Polynomial & Rational Expressions — 700-800

Polynomial & Rational Expressions: The 700-800 Patterns

Part 1 of 3 — The Archetypes Hard-Tier Items Are Built From

Every hard rational item begins with the same instruction, whether or not it is written down: factor everything first. Almost nothing at this level survives factoring intact.

Archetype 1: Holes vs. Vertical Asymptotes

A rational function is undefined wherever the denominator is zero — but why it is undefined depends on whether the factor cancels.

• The factor cancels $\rightarrow$ hole (removable discontinuity).
• The factor survives in the denominator $\rightarrow$ vertical asymptote.

Worked example. $f(x) = \frac{x^{2} - 7x + 12}{x^{2} - 9} = \frac{(x-3)(x-4)}{(x-3)(x+3)}$. The $(x-3)$ cancels, so there is a hole at $x = 3$, and $x = -3$ is a vertical asymptote. The simplified rule is $\frac{x-4}{x+3}$, so the hole's height is $\frac{3-4}{3+3} = -\frac{1}{6}$ and the hole sits at $\left(3, -\frac{1}{6}\right)$.

The 700-800 twist: the question never asks "where is the hole." It asks for the sum of the coordinates of the hole, or the sum of the hole's $x$-value and the asymptote's $x$-value. That phrasing exists so that the $x$-coordinate alone, the $y$-coordinate alone, and the asymptote alone can all be answer choices.

Two related features are also planted as choices: the horizontal asymptote (ratio of leading coefficients when the degrees match) and, when the cancellation leaves a polynomial, the $y$-intercept of the resulting line. In $\frac{2x^{2}-5x-3}{x-3} = \frac{(2x+1)(x-3)}{x-3} = 2x + 1$, the graph is a line with a hole at $(3, 7)$: the $y$-intercept is $1$, but $7$ and $3$ are both sitting in the options.

Archetype 2: Extraneous Solutions

Clearing denominators can manufacture roots that the original equation forbids. Write the excluded values before you solve, not after.

Worked example. $\frac{2x}{x-3} = 1 + \frac{6}{x-3}$. Note $x \ne 3$. Multiply by $x-3$: $2x = (x-3) + 6$, so $x = 3$ — the one value the domain bans. Every candidate is eliminated, so the answer is no solution.

This archetype has exactly three planted outcomes: keeping the extraneous root, keeping both a valid root and an extraneous one, and over-discarding (declaring "no solution" when a perfectly legal root like $x = 0$ survives). Notice that $x = 0$ is almost never excluded — a denominator of $x - 1$ or $x + 2$ is fine at zero — yet students discard it by reflex.

Archetype 3: Context Models Built on a Rational Expression

Three families cover nearly all of them, and each has a signature "answer one step past the algebra" ask.

Average cost / average time. $A(x) = \frac{F + vx}{x}$, where $F$ is fixed and $v$ is per-unit. Setting $A(x) = k$ and clearing gives a linear equation — no quadratic needed. The ask is usually how many ADDITIONAL units, so you solve twice and subtract. Both production levels and their sum are choices.

Work and rate. Rates add: $\frac{1}{x} + \frac{1}{x+3} = \frac{1}{2}$. Clearing gives a quadratic; reject the negative root as a time. The ask is then for the other worker: if you solved for $x$, the question wants $x + 3$. The negative root and the given combined time are both planted.

Round trips and average speed. $\frac{d}{r} + \frac{d}{r+k} = T$. Once you have both leg speeds, average speed is total distance over total time — never the mean of the two speeds. For $10$ and $15$ mph over equal distances the average is $12$, not $12.5$, and $12.5$ is the item's whole reason for existing.

Mixture and concentration. $C(x) = \frac{a + x}{b + x}$ — pure substance added to both numerator and denominator. Solve for $x$, then check whether the ask is $x$ (the amount added), $a + x$ (the final amount of solute), or $b + x$ (the final total volume). All three are choices.

Archetype 4: Algebraic Manipulation Under Time Pressure

• Combining into one fraction. $\frac{2}{x+1} + \frac{3}{x-1} = \frac{2(x-1) + 3(x+1)}{x^{2}-1} = \frac{5x+1}{x^{2}-1}$, then the ask is $a + b$. Pairing each numerator with the wrong factor gives $\frac{5x-1}{x^{2}-1}$ and a beautifully wrong $a + b$.
• Complex fractions. $\frac{\frac{1}{x} - \frac{1}{3}}{x-3}$: combine the top into $\frac{3-x}{3x}$, then use $3 - x = -(x-3)$ to cancel, leaving $-\frac{1}{3x}$. That sign flip is the entire item.
• Negative exponents. Rewrite $x^{-1} + y^{-1}$ as $\frac{x+y}{xy}$ and $x^{-2} - y^{-2}$ as $\frac{(y-x)(y+x)}{x^{2}y^{2}}$ before doing anything else.
• Factor theorem. "$\frac{x^{2}+kx-15}{x-3}$ has no remainder" means the numerator is $0$ at $x = 3$. Solve for $k$, then answer whatever is actually asked about the resulting quotient — $k$ itself is a choice.

Polynomial & Rational Expressions: Traps & Speed

Part 2 of 3 — Distractor Species and Fast Routes

Species 1: The Other Coordinate, The Other Feature

Rational-graph items generate a small family of numbers — hole $x$, hole $y$, vertical asymptote, horizontal asymptote, $y$-intercept, $x$-intercept — and then ask for a combination of two of them. The options are simply the family members plus the wrong combination.

For $f(x) = \frac{3x^{2}-12}{x^{2}+x-6} = \frac{3(x-2)(x+2)}{(x+3)(x-2)}$: the hole is at $x = 2$ with height $\frac{3(4)}{5} = \frac{12}{5}$, the vertical asymptote is $x = -3$, and the horizontal asymptote is $y = 3$. A question asking for the hole's $y$-coordinate has $2$, $-3$, and $3$ waiting for you. Label every feature you compute — "hole $x$," "VA" — so you can match the label to the ask.

The combination trap is its own species: when the ask is a sum and one value is negative, subtracting instead of adding produces a clean-looking option. $5 + \left(-\frac{5}{2}\right) = \frac{5}{2}$, but $\frac{15}{2}$ is on the list.

Species 2: The Extraneous Root Kept (or the Valid Root Thrown Away)

Both directions are planted on every equation item. The fix is mechanical: the first thing you write is the excluded values. $x \ne 1$, $x \ne -1$. Then solve, then cross-check the candidate list against it. Zero is legal unless a bare $x$ sits in a denominator.

Species 3: The Intermediate Root

Work-rate and per-person-cost items are two-stage: solve a quadratic for $x$, then convert $x$ into the requested quantity. Both stages produce numbers, and the stage-one number is always a choice.

• Pumps: you solve for pump A's time; the question wants pump B's, $x + 3$.
• Bus charter: you solve for the original headcount $n$; the question wants the new cost per person, $\frac{600}{n+5}$.
• Defect model: you solve for both training times; the question wants the elapsed weeks between them.

The rejected negative root is also planted, as is the value handed to you in the stem (the combined time, the flat fee).

Species 4: Averaging What Cannot Be Averaged

Two equal-distance legs at $10$ and $15$ mph do not average $12.5$ mph. Average speed is $\frac{\text{total distance}}{\text{total time}} = \frac{60}{5} = 12$. Similarly, when total distance rises $20\%$ and fuel rises $25\%$, fuel economy scales by $\frac{1.20}{1.25} = 0.96$ — a $4\%$ decrease. The three planted errors are the reciprocal ratio ($+4\%$), the subtracted percents ($5\%$), and both at once.

Species 5: The Sign Flip in a Difference

$3 - x = -(x - 3)$, $y^{2} - x^{2} = -(x^{2} - y^{2})$. In a complex fraction, that single negative is usually the only thing separating the right answer from the top distractor. Whenever a numerator and a denominator contain the same two terms in opposite order, write the negative sign out explicitly rather than cancelling in your head.

Speed Techniques

1. Factor before you read the question again. With the expression factored, holes, asymptotes, intercepts, and cancellations are all visible at once, and you can answer whichever of them is asked in five seconds.

2. Clear denominators in one multiplication. Identify the LCD (usually the difference of squares already sitting in the problem: $x^{2} - 1 = (x-1)(x+1)$, $x^{2} - 4 = (x-2)(x+2)$) and multiply every term, including the lone constant on the right. Skipping the constant is the single most common clearing error and always has a dedicated wrong answer.

3. When a fraction equals zero, only the numerator matters. $\frac{-2x+7}{(x-2)(x+1)} = 0$ needs $-2x + 7 = 0$, so $x = \frac{7}{2}$ — after confirming it is not an excluded value. Do not solve the denominator.

4. Inequalities with a guaranteed-positive denominator are safe. In context, $p > 0$ or $x \ge 1$, so you may multiply through by $p$ without flipping the sign. $\frac{45 + 4p}{p} \le 4.5$ becomes $45 \le 0.5p$, so $p \ge 90$. Outside of context, do not multiply by a variable of unknown sign.

5. Counting integer solutions: simplify, solve, count inclusively, then remove excluded values. From $\frac{n^{2}-100}{n-10} = n + 10$, the condition $30 \le n + 10 \le 40$ gives $20 \le n \le 30$, and the count from $20$ to $30$ is $30 - 20 + 1 = 11$ (the excluded value $n = 10$ lies outside this range, so nothing is removed). Two errors are planted: dropping the $+1$, and forgetting to remove the excluded value when it falls inside the range.

6. Verify a messy root by substitution, not by re-deriving. If you get $x = \frac{13}{2}$, plugging it back into the original equation on your calculator takes fifteen seconds and catches every distribution error at once.

Pacing: a factor-and-read item (hole, asymptote, intercept) should take $45$ seconds. An equation with extraneous-root checking, about $75$. A two-stage context model — quadratic, then convert — is a full $2$ minutes, and it is worth it, because that is precisely where the intermediate-value trap lives.

Polynomial & Rational Expressions: Timed Drill

Part 3 of 3 — Four Questions at Full Difficulty

Target about 90 seconds per question. Before you compute anything on a rational item, do two things that cost five seconds each and save the item:

1. Factor every numerator and denominator you can see.
2. Write the excluded values in the margin: $x \ne \ldots$

Then solve, and finish by asking the only question that matters at this level: is the number I computed the one the last sentence names? On two of the four items below, the value your algebra produces first is an answer choice — and it is the wrong one.