on paper, keeping the same excluded values. Set it equal to C/(x − 1) + D/(x + 2), combine the fractions and compare coefficients. You should obtain C + D = 7 and 2C − D = −1. Adding gives 3C = 6, hence C = 2 and D = 5. Recombining produces 2(x + 2) + 5(x − 1) = 7x − 1.
Explain what changed and what did not. The factors still determine the denominator pairing and domain, while the new numerator changes the coefficient system. If you can reconstruct that chain and justify the excluded-root shortcut, you have learned a reusable method. If you only remember the two constants from the first problem, return to the common-denominator line and identify its missing factors. Finish your homework with the original restrictions, an exact decomposition and a check you actually performed.
