Solve left(x+1right)^2-25=0

$${ \left(x+1 \right) }^{ 2 } -25=0$$

Solve for x

$$x=4$$

$$x=-6$$

Steps Using Factoring

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

$$x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$x^{2}+2x-24=0$$

To solve the equation, factor x^{2}+2x-24 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.

$$a+b=2 ab=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

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

$$\left(x-4\right)\left(x+6\right)$$

To find equation solutions, solve x-4=0 and x+6=0.

$$x=4 x=-6$$

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}.

$$x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$x^{2}+2x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-24\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

Rewrite x^{2}+2x-24 as \left(x^{2}-4x\right)+\left(6x-24\right).

$$\left(x^{2}-4x\right)+\left(6x-24\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-4\right)+6\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+6\right)$$

To find equation solutions, solve x-4=0 and x+6=0.

$$x=4 x=-6$$

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}.

$$x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$x^{2}+2x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-24\right)}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$x=\frac{-2±\sqrt{4+96}}{2}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-2±10}{2}$$

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

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

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

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=4 x=-6$$

Steps for Completing the Square

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

$$x^{2}+2x+1-25=0$$

Subtract 25 from 1 to get -24.

$$x^{2}+2x-24=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$x^{2}+2x=24$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=24+1^{2}$$

Square 1.

$$x^{2}+2x+1=24+1$$

Add 24 to 1.

$$x^{2}+2x+1=25$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+1=5 x+1=-5$$

Subtract 1 from both sides of the equation.

$$x=4 x=-6$$