Solve x=-3;f(x)=x^3+5x^2-9x-45

$$x = - 3 ; f ( x ) = x ^ { 3 } + 5 x ^ { 2 } - 9 x - 45$$

Solve for x, f

$$x=-3$$

$$f=0$$

Solution Steps

Consider the second equation. Insert the known values of variables into the equation.

$$f\left(-3\right)=\left(-3\right)^{3}+5\left(-3\right)^{2}-9\left(-3\right)-45$$

Calculate -3 to the power of 3 and get -27.

$$f\left(-3\right)=-27+5\left(-3\right)^{2}-9\left(-3\right)-45$$

Calculate -3 to the power of 2 and get 9.

$$f\left(-3\right)=-27+5\times 9-9\left(-3\right)-45$$

Multiply 5 and 9 to get 45.

$$f\left(-3\right)=-27+45-9\left(-3\right)-45$$

Add -27 and 45 to get 18.

$$f\left(-3\right)=18-9\left(-3\right)-45$$

Multiply -9 and -3 to get 27.

$$f\left(-3\right)=18+27-45$$

Add 18 and 27 to get 45.

$$f\left(-3\right)=45-45$$

Subtract 45 from 45 to get 0.

$$f\left(-3\right)=0$$

Divide both sides by -3. Zero divided by any non-zero number gives zero.

$$f=0$$

The system is now solved.

$$x=-3 f=0$$