Solve P(x)=-6x^2+8x-2

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

Factor

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

Solution Steps

Factor out 2.

$$2\left(-3x^{2}+4x-1\right)$$

Consider -3x^{2}+4x-1. Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=4 ab=-3\left(-1\right)=3$$

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

$$a=3 b=1$$

Rewrite -3x^{2}+4x-1 as \left(-3x^{2}+3x\right)+\left(x-1\right).

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

Factor out 3x in the first and -1 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square 8.

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

Multiply -4 times -6.

$$x=\frac{-8±\sqrt{64+24\left(-2\right)}}{2\left(-6\right)}$$

Multiply 24 times -2.

$$x=\frac{-8±\sqrt{64-48}}{2\left(-6\right)}$$

Add 64 to -48.

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

Take the square root of 16.

$$x=\frac{-8±4}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-8±4}{-12}$$

Now solve the equation x=\frac{-8±4}{-12} when ± is plus. Add -8 to 4.

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

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-8±4}{-12} when ± is minus. Subtract 4 from -8.

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

Divide -12 by -12.

$$x=1$$

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

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

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

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.

$$x ^ 2 -\frac{4}{3}x +\frac{1}{3} = 0$$

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

$$r + s = \frac{4}{3} rs = \frac{1}{3}$$

Two numbers r and s sum up to \frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{3} = \frac{2}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}

$$(\frac{2}{3} - u) (\frac{2}{3} + u) = \frac{1}{3}$$

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

$$\frac{4}{9} - u^2 = \frac{1}{3}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = \frac{1}{3}-\frac{4}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =\frac{2}{3} - \frac{1}{3} = 0.333 s = \frac{2}{3} + \frac{1}{3} = 1$$

Evaluate

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