Solve (p+2)^2=8(p+2)-15

$$( p + 2 ) ^ { 2 } = 8 ( p + 2 ) - 15$$

Solve for p

$$p=1$$

$$p=3$$

Steps Using Factoring

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

$$p^{2}+4p+4=8\left(p+2\right)-15$$

Use the distributive property to multiply 8 by p+2.

$$p^{2}+4p+4=8p+16-15$$

Subtract 15 from 16 to get 1.

$$p^{2}+4p+4=8p+1$$

Subtract 8p from both sides.

$$p^{2}+4p+4-8p=1$$

Combine 4p and -8p to get -4p.

$$p^{2}-4p+4=1$$

Subtract 1 from both sides.

$$p^{2}-4p+4-1=0$$

Subtract 1 from 4 to get 3.

$$p^{2}-4p+3=0$$

To solve the equation, factor p^{2}-4p+3 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

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

$$\left(p-3\right)\left(p-1\right)$$

To find equation solutions, solve p-3=0 and p-1=0.

$$p=3 p=1$$

Steps Using Factoring By Grouping

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

$$p^{2}+4p+4=8\left(p+2\right)-15$$

Use the distributive property to multiply 8 by p+2.

$$p^{2}+4p+4=8p+16-15$$

Subtract 15 from 16 to get 1.

$$p^{2}+4p+4=8p+1$$

Subtract 8p from both sides.

$$p^{2}+4p+4-8p=1$$

Combine 4p and -8p to get -4p.

$$p^{2}-4p+4=1$$

Subtract 1 from both sides.

$$p^{2}-4p+4-1=0$$

Subtract 1 from 4 to get 3.

$$p^{2}-4p+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite p^{2}-4p+3 as \left(p^{2}-3p\right)+\left(-p+3\right).

$$\left(p^{2}-3p\right)+\left(-p+3\right)$$

Factor out p in the first and -1 in the second group.

$$p\left(p-3\right)-\left(p-3\right)$$

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

$$\left(p-3\right)\left(p-1\right)$$

To find equation solutions, solve p-3=0 and p-1=0.

$$p=3 p=1$$

Steps Using the Quadratic Formula

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

$$p^{2}+4p+4=8\left(p+2\right)-15$$

Use the distributive property to multiply 8 by p+2.

$$p^{2}+4p+4=8p+16-15$$

Subtract 15 from 16 to get 1.

$$p^{2}+4p+4=8p+1$$

Subtract 8p from both sides.

$$p^{2}+4p+4-8p=1$$

Combine 4p and -8p to get -4p.

$$p^{2}-4p+4=1$$

Subtract 1 from both sides.

$$p^{2}-4p+4-1=0$$

Subtract 1 from 4 to get 3.

$$p^{2}-4p+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3}}{2}$$

Square -4.

$$p=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2}$$

Multiply -4 times 3.

$$p=\frac{-\left(-4\right)±\sqrt{16-12}}{2}$$

Add 16 to -12.

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

Take the square root of 4.

$$p=\frac{-\left(-4\right)±2}{2}$$

The opposite of -4 is 4.

$$p=\frac{4±2}{2}$$

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

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

Divide 6 by 2.

$$p=3$$

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

$$p=\frac{2}{2}$$

Divide 2 by 2.

$$p=1$$

The equation is now solved.

$$p=3 p=1$$

Steps for Completing the Square

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

$$p^{2}+4p+4=8\left(p+2\right)-15$$

Use the distributive property to multiply 8 by p+2.

$$p^{2}+4p+4=8p+16-15$$

Subtract 15 from 16 to get 1.

$$p^{2}+4p+4=8p+1$$

Subtract 8p from both sides.

$$p^{2}+4p+4-8p=1$$

Combine 4p and -8p to get -4p.

$$p^{2}-4p+4=1$$

Factor p^{2}-4p+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$p-2=1 p-2=-1$$

Add 2 to both sides of the equation.

$$p=3 p=1$$