Question

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

Answer 1

Since, it is given that $x=-\frac{1}{3}$ is the zero of the polynomial $p\left(x\right)=27x^3-a{x}^2-x+3$, therefore, $p\left(x\right)=0$ when $x=-\frac{1}{3}$.

We now substitute $x=-\frac{1}{3}$ in the polynomial $p\left(x\right)=27x^3-a{x}^2-x+3$ as shown below:

$p(-\frac{1}{3})=27{(-\frac{1}{3})}^3-a{(-\frac{1}{3})}^2-(-\frac{1}{3})+3\Rightarrow 0=(27\times -\frac{1}{27})-(a\times \frac{1}{9})+\frac{1}{3}+3\Rightarrow 0=-1-\frac{a}{9}+\frac{1}{3}+3\Rightarrow 0=\frac{-9-a+3+27}{9}\Rightarrow 0\times 9=-9-a+3+27\Rightarrow 0=-a+21\Rightarrow a=21$

Hence, $a=21$.