Polynomials & Factoring at the 700-800 Level

Hard polynomial items are built from four archetypes, and none of them is really a factoring exercise.

• Factor theorem with a parameter. "$x + 3$ is a factor of $p(x) = 2x^{3} + ax^{2} - 13x + 6$" means $p(-3) = 0$, so one substitution finds $a$. The stem then asks about the OTHER zeros, and the parameter you just computed sits in the choices. It is never the answer; it is the toll you paid to start.
• Vieta sees through the parameter. For $ax^{3} + bx^{2} + cx + d$, the sum of the zeros is $-\frac{b}{a}$ and the product is $-\frac{d}{a}$ — both independent of the planted $k$. Spotting the decoy turns a three-minute item into a fifteen-second one.
• Special patterns. Difference of squares twice ($81x^{4} - 16$, where only the minus factor has real zeros); difference of cubes, so $\frac{x^{3} - 27}{x - 3}$ IS the quadratic factor; and $P(1) =$ the sum of the coefficients. The remainder on division by $x - c$ is $P(c)$ — never long-divide to find a remainder.
• Factored cubics in context. Zeros read straight off the factors, but $x$ is measured in HUNDREDS and the stem wants a difference, not a value.

The four standard distractors: the parameter itself, the sign-read of factors ($2x + 5$ has zero $-\frac{5}{2}$), the wrong Vieta relation, and the unconverted number. The algebra is the easy half; the last sentence is the hard half.