Solve {l}{(x^2+2x)^2}{-7(x^2+2x)}{-8}

$$\left. \begin{array} { l } { ( x ^ { 2 } + 2 x ) ^ { 2 } } \\ { - 7 ( x ^ { 2 } + 2 x ) } \\ { - 8 } \end{array} \right.$$

Evaluate

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

Solution Steps

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

$$\left(x^{2}\right)^{2}+4x^{2}x+4x^{2}-7\left(x^{2}+2x\right)-8$$

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

$$x^{4}+4x^{2}x+4x^{2}-7\left(x^{2}+2x\right)-8$$

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

$$x^{4}+4x^{3}+4x^{2}-7\left(x^{2}+2x\right)-8$$

Use the distributive property to multiply -7 by x^{2}+2x.

$$x^{4}+4x^{3}+4x^{2}-7x^{2}-14x-8$$

Combine 4x^{2} and -7x^{2} to get -3x^{2}.

$$x^{4}+4x^{3}-3x^{2}-14x-8$$

Expand

$$x^{4}+4x^{3}-3x^{2}-14x-8$$

Solution Steps

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

$$\left(x^{2}\right)^{2}+4x^{2}x+4x^{2}-7\left(x^{2}+2x\right)-8$$

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

$$x^{4}+4x^{2}x+4x^{2}-7\left(x^{2}+2x\right)-8$$

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

$$x^{4}+4x^{3}+4x^{2}-7\left(x^{2}+2x\right)-8$$

Use the distributive property to multiply -7 by x^{2}+2x.

$$x^{4}+4x^{3}+4x^{2}-7x^{2}-14x-8$$

Combine 4x^{2} and -7x^{2} to get -3x^{2}.

$$x^{4}+4x^{3}-3x^{2}-14x-8$$