Solve (2x-x^{2})^2+3x^3(1-x)-2x^2(2-x^2-1/2x)

$$( 2 x - x ^ { 2 } ) ^ { 2 } + 3 x ^ { 3 } ( 1 - x ) - 2 x ^ { 2 } ( 2 - x ^ { 2 } - \frac { 1 } { 2 } x )$$

Evaluate

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Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-x^{2}\right)^{2}.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

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

Combine -4x^{3} and 3x^{3} to get -x^{3}.

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

Combine x^{4} and -3x^{4} to get -2x^{4}.

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

Use the distributive property to multiply -2x^{2} by 2-x^{2}-\frac{1}{2}x.

$$4x^{2}-x^{3}-2x^{4}-4x^{2}+2x^{4}+x^{3}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-x^{3}-2x^{4}+2x^{4}+x^{3}$$

Combine -2x^{4} and 2x^{4} to get 0.

$$-x^{3}+x^{3}$$

Combine -x^{3} and x^{3} to get 0.

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Factor

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