Solve {l}{5x^2+15x}{+30}

$$\left. \begin{array} { l } { 5 x ^ { 2 } + 15 x } \\ { + 30 } \end{array} \right.$$

Factor

$$5\left(x^{2}+3x+6\right)$$

Solution Steps

Factor out 5. Polynomial x^{2}+3x+6 is not factored since it does not have any rational roots.

$$5\left(x^{2}+3x+6\right)$$

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.

$$5x^{2}+15x+30=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 5\times 30}}{2\times 5}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 5\times 30}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-15±\sqrt{225-20\times 30}}{2\times 5}$$

Multiply -20 times 30.

$$x=\frac{-15±\sqrt{225-600}}{2\times 5}$$

Add 225 to -600.

$$x=\frac{-15±\sqrt{-375}}{2\times 5}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$5x^{2}+15x+30$$

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 5

$$x ^ 2 +3x +6 = 0$$

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

$$r + s = -3 rs = 6$$

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 = 6

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = 6$$

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

$$\frac{9}{4} - u^2 = 6$$

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

$$-u^2 = 6-\frac{9}{4} = \frac{15}{4}$$

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

$$u^2 = -\frac{15}{4} u = \pm\sqrt{-\frac{15}{4}} = \pm \frac{\sqrt{15}}{2}i$$

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

$$r =-\frac{3}{2} - \frac{\sqrt{15}}{2}i = -1.500 - 1.936i s = -\frac{3}{2} + \frac{\sqrt{15}}{2}i = -1.500 + 1.936i$$

Evaluate

$$5\left(x^{2}+3x+6\right)$$