Solve (8x^3+2x^2-7)+(-6x^2-3x+1)

$$( 8 x ^ { 3 } + 2 x ^ { 2 } - 7 ) + ( - 6 x ^ { 2 } - 3 x + 1 )$$

Evaluate

$$8x^{3}-4x^{2}-3x-6$$

Solution Steps

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$8x^{3}-4x^{2}-7-3x+1$$

Add -7 and 1 to get -6.

$$8x^{3}-4x^{2}-6-3x$$

Differentiate w.r.t. x

$$24x^{2}-8x-3$$

Steps Using Derivative Rule for Sum

Combine 2x^{2} and -6x^{2} to get -4x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{3}-4x^{2}-7-3x+1)$$

Add -7 and 1 to get -6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{3}-4x^{2}-6-3x)$$

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}.

$$3\times 8x^{3-1}+2\left(-4\right)x^{2-1}-3x^{1-1}$$

Multiply 3 times 8.

$$24x^{3-1}+2\left(-4\right)x^{2-1}-3x^{1-1}$$

Subtract 1 from 3.

$$24x^{2}+2\left(-4\right)x^{2-1}-3x^{1-1}$$

Multiply 2 times -4.

$$24x^{2}-8x^{2-1}-3x^{1-1}$$

Subtract 1 from 2.

$$24x^{2}-8x^{1}-3x^{1-1}$$

Subtract 1 from 1.

$$24x^{2}-8x^{1}-3x^{0}$$

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

$$24x^{2}-8x-3x^{0}$$

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

$$24x^{2}-8x-3$$