Question

If $x=\frac{-1}{3}$ is the zero of the polynomial $p\left(x\right)=27x^3-a{x}^2-x+3$, then value of $a.$

Answer 1

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

$x=\frac{-1}{3}$ is a zero of polynomial of p(x)

So $p\left(\frac{-1}{3}\right)=0$

$\Rightarrow 27\times (\frac{-1}{3}{)}^3-a(\frac{-1}{3}{)}^2-\left(\frac{-1}{3}\right)+3=0$

$\Rightarrow \frac{-27}{27}-\frac{a}{9}+\frac{1}{3}+3=0$

$\Rightarrow -1-\frac{a}{9}+3+\frac{1}{3}=0$

$\Rightarrow 2+\frac{1}{3}=\frac{a}{9}$

$\Rightarrow \frac{7}{3}=\frac{a}{9}$

$\Rightarrow 7\times \frac{9}{3}=a$

$\Rightarrow 7\times 3=21=a$

So at $a=21,x=\frac{-1}{3}$ is a zero of polynomial of $p\left(x\right)=27x^3-a{x}^2-x+3$