Solve left(x+3right)^2-25=0

$${ \left(x+3 \right) }^{ 2 } -25=0$$

Solve for x

$$x=2$$

$$x=-8$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-25=0$$

Subtract 25 from 9 to get -16.

$$x^{2}+6x-16=0$$

To solve the equation, factor x^{2}+6x-16 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.

$$a+b=6 ab=-16$$

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 -16.

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

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

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

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

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

$$x=2 x=-8$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-25=0$$

Subtract 25 from 9 to get -16.

$$x^{2}+6x-16=0$$

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.

$$a+b=6 ab=1\left(-16\right)=-16$$

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 -16.

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

Calculate the sum for each pair.

$$-1+16=15 -2+8=6 -4+4=0$$

The solution is the pair that gives sum 6.

$$a=-2 b=8$$

Rewrite x^{2}+6x-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(x+3\right)^{2}.

$$x^{2}+6x+9-25=0$$

Subtract 25 from 9 to get -16.

$$x^{2}+6x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-16\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-6±\sqrt{36+64}}{2}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-6±10}{2}$$

Now solve the equation x=\frac{-6±10}{2} when ± is plus. Add -6 to 10.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-6±10}{2} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

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(x+3\right)^{2}.

$$x^{2}+6x+9-25=0$$

Subtract 25 from 9 to get -16.

$$x^{2}+6x-16=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$x^{2}+6x=16$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=16+3^{2}$$

Square 3.

$$x^{2}+6x+9=16+9$$

Add 16 to 9.

$$x^{2}+6x+9=25$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{25}$$

Simplify.

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

Subtract 3 from both sides of the equation.

$$x=2 x=-8$$