Question 3
One solution of $z^3-z^2-7z-2=0$ is $z=-2$.
Which equation could be used to determine the remaining solutions?
That’s it. Step through the hints below to see why. In 2023, 69.8% of Queensland students chose (A).
Not quite. Multiply it out: $(z+2)(z^2-3z+1)=z^3-z^2-5z+2$. The last two terms don’t match. In 2023, 16.4% of Queensland students chose (B).
Not quite. $(z+2)(z^2-z-1)=z^3+z^2-3z-2$, so the $z^2$ and $z$ terms are wrong. In 2023, 8.0% of Queensland students chose (C).
Not quite. $(z+2)(z^2-z+1)=z^3+z^2-z+2$, which doesn’t match the cubic. In 2023, 5.1% of Queensland students chose (D).
Black: $y=z^3-z^2-7z-2$ for real $z$. Blue dashed: the quadratic you are testing.
The product is the original cubic, and the blue curve crosses the axis at the cubic’s other two roots.
The product isn’t the original cubic, and the blue curve misses the cubic’s other roots.
$z=-2$ is a solution, so $z+2$ is a factor of the cubic. The other factor is a quadratic, and its solutions are the remaining ones.
Write the cubic as $(z+2)(z^2+bz+c)$ and match coefficients, or use long division.
$$(z+2)(z^2+bz+c)=z^3+(b+2)z^2+(c+2b)z+2c$$
Match with $z^3-z^2-7z-2$: the constant gives $2c=-2$, so $c=-1$. The $z^2$ term gives $b+2=-1$, so $b=-3$. Check the $z$ term: $c+2b=-1-6=-7$.
Since $z=-2$ is a solution, $z+2$ is a factor. Dividing it out leaves a quadratic.
$$z^3-z^2-7z-2=(z+2)(z^2-3z-1)$$
So the remaining solutions come from $z^2-3z-1=0$.
How Queensland students answered in 2023
From the QCAA Specialist Mathematics subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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