How do you factor $x + 1 + 2 x^{2}$ $x+1+2x^2$ ?

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How do you factor $x + 1 + 2 x^{2}$ $x+1+2x^2$ ?

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Answer 1 · 2018-06-11T09:38:07

$x + 1 + 2 \cdot x^{2} = (x + (\frac{1}{2} \pm \frac{\sqrt{2}}{2})) \cdot (x - (\frac{1}{2} \pm \frac{\sqrt{2}}{2})$ $x+1+2*x^2=(x+(1/2+-sqrt(2)/(2)))*(x-(1/2+-sqrt(2)/(2))$

Explanation:

The Quadratic Formula: For $a x^{2} + b x + c = 0$ $ax^2" + bx + c = 0$ , the values of x (factors) which are the solutions of the equation are given by:
$x = \frac{b \pm \sqrt{4 \cdot a \cdot c}}{2 \cdot a}$ $x=(b+-sqrt(4*a*c))/(2*a)$

Here, a=2, b=1 and c=1, so
$x = \frac{1 \pm \sqrt{4 \cdot 2 \cdot 1}}{2 \cdot 2}$ $x=(1+-sqrt(4*2*1))/(2*2)$

$= \frac{1 \pm 2 \cdot \sqrt{2}}{4}$ $=(1+-2*sqrt(2))/(4)$

$= \frac{1}{2} \pm \frac{\sqrt{2}}{2}$ $=1/2+-sqrt(2)/(2)$

You can then factor as follows, using the two factors found:

$\to x + 1 + 2 \cdot x^{2} = (x + (\frac{1}{2} \pm \frac{\sqrt{2}}{2})) \cdot (x - (\frac{1}{2} \pm \frac{\sqrt{2}}{2})$ $-> x+1+2*x^2=(x+(1/2+-sqrt(2)/(2)))*(x-(1/2+-sqrt(2)/(2))$

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