Solve 8x^2-14x+5

$$8 { x }^{ 2 } -14x+5$$

Factor

$$\left(2x-1\right)\left(4x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-14 ab=8\times 5=40$$

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 -14.

$$a=-10 b=-4$$

Rewrite 8x^{2}-14x+5 as \left(8x^{2}-10x\right)+\left(-4x+5\right).

$$\left(8x^{2}-10x\right)+\left(-4x+5\right)$$

Factor out 2x in the first and -1 in the second group.

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

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(2x-1\right)$$

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.

$$8x^{2}-14x+5=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 8\times 5}}{2\times 8}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 8\times 5}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-14\right)±\sqrt{196-32\times 5}}{2\times 8}$$

Multiply -32 times 5.

$$x=\frac{-\left(-14\right)±\sqrt{196-160}}{2\times 8}$$

Add 196 to -160.

$$x=\frac{-\left(-14\right)±\sqrt{36}}{2\times 8}$$

Take the square root of 36.

$$x=\frac{-\left(-14\right)±6}{2\times 8}$$

The opposite of -14 is 14.

$$x=\frac{14±6}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{14±6}{16}$$

Now solve the equation x=\frac{14±6}{16} when ± is plus. Add 14 to 6.

$$x=\frac{20}{16}$$

Reduce the fraction \frac{20}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{14±6}{16} when ± is minus. Subtract 6 from 14.

$$x=\frac{8}{16}$$

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

$$x=\frac{1}{2}$$

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

$$8x^{2}-14x+5=8\left(x-\frac{5}{4}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-14x+5=8\times \frac{4x-5}{4}\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-14x+5=8\times \frac{4x-5}{4}\times \frac{2x-1}{2}$$

Multiply \frac{4x-5}{4} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-14x+5=8\times \frac{\left(4x-5\right)\left(2x-1\right)}{4\times 2}$$

Multiply 4 times 2.

$$8x^{2}-14x+5=8\times \frac{\left(4x-5\right)\left(2x-1\right)}{8}$$

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

$$8x^{2}-14x+5=\left(4x-5\right)\left(2x-1\right)$$

Evaluate

$$\left(2x-1\right)\left(4x-5\right)$$