How do you solve 0=4 $x^{3}$ $x^3$ -20 $x$ $x$ +16?

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Answer 1 · 2016-04-30T13:00:34

${ (x=1), (x=-1/2+sqrt(17)/2), (x=-1/2-sqrt(17)/2) :}$

First note that all of the terms are divisible by $4$ , so we can separate that out to get:

$0 = 4x^3-20x+16 = 4(x^3-5x+4)$

Next note that the sum of the coefficients of the simplified cubic is zero. that is: $1-5+4 = 0$ . Hence $x=1$ is a zero and $(x-1)$ a factor:

$x^3-5x+4 = (x-1)(x^2+x-4)$

The remaining quadratic factor is in the form $ax^2+bx+c$ with $a=1$ , $b=1$ and $c=-4$ . This has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(1^2-(4*1*-4)))/(2*1)$

$=(-1+-sqrt(17))/2 = -1/2+-sqrt(17)/2$

How do you solve 0=4x^3x3x^3-20xxx+16?