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

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

Factor

$$3\left(x-\left(-\frac{\sqrt{33}}{6}-\frac{1}{2}\right)\right)\left(x-\left(\frac{\sqrt{33}}{6}-\frac{1}{2}\right)\right)$$

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.

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

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.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-3±\sqrt{9-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$x=\frac{-3±\sqrt{9+24}}{2\times 3}$$

Add 9 to 24.

$$x=\frac{-3±\sqrt{33}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-3±\sqrt{33}}{6}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{6} when ± is plus. Add -3 to \sqrt{33}.

$$x=\frac{\sqrt{33}-3}{6}$$

Divide -3+\sqrt{33} by 6.

$$x=\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Now solve the equation x=\frac{-3±\sqrt{33}}{6} when ± is minus. Subtract \sqrt{33} from -3.

$$x=\frac{-\sqrt{33}-3}{6}$$

Divide -3-\sqrt{33} by 6.

$$x=-\frac{\sqrt{33}}{6}-\frac{1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{2}+\frac{\sqrt{33}}{6} for x_{1} and -\frac{1}{2}-\frac{\sqrt{33}}{6} for x_{2}.

$$3x^{2}+3x-2=3\left(x-\left(\frac{\sqrt{33}}{6}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{33}}{6}-\frac{1}{2}\right)\right)$$

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.This is achieved by dividing both sides of the equation by 3

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

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

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

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

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

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

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

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

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

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{2}{3}-\frac{1}{4} = -\frac{11}{12}$$

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

$$u^2 = \frac{11}{12} u = \pm\sqrt{\frac{11}{12}} = \pm \frac{\sqrt{11}}{\sqrt{12}}$$

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

$$r =-\frac{1}{2} - \frac{\sqrt{11}}{\sqrt{12}} = -1.457 s = -\frac{1}{2} + \frac{\sqrt{11}}{\sqrt{12}} = 0.457$$

Evaluate

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