Solve 8x^2-26x+15

$$8 { x }^{ 2 } -26x+15$$

Factor

$$\left(2x-5\right)\left(4x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-26 ab=8\times 15=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -26.

$$a=-20 b=-6$$

Rewrite 8x^{2}-26x+15 as \left(8x^{2}-20x\right)+\left(-6x+15\right).

$$\left(8x^{2}-20x\right)+\left(-6x+15\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(2x-5\right)-3\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(4x-3\right)$$

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.

$$8x^{2}-26x+15=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 8\times 15}}{2\times 8}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 8\times 15}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-26\right)±\sqrt{676-32\times 15}}{2\times 8}$$

Multiply -32 times 15.

$$x=\frac{-\left(-26\right)±\sqrt{676-480}}{2\times 8}$$

Add 676 to -480.

$$x=\frac{-\left(-26\right)±\sqrt{196}}{2\times 8}$$

Take the square root of 196.

$$x=\frac{-\left(-26\right)±14}{2\times 8}$$

The opposite of -26 is 26.

$$x=\frac{26±14}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{26±14}{16}$$

Now solve the equation x=\frac{26±14}{16} when ± is plus. Add 26 to 14.

$$x=\frac{40}{16}$$

Reduce the fraction \frac{40}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{26±14}{16} when ± is minus. Subtract 14 from 26.

$$x=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and \frac{3}{4} for x_{2}.

$$8x^{2}-26x+15=8\left(x-\frac{5}{2}\right)\left(x-\frac{3}{4}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-26x+15=8\times \frac{2x-5}{2}\left(x-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-26x+15=8\times \frac{2x-5}{2}\times \frac{4x-3}{4}$$

Multiply \frac{2x-5}{2} times \frac{4x-3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-26x+15=8\times \frac{\left(2x-5\right)\left(4x-3\right)}{2\times 4}$$

Multiply 2 times 4.

$$8x^{2}-26x+15=8\times \frac{\left(2x-5\right)\left(4x-3\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8x^{2}-26x+15=\left(2x-5\right)\left(4x-3\right)$$

Evaluate

$$\left(2x-5\right)\left(4x-3\right)$$