How do you factor $2 x^{2} - 5 x + 1$ $2x^2-5x+1$ ?

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How do you factor $2 x^{2} - 5 x + 1$ $2x^2-5x+1$ ?

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Answer 1 · 2017-03-02T08:18:53

$2 x^{2} - 5 x + 1 = 2 (x - \frac{5 + \sqrt{17}}{4}) (x - \frac{5 - \sqrt{17}}{4})$ $2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)$

Explanation:

$2 x^{2} - 5 x + 1 = 2 (x^{2} - \frac{5}{2} x + \frac{1}{2})$ $2 x^2 - 5 x + 1 = 2(x^2 - 5/2 x + 1/2)$

Consider $x^{2} - \frac{5}{2} x + \frac{1}{2} = 0$ $x^2 - 5/2 x + 1/2 = 0$ and solve by completing a square
$x^{2} - \frac{5}{2} x + \frac{1}{2} = 0$ $x^2 - 5/2 x + 1/2 = 0$

${(x - \frac{5}{4})}^{2} - {(\frac{5}{4})}^{2} + \frac{1}{2} = 0$ $(x -5/4)^2 -(5/4)^2 + 1/2 = 0$

${(x - \frac{5}{4})}^{2} - \frac{25}{16} + \frac{1}{2} = 0$ $(x -5/4)^2 -25/16 + 1/2 = 0$

${(x - \frac{5}{4})}^{2} - \frac{17}{16} = 0$ $(x -5/4)^2 -17/16 = 0$

${(x - \frac{5}{4})}^{2} = \frac{17}{16}$ $(x -5/4)^2 = 17/16$

$(x - \frac{5}{4}) = \pm \sqrt{\frac{17}{16}} = \pm \frac{\sqrt{17}}{4}$ $(x -5/4) = +-sqrt (17/16) = +-sqrt 17/4$

$x = \frac{5}{4} \pm \frac{\sqrt{17}}{4} = \frac{5 \pm \sqrt{17}}{4}$ $x = 5/4 +-sqrt 17/4 = (5+- sqrt 17)/4$

$(x^{2} - \frac{5}{2} x + \frac{1}{2}) = (x - \frac{5 + \sqrt{17}}{4}) (x - \frac{5 - \sqrt{17}}{4})$ $(x^2 - 5/2 x + 1/2) = (x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)$

therefore

$2 x^{2} - 5 x + 1 = 2 (x - \frac{5 + \sqrt{17}}{4}) (x - \frac{5 - \sqrt{17}}{4})$ $2 x^2 - 5 x + 1 = 2(x - (5 + sqrt 17)/4)(x - (5- sqrt 17)/4)$

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How do you factor $2 x^{2} - 5 x + 1$ $2x^2-5x+1$ ?

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