Solve 2x^2-8x+16=

$$2 { x }^{ 2 } -8x+16=$$

Factor

$$2\left(x^{2}-4x+8\right)$$

Solution Steps

Factor out 2. Polynomial x^{2}-4x+8 is not factored since it does not have any rational roots.

$$2\left(x^{2}-4x+8\right)$$

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.

$$2x^{2}-8x+16=0$$

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.

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

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 16}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 16}}{2\times 2}$$

Multiply -8 times 16.

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

Add 64 to -128.

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

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

$$2x^{2}-8x+16$$

Evaluate

$$2\left(x^{2}-4x+8\right)$$