Solve x^2+20x+96

$${ x }^{ 2 } +20x+96$$

Factor

$$\left(x+8\right)\left(x+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+96. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\times 96=96$$

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 96.

$$1,96 2,48 3,32 4,24 6,16 8,12$$

Calculate the sum for each pair.

$$1+96=97 2+48=50 3+32=35 4+24=28 6+16=22 8+12=20$$

The solution is the pair that gives sum 20.

$$a=8 b=12$$

Rewrite x^{2}+20x+96 as \left(x^{2}+8x\right)+\left(12x+96\right).

$$\left(x^{2}+8x\right)+\left(12x+96\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(x+8\right)+12\left(x+8\right)$$

Factor out common term x+8 by using distributive property.

$$\left(x+8\right)\left(x+12\right)$$

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.

$$x^{2}+20x+96=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 96}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 96}}{2}$$

Multiply -4 times 96.

$$x=\frac{-20±\sqrt{400-384}}{2}$$

Add 400 to -384.

$$x=\frac{-20±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-20±4}{2}$$

Now solve the equation x=\frac{-20±4}{2} when ± is plus. Add -20 to 4.

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

Divide -16 by 2.

$$x=-8$$

Now solve the equation x=\frac{-20±4}{2} when ± is minus. Subtract 4 from -20.

$$x=-\frac{24}{2}$$

Divide -24 by 2.

$$x=-12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -8 for x_{1} and -12 for x_{2}.

$$x^{2}+20x+96=\left(x-\left(-8\right)\right)\left(x-\left(-12\right)\right)$$

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

$$x^{2}+20x+96=\left(x+8\right)\left(x+12\right)$$

Evaluate

$$\left(x+8\right)\left(x+12\right)$$