How do you find the roots of $x^{3} - 6 x^{2} + 13 x - 10 = 0$ $x^3-6x^2+13x-10=0$ ?

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How do you find the roots of $x^{3} - 6 x^{2} + 13 x - 10 = 0$ $x^3-6x^2+13x-10=0$ ?

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Answer 1 · 2016-09-15T17:04:42

$x = 2$ $x=2$

Explanation:

$x^{3} - 6 x^{2} + 13 x - 10 = 0$ $x^3-6x^2+13x-10=0$

$x^{3} - 3 {(x)}^{2} (2) + 3 {(2)}^{2} x + x - 2^{3} - 2 = 0$ $x^3-3(x)^2(2)+3(2)^2x+x-2^3-2=0$

$(x^{3} - 3 {(x)}^{2} (2) + 3 x {(2)}^{2} - 2^{3}) + x - 2 = 0$ $(x^3-3(x)^2(2)+3x(2)^2-2^3)+x-2=0$

We can factorize using the polynomial identity that follows:
${(a - b)}^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} + b^{3}$ $(a-b)^3= a^3-3a^2b+3ab^2+b^3$

where in our case $a = x$ $a=x$ and $b = 2$ $b=2$

So,
${(x - 2)}^{3} + (x - 2) = 0$ $(x-2)^3+(x-2)=0$ taking $x - 2$ $x-2$ as common factor
$(x - 2) ({(x - 2)}^{2} + 1) = 0$ $(x-2)((x-2)^2+1)=0$
$(x - 2) (x^{2} - 4 x + 4 + 1) = 0$ $(x-2)(x^2-4x+4+1)=0$
$(x - 2) (x^{2} - 4 x + 5) = 0$ $(x-2)(x^2-4x+5)=0$

$x - 2 = 0$ $x-2=0$ then $x = 2$ $x=2$
Or
$x^{2} - 4 x + 5 = 0$ $x^2-4x+5=0$
$δ = {(- 4)}^{2} - 4 (1) (5) = 16 - 20 = - 4 < 0$ $delta=(-4)^2-4(1)(5)=16-20=-4<0$
$δ < 0 \Rightarrow$ $delta <0rArr$ no root in R

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How do you find the roots of $x^{3} - 6 x^{2} + 13 x - 10 = 0$ $x^3-6x^2+13x-10=0$ ?

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Explanation:

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How do you find the roots of x3−6x2+13x−10=0x^3-6x^2+13x-10=0?

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How do you find the roots of $x^{3} - 6 x^{2} + 13 x - 10 = 0$ $x^3-6x^2+13x-10=0$ ?