find the Roots of this equation

Question

The roots of the equation $x^{4} -2x^{3} + x = 380$ are:

Though by trial and error one could solve the question, but I wanted to now if there is a particular method for solving the above equation ?

If you know the answers are integers, they will be about $\pm 380^{1/4}$ so try $\pm4, \pm5$, two of which are correct. That would help with factoring — Henry, Jun 27 '17 at 11:29
For a monic integer polynomial, the Rational Roots Test gives us that the only rational roots are in fact integer roots. So what Henry suggests is essentially decisive in getting all roots, once the "known" factors are divided out. That is, $x^4 - 2x^3 + x -380$ must be divisible by $x-5$, etc. — hardmath, Jun 27 '17 at 16:59

score 2 · Answer 1 · answered Jun 27 '17 at 11:27

2

HINT: note that $$x^4-2x^3+x-380=\left( x+4 \right) \left( x-5 \right) \left( {x}^{2}-x+19 \right)$$

answered Jun 27 '17 at 11:27

I know the answer, but how did you factorise it. I think there should be a proper way of getting its answer by adding and subtracting some terms of degree 2. Th is was a question in a particular exam from quadratic equations – ShiS Jun 27 '17 at 11:30
use the rational root theorem – Dr. Sonnhard Graubner Jun 27 '17 at 11:31
Can there be any other way of getting the answer – ShiS Jun 27 '17 at 11:35

score 0 · Accepted Answer · answered Jun 27 '17 at 12:15

0

I got the answer

$x^4-2x^3+x^2-x^2+x-380=0$

$(x^2-x)^2-(x^2-x)-380=0$

$ p = x^2-x$

Now quadratic equation can be solved to get the answer

answered Jun 27 '17 at 12:15

ShiS

2 Answers2