Solve g(x)=-2x^2+4x-8

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

Factor

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

Solution Steps

Factor out 2. Polynomial -x^{2}+2x-4 is not factored since it does not have any rational roots.

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

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.

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

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\left(-8\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\left(-8\right)}}{2\left(-2\right)}$$

Multiply 8 times -8.

$$x=\frac{-4±\sqrt{16-64}}{2\left(-2\right)}$$

Add 16 to -64.

$$x=\frac{-4±\sqrt{-48}}{2\left(-2\right)}$$

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

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

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.

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

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

$$r + s = 2 rs = 4$$

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-gzdabgg4ehffg0hf.b01.azurefd.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 = 4

$$(1 - u) (1 + u) = 4$$

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

$$1 - u^2 = 4$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = 4-1 = 3$$

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

$$u^2 = -3 u = \pm\sqrt{-3} = \pm \sqrt{3}i$$

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

$$r =1 - \sqrt{3}i s = 1 + \sqrt{3}i$$

Evaluate

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