Solve x^2+22x-104

$${ x }^{ 2 } +22x-104$$

Factor

$$\left(x-4\right)\left(x+26\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-104. To find a and b, set up a system to be solved.

$$a+b=22 ab=1\left(-104\right)=-104$$

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 -104.

$$-1,104 -2,52 -4,26 -8,13$$

Calculate the sum for each pair.

$$-1+104=103 -2+52=50 -4+26=22 -8+13=5$$

The solution is the pair that gives sum 22.

$$a=-4 b=26$$

Rewrite x^{2}+22x-104 as \left(x^{2}-4x\right)+\left(26x-104\right).

$$\left(x^{2}-4x\right)+\left(26x-104\right)$$

Factor out x in the first and 26 in the second group.

$$x\left(x-4\right)+26\left(x-4\right)$$

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

$$\left(x-4\right)\left(x+26\right)$$

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.

$$x^{2}+22x-104=0$$

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.

$$x=\frac{-22±\sqrt{22^{2}-4\left(-104\right)}}{2}$$

Square 22.

$$x=\frac{-22±\sqrt{484-4\left(-104\right)}}{2}$$

Multiply -4 times -104.

$$x=\frac{-22±\sqrt{484+416}}{2}$$

Add 484 to 416.

$$x=\frac{-22±\sqrt{900}}{2}$$

Take the square root of 900.

$$x=\frac{-22±30}{2}$$

Now solve the equation x=\frac{-22±30}{2} when ± is plus. Add -22 to 30.

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

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-22±30}{2} when ± is minus. Subtract 30 from -22.

$$x=-\frac{52}{2}$$

Divide -52 by 2.

$$x=-26$$

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 -26 for x_{2}.

$$x^{2}+22x-104=\left(x-4\right)\left(x-\left(-26\right)\right)$$

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

$$x^{2}+22x-104=\left(x-4\right)\left(x+26\right)$$

Evaluate

$$\left(x-4\right)\left(x+26\right)$$