Solve 2x^2+8x-128

$$2 x ^ { 2 } + 8 x - 128$$

Factor

$$2\left(x-\left(-2\sqrt{17}-2\right)\right)\left(x-\left(2\sqrt{17}-2\right)\right)$$

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.

$$2x^{2}+8x-128=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\left(-128\right)}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\left(-128\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\left(-128\right)}}{2\times 2}$$

Multiply -8 times -128.

$$x=\frac{-8±\sqrt{64+1024}}{2\times 2}$$

Add 64 to 1024.

$$x=\frac{-8±\sqrt{1088}}{2\times 2}$$

Take the square root of 1088.

$$x=\frac{-8±8\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±8\sqrt{17}}{4}$$

Now solve the equation x=\frac{-8±8\sqrt{17}}{4} when ± is plus. Add -8 to 8\sqrt{17}.

$$x=\frac{8\sqrt{17}-8}{4}$$

Divide -8+8\sqrt{17} by 4.

$$x=2\sqrt{17}-2$$

Now solve the equation x=\frac{-8±8\sqrt{17}}{4} when ± is minus. Subtract 8\sqrt{17} from -8.

$$x=\frac{-8\sqrt{17}-8}{4}$$

Divide -8-8\sqrt{17} by 4.

$$x=-2\sqrt{17}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2+2\sqrt{17} for x_{1} and -2-2\sqrt{17} for x_{2}.

$$2x^{2}+8x-128=2\left(x-\left(2\sqrt{17}-2\right)\right)\left(x-\left(-2\sqrt{17}-2\right)\right)$$

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 2

$$x ^ 2 +4x -64 = 0$$

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

$$r + s = -4 rs = -64$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -64

$$(-2 - u) (-2 + u) = -64$$

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

$$4 - u^2 = -64$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -64-4 = -68$$

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

$$u^2 = 68 u = \pm\sqrt{68} = \pm \sqrt{68}$$

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

$$r =-2 - \sqrt{68} = -10.246 s = -2 + \sqrt{68} = 6.246$$

Evaluate

$$2\left(x^{2}+4x-64\right)$$