Question 2
Given that $z=2-i$ is a root of $P(z)=z^{3}-7z^{2}+17z-15$, a factor of $P(z)$ is
That’s it. Step through the hints below to see why.
Not quite. You took the real root as $-3$ (product of roots $-15$), but the real factor is $z-3$.
Not quite. You wrote $z^2+(\text{sum})z+\text{product}$, a sign error in the middle term.
Not quite. You took $(2-i)(2+i)=4-1=3$ instead of $4+1=5$.
By the conjugate root theorem, $z=2+i$ is also a root. So two factors are
$$\big(z-(2-i)\big)\ \text{ and }\ \big(z-(2+i)\big)$$
Multiply them as a difference of two squares.
$$\begin{aligned}&\big(z-(2-i)\big)\big(z-(2+i)\big)\\&\quad=\big((z-2)+i\big)\big((z-2)-i\big)\\&\quad=(z-2)^2-i^2\\&\quad=z^2-4z+4+1\\&\quad=z^2-4z+5\end{aligned}$$
Regroup as $(a+b)(a-b)$ with $a=z-2$ and $b=i$.
$$\begin{aligned}&\big(z-(2-i)\big)\big(z-(2+i)\big)\\&\quad=(z-2)^2-i^2=z^2-4z+5\end{aligned}$$
A factor of $P(z)$ is $z^2-4z+5$.
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.
Spot a mistake on this page? Report it