Solve x^2-13x+40=

$${ x }^{ 2 } -13x+40=$$

Factor

$$\left(x-8\right)\left(x-5\right)$$

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.

$$a+b=-13 ab=1\times 40=40$$

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 -13.

$$a=-8 b=-5$$

Rewrite x^{2}-13x+40 as \left(x^{2}-8x\right)+\left(-5x+40\right).

$$\left(x^{2}-8x\right)+\left(-5x+40\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-8\right)-5\left(x-8\right)$$

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

$$\left(x-8\right)\left(x-5\right)$$

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.

$$x^{2}-13x+40=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 40}}{2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-13\right)±\sqrt{169-160}}{2}$$

Add 169 to -160.

$$x=\frac{-\left(-13\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-13\right)±3}{2}$$

The opposite of -13 is 13.

$$x=\frac{13±3}{2}$$

Now solve the equation x=\frac{13±3}{2} when ± is plus. Add 13 to 3.

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

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{13±3}{2} when ± is minus. Subtract 3 from 13.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

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 5 for x_{2}.

$$x^{2}-13x+40=\left(x-8\right)\left(x-5\right)$$

Evaluate

$$\left(x-8\right)\left(x-5\right)$$