Solve (x^3+2)*[(4x^2+2)-(2x^2+x+1)]=I

$$( x ^ { 3 } + 2 ) \cdot [ ( 4 x ^ { 2 } + 2 ) - ( 2 x ^ { 2 } + x + 1 ) ] = I$$

Solve for I

$$I=\left(2x^{2}-x+1\right)\left(x^{3}+2\right)$$

Solution Steps

To find the opposite of 2x^{2}+x+1, find the opposite of each term.

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

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

$$\left(x^{3}+2\right)\left(2x^{2}+2-x-1\right)=I$$

Subtract 1 from 2 to get 1.

$$\left(x^{3}+2\right)\left(2x^{2}+1-x\right)=I$$

Use the distributive property to multiply x^{3}+2 by 2x^{2}+1-x.

$$2x^{5}+x^{3}-x^{4}+4x^{2}+2-2x=I$$

Swap sides so that all variable terms are on the left hand side.

$$I=2x^{5}+x^{3}-x^{4}+4x^{2}+2-2x$$