Solve f(x)=3x^2+9x+2

$$f ( x ) = 3 x ^ { 2 } + 9 x + 2$$

Factor

$$3\left(x-\left(-\frac{\sqrt{57}}{6}-\frac{3}{2}\right)\right)\left(x-\left(\frac{\sqrt{57}}{6}-\frac{3}{2}\right)\right)$$

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.

$$3x^{2}+9x+2=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 3\times 2}}{2\times 3}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-9±\sqrt{81-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-9±\sqrt{81-24}}{2\times 3}$$

Add 81 to -24.

$$x=\frac{-9±\sqrt{57}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-9±\sqrt{57}}{6}$$

Now solve the equation x=\frac{-9±\sqrt{57}}{6} when ± is plus. Add -9 to \sqrt{57}.

$$x=\frac{\sqrt{57}-9}{6}$$

Divide -9+\sqrt{57} by 6.

$$x=\frac{\sqrt{57}}{6}-\frac{3}{2}$$

Now solve the equation x=\frac{-9±\sqrt{57}}{6} when ± is minus. Subtract \sqrt{57} from -9.

$$x=\frac{-\sqrt{57}-9}{6}$$

Divide -9-\sqrt{57} by 6.

$$x=-\frac{\sqrt{57}}{6}-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{2}+\frac{\sqrt{57}}{6} for x_{1} and -\frac{3}{2}-\frac{\sqrt{57}}{6} for x_{2}.

$$3x^{2}+9x+2=3\left(x-\left(\frac{\sqrt{57}}{6}-\frac{3}{2}\right)\right)\left(x-\left(-\frac{\sqrt{57}}{6}-\frac{3}{2}\right)\right)$$

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

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

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

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

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

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

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

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

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

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

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

$$-u^2 = \frac{2}{3}-\frac{9}{4} = -\frac{19}{12}$$

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

$$u^2 = \frac{19}{12} u = \pm\sqrt{\frac{19}{12}} = \pm \frac{\sqrt{19}}{\sqrt{12}}$$

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

$$r =-\frac{3}{2} - \frac{\sqrt{19}}{\sqrt{12}} = -2.758 s = -\frac{3}{2} + \frac{\sqrt{19}}{\sqrt{12}} = -0.242$$

Evaluate

$$3x^{2}+9x+2$$