Solve (8x^2-6x+1)

$$( 8 x ^ { 2 } - 6 x + 1 )$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=8\times 1=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite 8x^{2}-6x+1 as \left(8x^{2}-4x\right)+\left(-2x+1\right).

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

Factor out 4x in the first and -1 in the second group.

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

Factor out common term 2x-1 by using distributive property.

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

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.

$$8x^{2}-6x+1=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32}}{2\times 8}$$

Add 36 to -32.

$$x=\frac{-\left(-6\right)±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

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

The opposite of -6 is 6.

$$x=\frac{6±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±2}{16}$$

Now solve the equation x=\frac{6±2}{16} when ± is plus. Add 6 to 2.

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

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

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

Now solve the equation x=\frac{6±2}{16} when ± is minus. Subtract 2 from 6.

$$x=\frac{4}{16}$$

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

$$x=\frac{1}{4}$$

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

$$8x^{2}-6x+1=8\left(x-\frac{1}{2}\right)\left(x-\frac{1}{4}\right)$$

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

$$8x^{2}-6x+1=8\times \frac{2x-1}{2}\left(x-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-6x+1=8\times \frac{2x-1}{2}\times \frac{4x-1}{4}$$

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

$$8x^{2}-6x+1=8\times \frac{\left(2x-1\right)\left(4x-1\right)}{2\times 4}$$

Multiply 2 times 4.

$$8x^{2}-6x+1=8\times \frac{\left(2x-1\right)\left(4x-1\right)}{8}$$

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

$$8x^{2}-6x+1=\left(2x-1\right)\left(4x-1\right)$$

Evaluate

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