Solve 6x^2-16x+8

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

Factor

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

Solution Steps

Factor out 2.

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

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

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

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

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

Factor out 3x in the first and -2 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square -16.

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

Multiply -4 times 6.

$$x=\frac{-\left(-16\right)±\sqrt{256-24\times 8}}{2\times 6}$$

Multiply -24 times 8.

$$x=\frac{-\left(-16\right)±\sqrt{256-192}}{2\times 6}$$

Add 256 to -192.

$$x=\frac{-\left(-16\right)±\sqrt{64}}{2\times 6}$$

Take the square root of 64.

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

The opposite of -16 is 16.

$$x=\frac{16±8}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{16±8}{12}$$

Now solve the equation x=\frac{16±8}{12} when ± is plus. Add 16 to 8.

$$x=\frac{24}{12}$$

Divide 24 by 12.

$$x=2$$

Now solve the equation x=\frac{16±8}{12} when ± is minus. Subtract 8 from 16.

$$x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

Cancel out 3, the greatest common factor in 6 and 3.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 6

$$x ^ 2 -\frac{8}{3}x +\frac{4}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{8}{3} rs = \frac{4}{3}$$

Two numbers r and s sum up to \frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{3} = \frac{4}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{4}{3} - u s = \frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{4}{3}

$$(\frac{4}{3} - u) (\frac{4}{3} + u) = \frac{4}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{16}{9} - u^2 = \frac{4}{3}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = \frac{4}{3}-\frac{16}{9} = -\frac{4}{9}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{4}{9} u = \pm\sqrt{\frac{4}{9}} = \pm \frac{2}{3}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{4}{3} - \frac{2}{3} = 0.667 s = \frac{4}{3} + \frac{2}{3} = 2$$

Evaluate

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