Solve 2x^2-8x+6

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

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}.

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

Evaluate

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