Solve f(x)=-2x^2-10x+1

$$f ( x ) = - 2 x ^ { 2 } - 10 x + 1$$

Factor

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

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.

$$-2x^{2}-10x+1=0$$

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.

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

Square -10.

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

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8}}{2\left(-2\right)}$$

Add 100 to 8.

$$x=\frac{-\left(-10\right)±\sqrt{108}}{2\left(-2\right)}$$

Take the square root of 108.

$$x=\frac{-\left(-10\right)±6\sqrt{3}}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±6\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±6\sqrt{3}}{-4}$$

Now solve the equation x=\frac{10±6\sqrt{3}}{-4} when ± is plus. Add 10 to 6\sqrt{3}.

$$x=\frac{6\sqrt{3}+10}{-4}$$

Divide 10+6\sqrt{3} by -4.

$$x=\frac{-3\sqrt{3}-5}{2}$$

Now solve the equation x=\frac{10±6\sqrt{3}}{-4} when ± is minus. Subtract 6\sqrt{3} from 10.

$$x=\frac{10-6\sqrt{3}}{-4}$$

Divide 10-6\sqrt{3} by -4.

$$x=\frac{3\sqrt{3}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5-3\sqrt{3}}{2} for x_{1} and \frac{-5+3\sqrt{3}}{2} for x_{2}.

$$-2x^{2}-10x+1=-2\left(x-\frac{-3\sqrt{3}-5}{2}\right)\left(x-\frac{3\sqrt{3}-5}{2}\right)$$

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.

$$x ^ 2 +5x -\frac{1}{2} = 0$$

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

$$r + s = -5 rs = -\frac{1}{2}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}

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

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

$$\frac{25}{4} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -\frac{1}{2}-\frac{25}{4} = -\frac{27}{4}$$

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

$$u^2 = \frac{27}{4} u = \pm\sqrt{\frac{27}{4}} = \pm \frac{\sqrt{27}}{2}$$

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

$$r =-\frac{5}{2} - \frac{\sqrt{27}}{2} = -5.098 s = -\frac{5}{2} + \frac{\sqrt{27}}{2} = 0.098$$

Evaluate

$$1-10x-2x^{2}$$