Solve (3y^3-2y^2-7y)+(-4y^2+2y-5)

$$( 3 y ^ { 3 } - 2 y ^ { 2 } - 7 y ) + ( - 4 y ^ { 2 } + 2 y - 5 )$$

Evaluate

$$3y^{3}-6y^{2}-5y-5$$

Solution Steps

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

$$3y^{3}-6y^{2}-7y+2y-5$$

Combine -7y and 2y to get -5y.

$$3y^{3}-6y^{2}-5y-5$$

Differentiate w.r.t. y

$$\left(3y-5\right)\left(3y+1\right)$$

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}y}(3y^{3}-6y^{2}-7y+2y-5)$$

Combine -7y and 2y to get -5y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(3y^{3}-6y^{2}-5y-5)$$

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}.

$$3\times 3y^{3-1}+2\left(-6\right)y^{2-1}-5y^{1-1}$$

Multiply 3 times 3.

$$9y^{3-1}+2\left(-6\right)y^{2-1}-5y^{1-1}$$

Subtract 1 from 3.

$$9y^{2}+2\left(-6\right)y^{2-1}-5y^{1-1}$$

Multiply 2 times -6.

$$9y^{2}-12y^{2-1}-5y^{1-1}$$

Subtract 1 from 2.

$$9y^{2}-12y^{1}-5y^{1-1}$$

Subtract 1 from 1.

$$9y^{2}-12y^{1}-5y^{0}$$

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

$$9y^{2}-12y-5y^{0}$$

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

$$9y^{2}-12y-5$$