Solve left(-2x-8right)^2=-8x

$${ \left(-2x-8 \right) }^{ 2 } =-8x$$

Solve for x

$$x=-8$$

$$x=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-8\right)^{2}.

$$4x^{2}+32x+64=-8x$$

Add 8x to both sides.

$$4x^{2}+32x+64+8x=0$$

Combine 32x and 8x to get 40x.

$$4x^{2}+40x+64=0$$

Divide both sides by 4.

$$x^{2}+10x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 16=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 10.

$$a=2 b=8$$

Rewrite x^{2}+10x+16 as \left(x^{2}+2x\right)+\left(8x+16\right).

$$\left(x^{2}+2x\right)+\left(8x+16\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x+2\right)+8\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(x+8\right)$$

To find equation solutions, solve x+2=0 and x+8=0.

$$x=-2 x=-8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-8\right)^{2}.

$$4x^{2}+32x+64=-8x$$

Add 8x to both sides.

$$4x^{2}+32x+64+8x=0$$

Combine 32x and 8x to get 40x.

$$4x^{2}+40x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 40 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 4\times 64}}{2\times 4}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 4\times 64}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-40±\sqrt{1600-16\times 64}}{2\times 4}$$

Multiply -16 times 64.

$$x=\frac{-40±\sqrt{1600-1024}}{2\times 4}$$

Add 1600 to -1024.

$$x=\frac{-40±\sqrt{576}}{2\times 4}$$

Take the square root of 576.

$$x=\frac{-40±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-40±24}{8}$$

Now solve the equation x=\frac{-40±24}{8} when ± is plus. Add -40 to 24.

$$x=-\frac{16}{8}$$

Divide -16 by 8.

$$x=-2$$

Now solve the equation x=\frac{-40±24}{8} when ± is minus. Subtract 24 from -40.

$$x=-\frac{64}{8}$$

Divide -64 by 8.

$$x=-8$$

The equation is now solved.

$$x=-2 x=-8$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2x-8\right)^{2}.

$$4x^{2}+32x+64=-8x$$

Add 8x to both sides.

$$4x^{2}+32x+64+8x=0$$

Combine 32x and 8x to get 40x.

$$4x^{2}+40x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+40x=-64$$

Divide both sides by 4.

$$\frac{4x^{2}+40x}{4}=-\frac{64}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{40}{4}x=-\frac{64}{4}$$

Divide 40 by 4.

$$x^{2}+10x=-\frac{64}{4}$$

Divide -64 by 4.

$$x^{2}+10x=-16$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-16+5^{2}$$

Square 5.

$$x^{2}+10x+25=-16+25$$

Add -16 to 25.

$$x^{2}+10x+25=9$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+5=3 x+5=-3$$

Subtract 5 from both sides of the equation.

$$x=-2 x=-8$$