Solve x^2-22x+40

$${ x }^{ 2 } -22x+40$$

Factor

$$\left(x-20\right)\left(x-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-22 ab=1\times 40=40$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -22.

$$a=-20 b=-2$$

Rewrite x^{2}-22x+40 as \left(x^{2}-20x\right)+\left(-2x+40\right).

$$\left(x^{2}-20x\right)+\left(-2x+40\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-20\right)-2\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x-2\right)$$

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.

$$x^{2}-22x+40=0$$

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.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 40}}{2}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-22\right)±\sqrt{484-160}}{2}$$

Add 484 to -160.

$$x=\frac{-\left(-22\right)±\sqrt{324}}{2}$$

Take the square root of 324.

$$x=\frac{-\left(-22\right)±18}{2}$$

The opposite of -22 is 22.

$$x=\frac{22±18}{2}$$

Now solve the equation x=\frac{22±18}{2} when ± is plus. Add 22 to 18.

$$x=\frac{40}{2}$$

Divide 40 by 2.

$$x=20$$

Now solve the equation x=\frac{22±18}{2} when ± is minus. Subtract 18 from 22.

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

Divide 4 by 2.

$$x=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and 2 for x_{2}.

$$x^{2}-22x+40=\left(x-20\right)\left(x-2\right)$$

Evaluate

$$\left(x-20\right)\left(x-2\right)$$