Solve {l}{x^3+2x^2-9x-18}{-3x-5}

$$\left. \begin{array} { l } { x ^ { 3 } + 2 x ^ { 2 } - 9 x - 18 } \\ { - 3 x - 5 } \end{array} \right.$$

Evaluate

$$x^{3}+2x^{2}-12x-23$$

Solution Steps

Combine -9x and -3x to get -12x.

$$x^{3}+2x^{2}-12x-18-5$$

Subtract 5 from -18 to get -23.

$$x^{3}+2x^{2}-12x-23$$

Differentiate w.r.t. x

$$3x^{2}+4x-12$$

Steps Using Derivative Rule for Sum

Combine -9x and -3x to get -12x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+2x^{2}-12x-18-5)$$

Subtract 5 from -18 to get -23.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+2x^{2}-12x-23)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$3x^{3-1}+2\times 2x^{2-1}-12x^{1-1}$$

Subtract 1 from 3.

$$3x^{2}+2\times 2x^{2-1}-12x^{1-1}$$

Multiply 2 times 2.

$$3x^{2}+4x^{2-1}-12x^{1-1}$$

Subtract 1 from 2.

$$3x^{2}+4x^{1}-12x^{1-1}$$

Subtract 1 from 1.

$$3x^{2}+4x^{1}-12x^{0}$$

For any term t, t^{1}=t.

$$3x^{2}+4x-12x^{0}$$

For any term t except 0, t^{0}=1.

$$3x^{2}+4x-12$$