Solve f(x)=-3x^2+6x-2

$$f ( x ) = - 3 x ^ { 2 } + 6 x - 2$$

Factor

$$-3\left(x-\left(-\frac{\sqrt{3}}{3}+1\right)\right)\left(x-\left(\frac{\sqrt{3}}{3}+1\right)\right)$$

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.

$$-3x^{2}+6x-2=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\left(-2\right)}}{2\left(-3\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\left(-2\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-6±\sqrt{36+12\left(-2\right)}}{2\left(-3\right)}$$

Multiply 12 times -2.

$$x=\frac{-6±\sqrt{36-24}}{2\left(-3\right)}$$

Add 36 to -24.

$$x=\frac{-6±\sqrt{12}}{2\left(-3\right)}$$

Take the square root of 12.

$$x=\frac{-6±2\sqrt{3}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-6±2\sqrt{3}}{-6}$$

Now solve the equation x=\frac{-6±2\sqrt{3}}{-6} when ± is plus. Add -6 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-6}{-6}$$

Divide -6+2\sqrt{3} by -6.

$$x=-\frac{\sqrt{3}}{3}+1$$

Now solve the equation x=\frac{-6±2\sqrt{3}}{-6} when ± is minus. Subtract 2\sqrt{3} from -6.

$$x=\frac{-2\sqrt{3}-6}{-6}$$

Divide -6-2\sqrt{3} by -6.

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

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

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

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.

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

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

$$r + s = 2 rs = \frac{2}{3}$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

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

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

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

Simplify the expression by subtracting 1 on both sides

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

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

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

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

$$r =1 - \frac{1}{\sqrt{3}} = 0.423 s = 1 + \frac{1}{\sqrt{3}} = 1.577$$

Evaluate

$$-3x^{2}+6x-2$$