Solve (1)x^2-8x+15

$$( 1 ) x ^ { 2 } - 8 x + 15$$

Factor

$$\left(x-5\right)\left(x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-3\right)$$

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.

$$x^{2}-8x+15=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Multiply -4 times 15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-8\right)±2}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2}{2}$$

Now solve the equation x=\frac{8±2}{2} when ± is plus. Add 8 to 2.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{8±2}{2} when ± is minus. Subtract 2 from 8.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and 3 for x_{2}.

$$x^{2}-8x+15=\left(x-5\right)\left(x-3\right)$$

Evaluate

$$\left(x-5\right)\left(x-3\right)$$