How do you factor $x^{3} - 2 x^{2} + 3$ $x^3-2x^2+3$ ?

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How do you factor $x^{3} - 2 x^{2} + 3$ $x^3-2x^2+3$ ?

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Answer 1 · 2018-01-10T14:53:08

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

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

Explanation:

Given:

$f (x) = x^{3} - 2 x^{2} + 3$ $f(x) = x^3-2x^2+3$

Note that:

$f (- 1) = - 1 - 2 + 3 = 0$ $f(-1) = -1-2+3 = 0$

So $x = - 1$ $x=-1$ is a zero and $(x + 1)$ $(x+1)$ a factor:

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

The remaining quadratic has negative discriminant, so can only be factored further using complex coefficients:

$4 (x^{2} - 3 x + 3) = 4 x^{2} - 12 x + 12$ $4(x^2-3x+3) = 4x^2-12x+12$

$4 (x^{2} - 3 x + 3) = {(2 x)}^{2} - 2 (2 x) (3) + 3^{2} + 3$ $color(white)(4(x^2-3x+3)) = (2x)^2-2(2x)(3)+3^2+3$

$4 (x^{2} - 3 x + 3) = {(2 x - 3)}^{2} + {(\sqrt{3})}^{2}$ $color(white)(4(x^2-3x+3)) = (2x-3)^2+(sqrt(3))^2$

$4 (x^{2} - 3 x + 3) = {(2 x - 3)}^{2} - {(\sqrt{3} i)}^{2}$ $color(white)(4(x^2-3x+3)) = (2x-3)^2-(sqrt(3)i)^2$

$4 (x^{2} - 3 x + 3) = ((2 x - 3) - \sqrt{3} i) ((2 x - 3) + \sqrt{3} i)$ $color(white)(4(x^2-3x+3)) = ((2x-3)-sqrt(3)i)((2x-3)+sqrt(3)i)$

$4 (x^{2} - 3 x + 3) = (2 x - 3 - \sqrt{3} i) (2 x - 3 + \sqrt{3} i)$ $color(white)(4(x^2-3x+3)) = (2x-3-sqrt(3)i)(2x-3+sqrt(3)i)$

So:

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

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