Solve F(x)=x^3-15x^2+160x-120

$$F ( x ) = x ^ { 3 } - 15 x ^ { 2 } + 160 x - 120$$

Differentiate w.r.t. x

$$3x^{2}-30x+160$$

Steps Using Derivative Rule for Sum

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}.

$$3x^{3-1}+2\left(-15\right)x^{2-1}+160x^{1-1}$$

Subtract 1 from 3.

$$3x^{2}+2\left(-15\right)x^{2-1}+160x^{1-1}$$

Multiply 2 times -15.

$$3x^{2}-30x^{2-1}+160x^{1-1}$$

Subtract 1 from 2.

$$3x^{2}-30x^{1}+160x^{1-1}$$

Subtract 1 from 1.

$$3x^{2}-30x^{1}+160x^{0}$$

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

$$3x^{2}-30x+160x^{0}$$

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

$$3x^{2}-30x+160\times 1$$

For any term t, t\times 1=t and 1t=t.

$$3x^{2}-30x+160$$

Evaluate

$$x^{3}-15x^{2}+160x-120$$