Solve -15x^2-9x-6

$$- 15 x ^ { 2 } - 9 x - 6$$

Factor

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

Solution Steps

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

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

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.

$$-15x^{2}-9x-6=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-15\right)\left(-6\right)}}{2\left(-15\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-15\right)\left(-6\right)}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$x=\frac{-\left(-9\right)±\sqrt{81+60\left(-6\right)}}{2\left(-15\right)}$$

Multiply 60 times -6.

$$x=\frac{-\left(-9\right)±\sqrt{81-360}}{2\left(-15\right)}$$

Add 81 to -360.

$$x=\frac{-\left(-9\right)±\sqrt{-279}}{2\left(-15\right)}$$

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

$$-15x^{2}-9x-6$$

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.

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

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

$$r + s = -\frac{3}{5} rs = \frac{2}{5}$$

Two numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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>

$$r = -\frac{3}{10} - u s = -\frac{3}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}

$$(-\frac{3}{10} - u) (-\frac{3}{10} + u) = \frac{2}{5}$$

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

$$\frac{9}{100} - u^2 = \frac{2}{5}$$

Simplify the expression by subtracting \frac{9}{100} on both sides

$$-u^2 = \frac{2}{5}-\frac{9}{100} = \frac{31}{100}$$

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

$$u^2 = -\frac{31}{100} u = \pm\sqrt{-\frac{31}{100}} = \pm \frac{\sqrt{31}}{10}i$$

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

$$r =-\frac{3}{10} - \frac{\sqrt{31}}{10}i = -0.300 - 0.557i s = -\frac{3}{10} + \frac{\sqrt{31}}{10}i = -0.300 + 0.557i$$

Evaluate

$$-15x^{2}-9x-6$$