AS 91577 · 2025 paper · a remainder, a quadratic in k, a discriminant proof, a cube root, and a hyperbola
Question 1(a) Achieved
Practises Polynomial equations with complex roots
Five separate parts: a remainder, a quadratic solved in terms of a constant, a proof that a quadratic always has real roots, the cube roots of an imaginary number, and a locus that turns out to be a hyperbola.
When the polynomial 3x4 + px3 − 4x + 5 is divided by x − 2, the remainder is 21. Find the value of p.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 Dividing by x − 2 leaves a remainder of 21. What does the remainder theorem say that means?
The remainder is the polynomial’s value at x = 2, so f(2) = 21. No long division needed. -
2 Put x = 2 in and solve for p.
3(2)4 + p(2)3 − 4(2) + 5 = 21
48 + 8p − 8 + 5 = 21
8p = −24, so p = −3.
What the marker wanted
- Achieved The value of p, from f(2) = 21.
In our words, from NZQA’s 2025 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.