How do you factor completely $2 x^{4} + x^{3} + 2 x + 1$ $2x^4+x^3+2x+1$ ?

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How do you factor completely $2 x^{4} + x^{3} + 2 x + 1$ $2x^4+x^3+2x+1$ ?

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Answer 1 · 2018-04-21T12:31:36

$(2 x + 1) (x + 1) (x - \frac{1}{2} - \frac{1}{2} \sqrt{3} i) (x - \frac{1}{2} + \frac{1}{2} \sqrt{3} i)$ $(2x+1)(x+1)(x-1/2-1/2sqrt3i)(x-1/2+1/2sqrt3i)$

Explanation:

$factor by grouping$ $color(blue)"factor by grouping"$

$= x^{3} (2 x + 1) + 1 (2 x + 1)$ $=color(red)(x^3)(2x+1)color(red)(+1)(2x+1)$

$take out the common factor (2 x + 1)$ $"take out the "color(blue)"common factor "(2x+1)$

$= (2 x + 1) (x^{3} + 1)$ $=(2x+1)(color(red)(x^3+1))$

$x^{3} + 1 is a sum of cubes$ $x^3+1" is a "color(blue)"sum of cubes"$

$∙ x a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $•color(white)(x)a^3+b^3=(a+b)(a^2-ab+b^2)$

$\Rightarrow x^{3} + 1 = (x + 1) (x^{2} - x + 1)$ $rArrx^3+1=(x+1)(x^2-x+1)$

$we can factor x^{2} - x + 1 by solving x^{2} - x + 1 = 0$ $"we can factor "x^2-x+1" by solving "x^2-x+1=0$

$using the quadratic formula$ $"using the "color(blue)"quadratic formula"$

$with a = 1, b = - 1 and c = 1$ $"with "a=1,b=-1" and "c=1$

$\Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{3} i}{2} = \frac{1}{2} \pm \frac{1}{2} \sqrt{3} i$ $rArrx=(1+-sqrt(1-4))/2=(1+-sqrt3i)/2=1/2+-1/2sqrt3i$

$(x - (\frac{1}{2} + \frac{1}{2} \sqrt{3} i)) (x - (\frac{1}{2} - \frac{1}{2} \sqrt{3} i))$ $(x-(1/2+1/2sqrt3i))(x-(1/2-1/2sqrt3i))$

$= (x - \frac{1}{2} - \frac{1}{2} \sqrt{3} i) (x - \frac{1}{2} + \frac{1}{2} \sqrt{3} i)$ $=(x-1/2-1/2sqrt3i)(x-1/2+1/2sqrt3i)$

$\Rightarrow 2 x^{4} + x^{3} + 2 x + 1$ $rArr2x^4+x^3+2x+1$

$= (2 x + 1) (x + 1) (x - \frac{1}{2} - \frac{1}{2} \sqrt{3} i) (x - \frac{1}{2} + \frac{1}{2} \sqrt{3} i)$ $=(2x+1)(x+1)(x-1/2-1/2sqrt3i)(x-1/2+1/2sqrt3i)$

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How do you factor completely $2 x^{4} + x^{3} + 2 x + 1$ $2x^4+x^3+2x+1$ ?

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