Solve 2x^2-8x+6

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

Factor

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

Solution Steps

Factor out 2.

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

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

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 2}$$

Add 64 to -48.

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

Take the square root of 16.

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

The opposite of -8 is 8.

$$x=\frac{8±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±4}{4}$$

Now solve the equation x=\frac{8±4}{4} when ± is plus. Add 8 to 4.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{8±4}{4} when ± is minus. Subtract 4 from 8.

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

Divide 4 by 4.

$$x=1$$

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

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

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 2

$$x ^ 2 -4x +3 = 0$$

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

$$r + s = 4 rs = 3$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. 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>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(2 - u) (2 + u) = 3$$

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

$$4 - u^2 = 3$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 3-4 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =2 - 1 = 1 s = 2 + 1 = 3$$

Evaluate

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