Solve p(x)=-5x^2-10x-2

$$p ( x ) = - 5 x ^ { 2 } - 10 x - 2$$

Factor

$$-5\left(x-\left(-\frac{\sqrt{15}}{5}-1\right)\right)\left(x-\left(\frac{\sqrt{15}}{5}-1\right)\right)$$

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.

$$-5x^{2}-10x-2=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-5\right)\left(-2\right)}}{2\left(-5\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-5\right)\left(-2\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-10\right)±\sqrt{100+20\left(-2\right)}}{2\left(-5\right)}$$

Multiply 20 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100-40}}{2\left(-5\right)}$$

Add 100 to -40.

$$x=\frac{-\left(-10\right)±\sqrt{60}}{2\left(-5\right)}$$

Take the square root of 60.

$$x=\frac{-\left(-10\right)±2\sqrt{15}}{2\left(-5\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{15}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{10±2\sqrt{15}}{-10}$$

Now solve the equation x=\frac{10±2\sqrt{15}}{-10} when ± is plus. Add 10 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}+10}{-10}$$

Divide 10+2\sqrt{15} by -10.

$$x=-\frac{\sqrt{15}}{5}-1$$

Now solve the equation x=\frac{10±2\sqrt{15}}{-10} when ± is minus. Subtract 2\sqrt{15} from 10.

$$x=\frac{10-2\sqrt{15}}{-10}$$

Divide 10-2\sqrt{15} by -10.

$$x=\frac{\sqrt{15}}{5}-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1-\frac{\sqrt{15}}{5} for x_{1} and -1+\frac{\sqrt{15}}{5} for x_{2}.

$$-5x^{2}-10x-2=-5\left(x-\left(-\frac{\sqrt{15}}{5}-1\right)\right)\left(x-\left(\frac{\sqrt{15}}{5}-1\right)\right)$$

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.

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

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

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

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

$$r = -1 - u s = -1 + u$$

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

$$(-1 - u) (-1 + u) = \frac{2}{5}$$

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

$$1 - u^2 = \frac{2}{5}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{2}{5}-1 = -\frac{3}{5}$$

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

$$u^2 = \frac{3}{5} u = \pm\sqrt{\frac{3}{5}} = \pm \frac{\sqrt{3}}{\sqrt{5}}$$

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

$$r =-1 - \frac{\sqrt{3}}{\sqrt{5}} = -1.775 s = -1 + \frac{\sqrt{3}}{\sqrt{5}} = -0.225$$

Evaluate

$$-5x^{2}-10x-2$$