Question

obtain all zeroes of x^3+13x^2+32x+20,if one of its zero is -2

Answer 1

We have the cubic equation $x^{3}+13x^{2}+32x+20=0$.
Clearly, it has three roots. Suppose that the roots are $\alpha, \ \beta, \ \gamma$.
Then,
$\alpha+\beta+\gamma=-13 \\ \alpha \beta+\beta \gamma+\gamma \alpha=+32 \\ \alpha \beta \gamma=-20$.
Since one of the roots, say $\gamma$, is -2.
Then the above equations get modified into,
$\alpha+\beta=-11 \\ \alpha \beta=10$.
Then, $| \alpha-\beta |= \sqrt{(\alpha+\beta)^{2}-4\alpha\beta}=\sqrt{(-11)^{2}-4(10)}=\sqrt{81}=9$.
Solving them you get the remaining roots to be$\{-1, -10\}$.