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

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Answer 1 · 2016-09-15T17:04:42

$x=2$

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

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

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

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

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

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

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

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