<p>It is given that $\mathrm{f}(x)=3x^3+2x^2+16$. The remainder when $\mathrm{f}(x)$ is divided by $x-3a$, where $a$ is a constant is the same as the remainder when it is divided by $3x+2$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the possible values of $a$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Show with clear working, that $x=-2$ is a solution of $\mathrm{f}(x)=0$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Explain why $\mathrm{f}(x)=0$ has only one real root.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">Hence use your answers to parts (b) and (c) to solve the equation \[16y^3+2y-3=0\]<span class="inline-marks">[2]</span></div></li></ol>

It is given that $\mathrm{f}(x)=3x^3+2x^2+16$. The remainder when $\mathrm{f}(x)$ is divided by $x-3a$, where $a$ is a constant is the same as the remainder when it is divided by $3x+2$.

Show with clear working, that $x=-2$ is a solution of $\mathrm{f}(x)=0$.

Explain why $\mathrm{f}(x)=0$ has only one real root.

Hence use your answers to parts (b) and (c) to solve the equation $$16y^3+2y-3=0$$

Quadratic Equations and Inequalities - Nature of Roots of Quadratic Equations

Question 2495f62f