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Question 122
If $$x - \alpha, x - \beta$$ and $$x - \gamma$$ are factors of the polynomial $$3x^3 + 4x^2 + 2x + 5$$ then the polynomial having factors $$x - \frac{1}{\alpha}, x - \frac{1}{\beta}$$ and $$x - \frac{1}{\gamma}$$ is
Solution
As $$\left(x-\alpha\ \right)$$ is a factor of $$3x^3 + 4x^2 + 2x + 5$$,
$$3x^3 + 4x^2 + 2x + 5$$ = 0 at $$x=\alpha\ $$
I.e. $$3\alpha\ ^3+4\alpha\ ^2+2\alpha\ +5=0$$
Dividing both sides of equation by $$\alpha^3$$ , we get
$$\frac{5}{\alpha^3}+\frac{2}{\alpha^2}+\frac{4}{\alpha\ }+3=0$$
This result is equivalent to saying that if $$x\ =\ \frac{1}{\alpha\ }$$ , then
$$5x^3+2x^2+4x+3=0$$
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