Solve (x+1)^2=16

$$( x + 1 ) ^ { 2 } = 16$$

Solve for x

$$x=3$$

$$x=-5$$

Steps Using Factoring

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

$$x^{2}+2x+1=16$$

Subtract 16 from both sides.

$$x^{2}+2x+1-16=0$$

Subtract 16 from 1 to get -15.

$$x^{2}+2x-15=0$$

To solve the equation, factor x^{2}+2x-15 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.

$$a+b=2 ab=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

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

$$\left(x-3\right)\left(x+5\right)$$

To find equation solutions, solve x-3=0 and x+5=0.

$$x=3 x=-5$$

Steps Using Factoring By Grouping

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

$$x^{2}+2x+1=16$$

Subtract 16 from both sides.

$$x^{2}+2x+1-16=0$$

Subtract 16 from 1 to get -15.

$$x^{2}+2x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-15\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite x^{2}+2x-15 as \left(x^{2}-3x\right)+\left(5x-15\right).

$$\left(x^{2}-3x\right)+\left(5x-15\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-3\right)+5\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+5\right)$$

To find equation solutions, solve x-3=0 and x+5=0.

$$x=3 x=-5$$

Steps Using the Quadratic Formula

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

$$x^{2}+2x+1=16$$

Subtract 16 from both sides.

$$x^{2}+2x+1-16=0$$

Subtract 16 from 1 to get -15.

$$x^{2}+2x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-15\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-15\right)}}{2}$$

Multiply -4 times -15.

$$x=\frac{-2±\sqrt{4+60}}{2}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-2±8}{2}$$

Now solve the equation x=\frac{-2±8}{2} when ± is plus. Add -2 to 8.

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

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-2±8}{2} when ± is minus. Subtract 8 from -2.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=3 x=-5$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+1=4 x+1=-4$$

Subtract 1 from both sides of the equation.

$$x=3 x=-5$$