# Verify that 5, −2 and

Question:

Verify that $5,-2$ and $\frac{1}{3}$ are the zeros of the cubic polynomial $p(x)=3 x^{3}-10 x^{2}-27 x+10$ and verify the relation between its zeros and coefficients.

Solution:

$p(x)=\left(3 x^{3}-10 x^{2}-27 x+10\right)$

$p(5)=\left(3 \times 5^{3}-10 \times 5^{2}-27 \times 5+10\right)=(375-250-135+10)=0$

$p(-2)=\left[3 \times\left(-2^{3}\right)-10 \times\left(-2^{2}\right)-27 \times(-2)+10\right]=(-24-40+54+10)=0$

$\left.p\left(\frac{1}{3}\right)=\left\{3 \times\left(\frac{1}{3}\right)^{3}\right)^{3}-10 \times\left(\frac{1}{3}\right)^{2}-27 \times \frac{1}{3}+10\right\}=\left(3 \times \frac{1}{27}-10 \times \frac{1}{9}-9+10\right)$

$=\left(\frac{1}{9}-\frac{10}{9}+1\right)=\left(\frac{1-10+9}{9}\right)=\left(\frac{0}{9}\right)=0$

$\therefore 5,-2$ and $\frac{1}{3}$ are the zeroes of $p(x)$.

Let $\alpha=5, \beta=-2$ and $\gamma=\frac{1}{3}$. Then we have:

$(\alpha+\beta+\gamma)=\left(5-2+\frac{1}{3}\right)=\left(\frac{10}{3}\right)=\frac{-\left(\text { coefficient of } x^{2}\right)}{\left(\text { coefficient of } x^{3}\right)}$

$(\alpha \beta+\beta \gamma+\gamma \alpha)=\left(-10-\frac{2}{3}+\frac{5}{3}\right)=\frac{-27}{3}=\frac{\text { coefficient of } x}{\text { coefficient of } x^{3}}$

$\alpha \beta \gamma=\left\{5 \times(-2) \times \frac{1}{3}\right\}=\frac{-10}{3}=\frac{-(\text { constant term })}{\left(\text { coefficient of } x^{3}\right)}$