Polynomials & Factoring — 700-800

Polynomials & Factoring: The 700-800 Patterns

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

Archetype 1: The Factor Theorem With a Parameter

"$x + 3$ is a factor of $p(x) = 2x^{3} + ax^{2} - 13x + 6$" means $p(-3) = 0$ — one substitution finds $a$. Then divide out the known factor and work with the leftover quadratic. The question asks about the OTHER zeros (their sum, a product, a difference), and the parameter value itself is planted as a wrong answer.

Worked example. $p(-3) = -54 + 9a + 39 + 6 = 9a - 9 = 0$ gives $a = 1$. Dividing $2x^{3} + x^{2} - 13x + 6$ by $x + 3$ leaves $2x^{2} - 5x + 2$, so the other zeros sum to $\frac{5}{2}$ by Vieta — no need to find them.

Archetype 2: Vieta Sees Through the Parameter

For $ax^{3} + bx^{2} + cx + d$: sum of zeros $= -\frac{b}{a}$, product $= -\frac{d}{a}$. The killer setup: "$x - 1$ is a factor of $3x^{3} - 12x^{2} + kx - 9$; what is the sum of the zeros?" You CAN find $k$ — but the sum is $-\frac{-12}{3} = 4$ no matter what $k$ is. Spotting that the parameter is a decoy converts a 3-minute item into a 15-second one.

Archetype 3: Special Patterns

• Difference of squares, twice: $16x^{4} - 81 = (4x^{2} - 9)(4x^{2} + 9)$, and only the first factor has real zeros.
• Difference of cubes: $x^{3} - 8 = (x - 2)(x^{2} + 2x + 4)$, so $\frac{x^{3} - 8}{x - 2}$ IS the quadratic factor.
• Sum of coefficients: $P(1)$ = sum of the coefficients, in one substitution. Remainder on division by $x - c$ is $P(c)$ — never long-divide to find a remainder.

Archetype 4: Factored Cubics in Context

$C(x) = x(2x - 3)(x - 4)$ models something; zeros are read off the factors ($0$, $\frac{3}{2}$, $4$) — but the stem measures $x$ in HUNDREDS of units, asks for a difference rather than a value, or both. The algebra is the easy half; the last sentence is the hard half.

Polynomials & Factoring: Traps & Speed

Part 2 of 3 — Distractor Autopsy and Faster Routes

The Four Standard Distractors

1. The parameter itself. The value of $k$ or $a$ found in step one is always a choice. It is never the answer — it is the toll you paid to start.
2. The sign-read of factors. $(2x + 5)$ has zero $-\frac{5}{2}$, not $\frac{5}{2}$. Zeros listed with flipped signs form a full distractor set.
3. The wrong Vieta relation. Sum asked, product planted; product asked, sum planted. For cubics: sum $= -\frac{b}{a}$, sum of pairwise products $= \frac{c}{a}$, product $= -\frac{d}{a}$.
4. The unit conversion. "$x$ hundred units" means every zero and difference must be scaled by $100$ at the end. The unconverted number is always planted.

Speed Techniques

• $P(1)$ is the sum of the coefficients. "Remainder when divided by $x - 1$," "sum of coefficients of the quotient" — both collapse to evaluations at $1$. For a clean division $P(x) = D(x)Q(x)$: $Q(1) = \frac{P(1)}{D(1)}$.
• Remainder theorem over long division. Remainder of $P(x) \div (x - c)$ is $P(c)$. Long division is for finding QUOTIENTS only — and even then, synthetic division is faster.
• Vieta before dividing. Asked for a sum or product of remaining zeros? Take the total from Vieta, then subtract the known zero (for a sum) or divide by it (for a product). Division is only needed when you must know the zeros individually.
• Desmos exploit: graph the polynomial and read the zeros directly; for a parameter, add a slider and tune until the graph passes through the required zero. Exact fractions still deserve an algebra check.

Polynomials & Factoring: Timed Drill

Part 3 of 3 — Four Questions at Full Difficulty

Budget about 75 seconds per question. Opening moves: named factor $\rightarrow$ substitute its zero; remainder $\rightarrow$ evaluate, never divide; sum/product of zeros $\rightarrow$ Vieta; quartic or cubic constant patterns $\rightarrow$ look for difference of squares or cubes before anything else. Save the last 10 seconds to reread which quantity — and which UNIT — the stem wants.