Question

Find the zero of the polynomial $p\left(x\right)=x^3-5x^2-16x+80$, if one zero is equal to other zero but with opposite sign.

Answer 1

$p\left(x\right)=x^3-5x^2-16x+80$

Let $\alpha ,\beta ,\gamma$ are the zeroes of polynomial p(x) such that $\alpha +\beta =0$.

Then,

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

$\alpha +\beta +\gamma =5$

$0+\gamma =5$

$\gamma =5$

Also,

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

$\alpha \beta \gamma =-80$

$-5{\alpha }^2=-80$

${\alpha }^2=16$

$\alpha =\pm 4$

Now,

$\alpha +\beta =0$

$⟹\pm 4+\beta =0$

$\beta =\mp 4$

Thus, zeros of the given polynomial are $4,-4,5$.