Question 3
Let $P(z)$ be a cubic polynomial with real coefficients such that $P(3-2i)=0$.
Let $Q(z)=(z-i)\,P(z)$.
Which statement must be true?
Not quite. You negated the whole root, $-(3-2i)=-3+2i$, instead of taking the conjugate.
Not quite. You negated the real part instead of the imaginary part ($3-2i\to-3-2i$).
Not quite. This is the trap. $Q(i)=(i-i)P(i)=0$ is true, but you can’t then say “so $Q(-i)=0$ by conjugates”. That applies the conjugate root theorem to $Q$, and the coefficients of $Q$ are not all real because of the factor $z-i$.
That’s it. Step through the hints below to see why.
Apply the conjugate root theorem to $P$. $P$ has real coefficients and $P(3-2i)=0$, so the conjugate is also a root.
$$\overline{3-2i}=3+2i,\qquad P(3+2i)=0$$
The theorem needs real coefficients, and $P$ is given to have them. The conjugate flips the sign of the imaginary part only.
Evaluate $Q$ at $3+2i$.
$$\begin{aligned}Q(3+2i)&=\big((3+2i)-i\big)\,P(3+2i)\\&=(3+i)\times0\\&=0\end{aligned}$$
$Q$ is a product, and a product is zero when any factor is zero. The first hint gave $P(3+2i)=0$.
So (D) is true for every such $P$.
$P$ has real coefficients, so the conjugate of $3-2i$ is also a root: $P(3+2i)=0$. Then $Q(3+2i)=(3+i)\times0=0$.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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