Pre-Calculus · Rational Expressions
Pre-Calc Diagnostic: Simplify Rational Expressions (Stewart Problem 5)
Four rational-expression problems from Stewart's Pre-Calculus Diagnostic — factor-and-cancel a trinomial ratio, multiply two rationals with difference-of-squares cancellation, subtract with a common denominator and distribute the minus carefully, and reduce a complex fraction in two variables. The patterns every pre-calc and AP Calculus student must own, taught the way we teach them at SOMATH on the Upper West Side.
Direct answer: Problem 5 of Stewart's Pre-Calculus diagnostic asks you to simplify four rational expressions. The answers are (a) x + 2x − 2, x ≠ −1, 2; (b) x − 1x − 3, x ≠ −3, 3, −12; (c) 1x − 2, x ≠ −2, 2; and (d) −(x + y), x ≠ 0, y ≠ 0, x ≠ y. The hardest of the four is (d), the complex fraction in two variables, where you have to track a sign flip that turns y + x into −(x + y). Below is the full walkthrough we use at SOMATH on the Upper West Side.
The four rational-expression patterns every pre-calc student should own
Before we touch the algebra, here's the menu. When you see one of these structures, you should recognize the move immediately:
- Factor and cancel — when both numerator and denominator are polynomials, factor each completely, then cancel matching factors. List the domain restrictions from the ORIGINAL denominator.
- Multiply two rationals — factor everything, multiply across, then cancel. The restrictions come from EVERY denominator that appeared along the way.
- Add or subtract with a common denominator — rewrite each fraction over the LCD, combine numerators, then keep parentheses around the second numerator so the minus distributes correctly.
- Complex fraction — a fraction with fractions inside. Combine top and bottom separately, then divide by multiplying by the reciprocal.
Every part of Problem 5 uses one of these. Now let's work them.
(a) Factor and cancel: x2 + 3x + 2x2 − x − 2
Two trinomials. Factor each by reverse-FOIL. For the numerator, two numbers multiplying to +2 and summing to +3 are 1 and 2. For the denominator, two numbers multiplying to −2 and summing to −1 are −2 and +1:
x2 + 3x + 2x2 − x − 2 = (x + 1)(x + 2)(x − 2)(x + 1)
The (x + 1) appears in both top and bottom — cancel it. But before you cross it off, write down the domain restriction it represents: x ≠ −1. The denominator's other factor gives x ≠ 2.
= x + 2x − 2, x ≠ −1, 2
The restriction x ≠ −1 is the one students forget — because after cancelling, the simplified fraction looks fine at x = −1. The original wasn't. To make the two expressions truly equivalent, you have to carry the restriction forward.
(b) Multiply two rationals: 2x2 − x − 1x2 − 9 · x + 32x + 1
Factor everything before you multiply. The numerator 2x2 − x − 1 is non-monic — use the AC method. ac = 2 · (−1) = −2, so find two numbers multiplying to −2 and summing to −1: those are −2 and +1. Split and group:
2x2 − x − 1 = 2x2 − 2x + x − 1 = 2x(x − 1) + 1(x − 1) = (2x + 1)(x − 1)
The x2 − 9 is a difference of squares: (x − 3)(x + 3). Substitute everything back:
= (2x + 1)(x − 1)(x − 3)(x + 3) · x + 32x + 1
The (x + 3) on top right cancels the (x + 3) on the bottom left. The (2x + 1) on top left cancels the (2x + 1) on bottom right. What remains is:
= x − 1x − 3, x ≠ −3, 3, −12
Three restrictions: x = 3 and x = −3 from the original x2 − 9, and x = −1/2 from the original 2x + 1 in the second fraction's denominator. Every denominator you started with contributes a restriction — even ones you cancelled.
(c) Subtract with a common denominator: x2x2 − 4 − x + 1x + 2
Factor the first denominator: x2 − 4 = (x − 2)(x + 2). That's the LCD — it already contains the second denominator (x + 2). To match denominators, multiply the second fraction's top and bottom by (x − 2):
x2(x − 2)(x + 2) − (x + 1)(x − 2)(x + 2)(x − 2)
Now subtract the numerators — and keep the parentheses around the second numerator until you've distributed the minus to every term inside:
= x2 − (x + 1)(x − 2)(x − 2)(x + 2)
Expand the product inside: (x + 1)(x − 2) = x2 − 2x + x − 2 = x2 − x − 2. Substitute and distribute the minus:
= x2 − (x2 − x − 2)(x − 2)(x + 2) = x2 − x2 + x + 2(x − 2)(x + 2) = x + 2(x − 2)(x + 2)
Notice the second term inside the parentheses was −x, so distributing the minus gives +x; and the constant −2 becomes +2. The x2 terms cancel. Now the numerator x + 2 matches a factor in the denominator — cancel:
= 1x − 2, x ≠ −2, 2
If you wrote x2 − x2 − x − 2 instead (forgetting to flip every sign), the numerator collapses to −x − 2 = −(x + 2) and the final answer comes out −1/(x − 2) — wrong sign. This is the single most-graded mistake in pre-calc rational arithmetic.
(d) Complex fraction in two variables: yx − xy1y − 1x
A complex fraction is a fraction whose numerator and/or denominator themselves contain fractions. The cleanest approach: combine top and bottom separately over a common denominator, then divide.
The top is y/x − x/y. The common denominator is xy. Rewrite each piece:
yx − xy = y2xy − x2xy = y2 − x2xy
Similarly for the bottom 1/y − 1/x:
1y − 1x = xxy − yxy = x − yxy
The complex fraction is now the top divided by the bottom. Dividing fractions = multiply by the reciprocal:
= y2 − x2xy · xyx − y = y2 − x2x − y
The xy in numerator and denominator cancels. Now factor y2 − x2 — this is a difference of squares, but written with y first: y2 − x2 = (y − x)(y + x). So:
= (y − x)(y + x)x − y
Here's the sign-flip move. Notice y − x = −(x − y). Substitute and cancel:
= −(x − y)(y + x)x − y = −(y + x) = −(x + y), x ≠ 0, y ≠ 0, x ≠ y
Three restrictions: x ≠ 0 and y ≠ 0 come from the inner fractions; x ≠ y comes from the outer denominator x − y. All three must appear in the final answer.
The pattern behind all four
Stewart's Problem 5 follows the same diagnostic order in every part: (1) factor every polynomial completely before touching arithmetic, (2) for sums and differences, find the LCD and keep parentheses around every numerator until the very last step, (3) for complex fractions, simplify top and bottom separately, then multiply by the reciprocal, (4) write the domain restrictions from the original expression, not the simplified one. Students who run this checklist on every problem stop losing points on restrictions and stop bleeding sign errors in the subtractions.
Part (d) is the highest-yield insight here. The trick — recognizing that y − x and x − y differ by a sign — comes back in AP Calculus AB when you simplify limits of the form [f(x + h) − f(x)] / h, and it appears throughout partial-fraction integration in AP Calc BC.
How SOMATH teaches rational expressions on the Upper West Side
At SOMATH (226 W 79th St, UWS), we teach rational expressions with a four-step checklist we call FCDR: Factor everything, Cancel only matching factors, Distribute the minus when subtracting, Restrict the domain from the original. Every student writes those four letters in the margin before starting a problem, and they don't touch a single expression until they've factored.
If your high schooler is heading into pre-calc, AP Calc AB/BC, or SAT Math 2, the diagnostic Problem 5 is a great one-session topic: 10 minutes of factoring review, 40 minutes of all four parts, then a 30-minute mixed worksheet. Most students walk in able to do (a) and walk out able to do (d) — the complex fraction — without help.
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FAQ
Why do you have to write the domain restrictions when simplifying a rational expression?
Cancelling a factor like (x + 1) from numerator and denominator changes the function's domain. The original expression (x2+3x+2)/(x2−x−2) is undefined at x = −1 and x = 2. After cancelling you get (x+2)/(x−2), which only flags x = 2. To preserve equivalence you must list x ≠ −1, 2 alongside the simplified form. On the SHSAT, SAT, and AP Calc, the missing restriction is the most common point deduction on this topic.
How do you simplify a complex fraction (a fraction with fractions inside)?
Two methods. Method 1 — combine numerator and denominator separately over a common denominator, then divide (multiply by the reciprocal). Method 2 — multiply the entire expression top and bottom by the LCD of the inner fractions to clear them in one move. Method 1 shows more structure; Method 2 is faster on a timed test.
When subtracting two rational expressions, why is distributing the minus sign the biggest source of errors?
After you write both fractions over a common denominator, the numerator looks like A − (B), where B is a product like (x+1)(x−2). Students often write A − x2 + x + 2 instead of A − (x2 − x − 2) = A − x2 + x + 2 — getting the right answer by luck, OR writing A − x2 − x − 2 and losing both signs inside the parentheses. The fix: ALWAYS keep the parentheses until you have distributed the minus to every term.
Why does y + x become −(x + y) in part (d)?
Look at the final fraction: the denominator x − y carries the minus that flips the sign. Specifically, (y2 − x2) / (x − y) = −(x2 − y2) / (x − y) = −(x − y)(x + y) / (x − y) = −(x + y). The minus comes from the fact that y2 − x2 and x2 − y2 are negatives of each other.
Does Stewart's Problem 5 appear on the AP Calculus exam?
Indirectly, yes. AP Calculus AB and BC both rely on rational-expression simplification as a prerequisite — limits, derivatives of quotients, and integrals of partial fractions all require fluent factor-and-cancel. Stewart's Problem 5 is the diagnostic version of those skills. A student who can do all four parts of Problem 5 quickly is ready for the algebra inside Calculus.