Solve f(x)=8x^2+160x-4

$$f ( x ) = 8 x ^ { 2 } + 160 x - 4$$

Factor

$$8\left(x-\left(-\frac{\sqrt{402}}{2}-10\right)\right)\left(x-\left(\frac{\sqrt{402}}{2}-10\right)\right)$$

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.

$$8x^{2}+160x-4=0$$

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.

$$x=\frac{-160±\sqrt{160^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\times 8\left(-4\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-160±\sqrt{25600-32\left(-4\right)}}{2\times 8}$$

Multiply -32 times -4.

$$x=\frac{-160±\sqrt{25600+128}}{2\times 8}$$

Add 25600 to 128.

$$x=\frac{-160±\sqrt{25728}}{2\times 8}$$

Take the square root of 25728.

$$x=\frac{-160±8\sqrt{402}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-160±8\sqrt{402}}{16}$$

Now solve the equation x=\frac{-160±8\sqrt{402}}{16} when ± is plus. Add -160 to 8\sqrt{402}.

$$x=\frac{8\sqrt{402}-160}{16}$$

Divide -160+8\sqrt{402} by 16.

$$x=\frac{\sqrt{402}}{2}-10$$

Now solve the equation x=\frac{-160±8\sqrt{402}}{16} when ± is minus. Subtract 8\sqrt{402} from -160.

$$x=\frac{-8\sqrt{402}-160}{16}$$

Divide -160-8\sqrt{402} by 16.

$$x=-\frac{\sqrt{402}}{2}-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -10+\frac{\sqrt{402}}{2} for x_{1} and -10-\frac{\sqrt{402}}{2} for x_{2}.

$$8x^{2}+160x-4=8\left(x-\left(\frac{\sqrt{402}}{2}-10\right)\right)\left(x-\left(-\frac{\sqrt{402}}{2}-10\right)\right)$$

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 8

$$x ^ 2 +20x -\frac{1}{2} = 0$$

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

$$r + s = -20 rs = -\frac{1}{2}$$

Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -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>

$$r = -10 - u s = -10 + u$$

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

$$(-10 - u) (-10 + u) = -\frac{1}{2}$$

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

$$100 - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = -\frac{1}{2}-100 = -\frac{201}{2}$$

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

$$u^2 = \frac{201}{2} u = \pm\sqrt{\frac{201}{2}} = \pm \frac{\sqrt{201}}{\sqrt{2}}$$

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

$$r =-10 - \frac{\sqrt{201}}{\sqrt{2}} = -20.025 s = -10 + \frac{\sqrt{201}}{\sqrt{2}} = 0.025$$

Evaluate

$$8x^{2}+160x-4$$