Solve 4(x+5)^2=16

$$4 ( x + 5 ) ^ { 2 } = 16$$

Solve for x

$$x=-3$$

$$x=-7$$

Steps Using Factoring

Divide both sides by 4.

$$\left(x+5\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x+5\right)^{2}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=4$$

Subtract 4 from both sides.

$$x^{2}+10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}+10x+21=0$$

To solve the equation, factor x^{2}+10x+21 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=10 ab=21$$

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 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 10.

$$a=3 b=7$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x+3\right)\left(x+7\right)$$

To find equation solutions, solve x+3=0 and x+7=0.

$$x=-3 x=-7$$

Steps Using Factoring By Grouping

Divide both sides by 4.

$$\left(x+5\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x+5\right)^{2}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=4$$

Subtract 4 from both sides.

$$x^{2}+10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}+10x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 21=21$$

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 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 10.

$$a=3 b=7$$

Rewrite x^{2}+10x+21 as \left(x^{2}+3x\right)+\left(7x+21\right).

$$\left(x^{2}+3x\right)+\left(7x+21\right)$$

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

$$x\left(x+3\right)+7\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(x+7\right)$$

To find equation solutions, solve x+3=0 and x+7=0.

$$x=-3 x=-7$$

Steps Using the Quadratic Formula

Divide both sides by 4.

$$\left(x+5\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x+5\right)^{2}=4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=4$$

Subtract 4 from both sides.

$$x^{2}+10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}+10x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 21}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 21}}{2}$$

Multiply -4 times 21.

$$x=\frac{-10±\sqrt{100-84}}{2}$$

Add 100 to -84.

$$x=\frac{-10±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-10±4}{2}$$

Now solve the equation x=\frac{-10±4}{2} when ± is plus. Add -10 to 4.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

Now solve the equation x=\frac{-10±4}{2} when ± is minus. Subtract 4 from -10.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=-3 x=-7$$

Steps for Completing the Square

Divide both sides by 4.

$$\left(x+5\right)^{2}=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$\left(x+5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+5=2 x+5=-2$$

Subtract 5 from both sides of the equation.

$$x=-3 x=-7$$