How do you solve $4x ^3 -13x^2 + 11x - 2 = 0$ ?

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Answer 1 · 2016-03-15T18:40:20

Lets look if there are any roots among the divisors of $-2$ which are $+-1$ , $+-2$ .
We can easily find out that $x=1$ and $x=2$ are roots.Hence

the original question can be written as follows

$4x^3 -13x^2 + 11x - 2 =(x-1)*(x-2)*(a*x+b)$

By equating the two sides we find out that $a=4$ , $b=-1$
hence

$4x^3 -13x^2 + 11x - 2 =(x-1)*(x-2)*(4x-1)$

Finally the roots are

$x_1=1$ , $x_2=2$ , $x_3=1/4$

How do you solve 4x ^3 -13x^2 + 11x - 2 = 04x ^3 -13x^2 + 11x - 2 = 0?