Solve b-(x-2)^2-4(x+1)^2

$$b - ( x - 2 ) ^ { 2 } - 4 ( x + 1 ) ^ { 2 }$$

Evaluate

$$-5x^{2}-4x+b-8$$

Solution Steps

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

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

To find the opposite of x^{2}-4x+4, find the opposite of each term.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$b-x^{2}+4x-4-4x^{2}-8x-4$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$b-5x^{2}+4x-4-8x-4$$

Combine 4x and -8x to get -4x.

$$b-5x^{2}-4x-4-4$$

Subtract 4 from -4 to get -8.

$$b-5x^{2}-4x-8$$

Expand

$$-5x^{2}-4x+b-8$$

Solution Steps

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

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

To find the opposite of x^{2}-4x+4, find the opposite of each term.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$b-x^{2}+4x-4-4x^{2}-8x-4$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$b-5x^{2}+4x-4-8x-4$$

Combine 4x and -8x to get -4x.

$$b-5x^{2}-4x-4-4$$

Subtract 4 from -4 to get -8.

$$b-5x^{2}-4x-8$$