Question

Show that the value of $x^3+6x^2-4x-9$ is equal to zero, when $x=-1$ .

Answer 1

We have to prove that :

At $x=-1$

$x^3+6x^2-4x-9=0$

Finding the value of given expression at the given value of variable:
So at $x=-1$

$$\begin{gathered} x^3+6x^2-4x-9={\left(-1\right)}^3+6{\left(-1\right)}^2-4\left(-1\right)-9 \\ =-1+6\left(1\right)+4-9 \\ =-1+6+4-9 \\ =-1-9+6+4 \\ =-10+10 \\ =0 \end{gathered}$$

Therefore, at $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}$

${\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{6}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}\mathit{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{9}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

Hence Proved