It is given that $\mathrm{f}(x) = px^6 + qx^4 + r$, where $p$, $q$, and $r$ are real constan

Question

It is given that $\mathrm{f}(x) = px^6 + qx^4 + r$, where $p$, $q$, and $r$ are real constants. Show that if $x = \alpha$ is a root of $\mathrm{f}(x) = 0$, then $x = -\alpha$ is also a root.[1] Given now that $x = 5$ and $x = \beta$ are roots of $\mathrm{f}(x) = 0$, where $\mathrm{Re}(\beta) \neq 0$ and $\mathrm{Im}(\beta) \neq 0$, write down all the remaining roots.[3] The complex number $z$ satisfies the equation[5] $$z^3 + (3 - a)z^2 - (2 + 6\mathrm{i})z - 6 = 0,$$ where $a$ is a complex number. It is given that one of the roots is $1 + \mathrm{i}$. Find $a$ and the other roots of the equation.

Accepted Answer

(a)(i) $f(-\alpha)=0$ (shown). (ii) $-5,\beta^*,-\beta,-\beta^*$. (b) $a=2\mathrm i$; other roots $-3,-1+\mathrm i$.

2025 SAJC P2 Q2

2025 SAJC P2 Q2

Solution:

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