Question

Find the zeros of the polynomial $f\left(x\right)=x^3-5x^2-16x+80$, if its two zeros are equal in magnitude but opposite in sign. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Let $\alpha$, $\beta$, $\gamma$ be the zeros of polynomial $f\left(x\right)$ such that $\alpha +\beta =0$

Then,

Sum of the zeros
$=-\frac{Coefficientof{x}^2}{Coefficientof{x}^3}$

$\Rightarrow \alpha +\beta +\gamma =-(-\frac{5}{1})$

$\Rightarrow \gamma =5$ $[\because \alpha +\beta =0]$

Product of the zeros
$=-\frac{Constantterm}{Coefficientof{x}^3}$

$\Rightarrow \alpha \beta \gamma =-\frac{80}{1}$

$\Rightarrow 5\alpha \beta =-80$ $[\because \gamma =5]$

$\Rightarrow \alpha \beta =-16$

$\Rightarrow -{\alpha }^2=-16$ $[\because \alpha +\beta =0\therefore \beta =-\alpha ]$

$\Rightarrow \alpha =\pm 4$

Now,
$\alpha +\beta =0and\alpha =\pm 4\Rightarrow \beta =\mp 4$

Hence, the zeros are 4, -4 and 5