Question

# If 3 is a zero of the polynomial $f\left( x \right)={{x}^{4}}-{{x}^{3}}-8{{x}^{2}}+kx+12$, then the value of k is,A. – 2B. 2 C. – 3D. $\dfrac{3}{2}$

Hint: We will start by using the fact that the zero of a polynomial $f\left( x \right)$ is the value of x for which $f\left( x \right)=0$. Then we will use this fact to find the value of k by using substituting x = 3 in $f\left( x \right)$ and equating it to zero.
Now, we have been given a polynomial as $f\left( x \right)={{x}^{4}}-{{x}^{3}}-8{{x}^{2}}+kx+12$. Now, we will use the fact that the zero of a polynomial $f\left( x \right)$ is the value of x for which the function $f\left( x \right)=0$. Therefore, using the same concept the value of $f\left( x \right)$ at x = 3 must be zero. Therefore, we have,
$f\left( 3 \right)=0$
Now, we have $f\left( x \right)={{x}^{4}}-{{x}^{3}}-8{{x}^{2}}+kx+12$
\begin{align} & \Rightarrow {{3}^{4}}-{{3}^{3}}-8\times {{3}^{2}}+k\times 3+12=0 \\ & \Rightarrow 81-27-8\times 9+3k+12=0 \\ & \Rightarrow 81-27-72+3k+12=0 \\ & \Rightarrow 81+12-27-72+3k=0 \\ & \Rightarrow 93-99+3k=0 \\ & \Rightarrow -6+3k=0 \\ & \Rightarrow 3k=6 \\ & \Rightarrow k=2 \\ \end{align}
Note: To solve this question we have used a fact that the zero of a polynomial is the value at which the function is zero while solving the equation by substituting the value of zero in $f\left( x \right)$. It is important to do calculations carefully to avoid mistakes.