Solve (x^2-x)^2+3(x^2-x)-40

$$( x ^ { 2 } - x ) ^ { 2 } + 3 ( x ^ { 2 } - x ) - 40$$

Evaluate

$$\left(x\left(x-1\right)\right)^{2}+3x\left(x-1\right)-40$$

Solution Steps

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

$$\left(x^{2}\right)^{2}-2x^{2}x+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{2}x+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{3}+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{3}+x^{2}+3x^{2}-3x-40$$

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

$$x^{4}-2x^{3}+4x^{2}-3x-40$$

Expand

$$x^{4}-2x^{3}+4x^{2}-3x-40$$

Solution Steps

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

$$\left(x^{2}\right)^{2}-2x^{2}x+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{2}x+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{3}+x^{2}+3\left(x^{2}-x\right)-40$$

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

$$x^{4}-2x^{3}+x^{2}+3x^{2}-3x-40$$

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

$$x^{4}-2x^{3}+4x^{2}-3x-40$$