Solve 3u^2+18u+5

$$3 u ^ { 2 } + 18 u + 5$$

Factor

$$3\left(u-\left(-\frac{\sqrt{66}}{3}-3\right)\right)\left(u-\left(\frac{\sqrt{66}}{3}-3\right)\right)$$

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.

$$3u^{2}+18u+5=0$$

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.

$$u=\frac{-18±\sqrt{18^{2}-4\times 3\times 5}}{2\times 3}$$

Square 18.

$$u=\frac{-18±\sqrt{324-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$u=\frac{-18±\sqrt{324-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$u=\frac{-18±\sqrt{324-60}}{2\times 3}$$

Add 324 to -60.

$$u=\frac{-18±\sqrt{264}}{2\times 3}$$

Take the square root of 264.

$$u=\frac{-18±2\sqrt{66}}{2\times 3}$$

Multiply 2 times 3.

$$u=\frac{-18±2\sqrt{66}}{6}$$

Now solve the equation u=\frac{-18±2\sqrt{66}}{6} when ± is plus. Add -18 to 2\sqrt{66}.

$$u=\frac{2\sqrt{66}-18}{6}$$

Divide -18+2\sqrt{66} by 6.

$$u=\frac{\sqrt{66}}{3}-3$$

Now solve the equation u=\frac{-18±2\sqrt{66}}{6} when ± is minus. Subtract 2\sqrt{66} from -18.

$$u=\frac{-2\sqrt{66}-18}{6}$$

Divide -18-2\sqrt{66} by 6.

$$u=-\frac{\sqrt{66}}{3}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\frac{\sqrt{66}}{3} for x_{1} and -3-\frac{\sqrt{66}}{3} for x_{2}.

$$3u^{2}+18u+5=3\left(u-\left(\frac{\sqrt{66}}{3}-3\right)\right)\left(u-\left(-\frac{\sqrt{66}}{3}-3\right)\right)$$

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

$$x ^ 2 +6x +\frac{5}{3} = 0$$

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

$$r + s = -6 rs = \frac{5}{3}$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}

$$(-3 - u) (-3 + u) = \frac{5}{3}$$

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

$$9 - u^2 = \frac{5}{3}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{5}{3}-9 = -\frac{22}{3}$$

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

$$u^2 = \frac{22}{3} u = \pm\sqrt{\frac{22}{3}} = \pm \frac{\sqrt{22}}{\sqrt{3}}$$

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

$$r =-3 - \frac{\sqrt{22}}{\sqrt{3}} = -5.708 s = -3 + \frac{\sqrt{22}}{\sqrt{3}} = -0.292$$

Evaluate

$$3u^{2}+18u+5$$