Solve 5y^3-10y^2+6+2y^3+4y-8

$$5 y ^ { 3 } - 10 y ^ { 2 } + 6 + 2 y ^ { 3 } + 4 y - 8$$

Evaluate

$$7y^{3}-10y^{2}+4y-2$$

Solution Steps

Combine 5y^{3} and 2y^{3} to get 7y^{3}.

$$7y^{3}-10y^{2}+6+4y-8$$

Subtract 8 from 6 to get -2.

$$7y^{3}-10y^{2}-2+4y$$

Differentiate w.r.t. y

$$\left(3y-2\right)\left(7y-2\right)$$

Steps Using Derivative Rule for Sum

Combine 5y^{3} and 2y^{3} to get 7y^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(7y^{3}-10y^{2}+6+4y-8)$$

Subtract 8 from 6 to get -2.

$$\frac{\mathrm{d}}{\mathrm{d}y}(7y^{3}-10y^{2}-2+4y)$$

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}.

$$3\times 7y^{3-1}+2\left(-10\right)y^{2-1}+4y^{1-1}$$

Multiply 3 times 7.

$$21y^{3-1}+2\left(-10\right)y^{2-1}+4y^{1-1}$$

Subtract 1 from 3.

$$21y^{2}+2\left(-10\right)y^{2-1}+4y^{1-1}$$

Multiply 2 times -10.

$$21y^{2}-20y^{2-1}+4y^{1-1}$$

Subtract 1 from 2.

$$21y^{2}-20y^{1}+4y^{1-1}$$

Subtract 1 from 1.

$$21y^{2}-20y^{1}+4y^{0}$$

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

$$21y^{2}-20y+4y^{0}$$

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

$$21y^{2}-20y+4\times 1$$

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

$$21y^{2}-20y+4$$