Solve 5(x+4)^2=2(x+1)^2-27

$$5 ( x + 4 ) ^ { 2 } = 2 ( x + 1 ) ^ { 2 } - 27$$

Solve for x

$$x=-7$$

$$x=-5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$5\left(x^{2}+8x+16\right)=2\left(x+1\right)^{2}-27$$

Use the distributive property to multiply 5 by x^{2}+8x+16.

$$5x^{2}+40x+80=2\left(x+1\right)^{2}-27$$

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

$$5x^{2}+40x+80=2\left(x^{2}+2x+1\right)-27$$

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

$$5x^{2}+40x+80=2x^{2}+4x+2-27$$

Subtract 27 from 2 to get -25.

$$5x^{2}+40x+80=2x^{2}+4x-25$$

Subtract 2x^{2} from both sides.

$$5x^{2}+40x+80-2x^{2}=4x-25$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}+40x+80=4x-25$$

Subtract 4x from both sides.

$$3x^{2}+40x+80-4x=-25$$

Combine 40x and -4x to get 36x.

$$3x^{2}+36x+80=-25$$

Add 25 to both sides.

$$3x^{2}+36x+80+25=0$$

Add 80 and 25 to get 105.

$$3x^{2}+36x+105=0$$

Divide both sides by 3.

$$x^{2}+12x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 35=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

Rewrite x^{2}+12x+35 as \left(x^{2}+5x\right)+\left(7x+35\right).

$$\left(x^{2}+5x\right)+\left(7x+35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x+5\right)+7\left(x+5\right)$$

Factor out common term x+5 by using distributive property.

$$\left(x+5\right)\left(x+7\right)$$

To find equation solutions, solve x+5=0 and x+7=0.

$$x=-5 x=-7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$5\left(x^{2}+8x+16\right)=2\left(x+1\right)^{2}-27$$

Use the distributive property to multiply 5 by x^{2}+8x+16.

$$5x^{2}+40x+80=2\left(x+1\right)^{2}-27$$

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

$$5x^{2}+40x+80=2\left(x^{2}+2x+1\right)-27$$

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

$$5x^{2}+40x+80=2x^{2}+4x+2-27$$

Subtract 27 from 2 to get -25.

$$5x^{2}+40x+80=2x^{2}+4x-25$$

Subtract 2x^{2} from both sides.

$$5x^{2}+40x+80-2x^{2}=4x-25$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}+40x+80=4x-25$$

Subtract 4x from both sides.

$$3x^{2}+40x+80-4x=-25$$

Combine 40x and -4x to get 36x.

$$3x^{2}+36x+80=-25$$

Add 25 to both sides.

$$3x^{2}+36x+80+25=0$$

Add 80 and 25 to get 105.

$$3x^{2}+36x+105=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 36 for b, and 105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\times 3\times 105}}{2\times 3}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 3\times 105}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-36±\sqrt{1296-12\times 105}}{2\times 3}$$

Multiply -12 times 105.

$$x=\frac{-36±\sqrt{1296-1260}}{2\times 3}$$

Add 1296 to -1260.

$$x=\frac{-36±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$x=\frac{-36±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-36±6}{6}$$

Now solve the equation x=\frac{-36±6}{6} when ± is plus. Add -36 to 6.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

Now solve the equation x=\frac{-36±6}{6} when ± is minus. Subtract 6 from -36.

$$x=-\frac{42}{6}$$

Divide -42 by 6.

$$x=-7$$

The equation is now solved.

$$x=-5 x=-7$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$5\left(x^{2}+8x+16\right)=2\left(x+1\right)^{2}-27$$

Use the distributive property to multiply 5 by x^{2}+8x+16.

$$5x^{2}+40x+80=2\left(x+1\right)^{2}-27$$

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

$$5x^{2}+40x+80=2\left(x^{2}+2x+1\right)-27$$

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

$$5x^{2}+40x+80=2x^{2}+4x+2-27$$

Subtract 27 from 2 to get -25.

$$5x^{2}+40x+80=2x^{2}+4x-25$$

Subtract 2x^{2} from both sides.

$$5x^{2}+40x+80-2x^{2}=4x-25$$

Combine 5x^{2} and -2x^{2} to get 3x^{2}.

$$3x^{2}+40x+80=4x-25$$

Subtract 4x from both sides.

$$3x^{2}+40x+80-4x=-25$$

Combine 40x and -4x to get 36x.

$$3x^{2}+36x+80=-25$$

Subtract 80 from both sides.

$$3x^{2}+36x=-25-80$$

Subtract 80 from -25 to get -105.

$$3x^{2}+36x=-105$$

Divide both sides by 3.

$$\frac{3x^{2}+36x}{3}=-\frac{105}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{36}{3}x=-\frac{105}{3}$$

Divide 36 by 3.

$$x^{2}+12x=-\frac{105}{3}$$

Divide -105 by 3.

$$x^{2}+12x=-35$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=-35+6^{2}$$

Square 6.

$$x^{2}+12x+36=-35+36$$

Add -35 to 36.

$$x^{2}+12x+36=1$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+6=1 x+6=-1$$

Subtract 6 from both sides of the equation.

$$x=-5 x=-7$$