Solve 32+12x+x^2

$$32 + 12 x + x ^ { 2 }$$

Factor

$$\left(x+4\right)\left(x+8\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+12x+32$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 32=32$$

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 32.

$$1,32 2,16 4,8$$

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$a=4 b=8$$

Rewrite x^{2}+12x+32 as \left(x^{2}+4x\right)+\left(8x+32\right).

$$\left(x^{2}+4x\right)+\left(8x+32\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x+4\right)+8\left(x+4\right)$$

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

$$\left(x+4\right)\left(x+8\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+12x+32=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-12±\sqrt{12^{2}-4\times 32}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 32}}{2}$$

Multiply -4 times 32.

$$x=\frac{-12±\sqrt{144-128}}{2}$$

Add 144 to -128.

$$x=\frac{-12±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-12±4}{2}$$

Now solve the equation x=\frac{-12±4}{2} when ± is plus. Add -12 to 4.

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

Divide -8 by 2.

$$x=-4$$

Now solve the equation x=\frac{-12±4}{2} when ± is minus. Subtract 4 from -12.

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

Divide -16 by 2.

$$x=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and -8 for x_{2}.

$$x^{2}+12x+32=\left(x-\left(-4\right)\right)\left(x-\left(-8\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}+12x+32=\left(x+4\right)\left(x+8\right)$$

Evaluate

$$\left(x+4\right)\left(x+8\right)$$