Solve 8x^2-16x+6

$$8 { x }^{ 2 } -16x+6$$

Factor

$$2\left(2x-3\right)\left(2x-1\right)$$

Solution Steps

Factor out 2.

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

Consider 4x^{2}-8x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=4\times 3=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite 4x^{2}-8x+3 as \left(4x^{2}-6x\right)+\left(-2x+3\right).

$$\left(4x^{2}-6x\right)+\left(-2x+3\right)$$

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

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

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$2\left(2x-3\right)\left(2x-1\right)$$

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.

$$8x^{2}-16x+6=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 8\times 6}}{2\times 8}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 8\times 6}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-16\right)±\sqrt{256-32\times 6}}{2\times 8}$$

Multiply -32 times 6.

$$x=\frac{-\left(-16\right)±\sqrt{256-192}}{2\times 8}$$

Add 256 to -192.

$$x=\frac{-\left(-16\right)±\sqrt{64}}{2\times 8}$$

Take the square root of 64.

$$x=\frac{-\left(-16\right)±8}{2\times 8}$$

The opposite of -16 is 16.

$$x=\frac{16±8}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{16±8}{16}$$

Now solve the equation x=\frac{16±8}{16} when ± is plus. Add 16 to 8.

$$x=\frac{24}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{16±8}{16} when ± is minus. Subtract 8 from 16.

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

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{3}{2} for x_{1} and \frac{1}{2} for x_{2}.

$$8x^{2}-16x+6=8\left(x-\frac{3}{2}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-16x+6=8\times \frac{2x-3}{2}\left(x-\frac{1}{2}\right)$$

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

$$8x^{2}-16x+6=8\times \frac{2x-3}{2}\times \frac{2x-1}{2}$$

Multiply \frac{2x-3}{2} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-16x+6=8\times \frac{\left(2x-3\right)\left(2x-1\right)}{2\times 2}$$

Multiply 2 times 2.

$$8x^{2}-16x+6=8\times \frac{\left(2x-3\right)\left(2x-1\right)}{4}$$

Cancel out 4, the greatest common factor in 8 and 4.

$$8x^{2}-16x+6=2\left(2x-3\right)\left(2x-1\right)$$

Evaluate

$$8x^{2}-16x+6$$