Solve x^2-21x+104

$${ x }^{ 2 } -21x+104$$

Factor

$$\left(x-13\right)\left(x-8\right)$$

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.

$$a+b=-21 ab=1\times 104=104$$

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 104.

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

Calculate the sum for each pair.

$$-1-104=-105 -2-52=-54 -4-26=-30 -8-13=-21$$

The solution is the pair that gives sum -21.

$$a=-13 b=-8$$

Rewrite x^{2}-21x+104 as \left(x^{2}-13x\right)+\left(-8x+104\right).

$$\left(x^{2}-13x\right)+\left(-8x+104\right)$$

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

$$x\left(x-13\right)-8\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(x-8\right)$$

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.

$$x^{2}-21x+104=0$$

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.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 104}}{2}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 104}}{2}$$

Multiply -4 times 104.

$$x=\frac{-\left(-21\right)±\sqrt{441-416}}{2}$$

Add 441 to -416.

$$x=\frac{-\left(-21\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-21\right)±5}{2}$$

The opposite of -21 is 21.

$$x=\frac{21±5}{2}$$

Now solve the equation x=\frac{21±5}{2} when ± is plus. Add 21 to 5.

$$x=\frac{26}{2}$$

Divide 26 by 2.

$$x=13$$

Now solve the equation x=\frac{21±5}{2} when ± is minus. Subtract 5 from 21.

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

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 13 for x_{1} and 8 for x_{2}.

$$x^{2}-21x+104=\left(x-13\right)\left(x-8\right)$$

Evaluate

$$\left(x-13\right)\left(x-8\right)$$