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what are the roots of find the roots of p(x)=x^4-4x^2-4x-1

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3 Answers

-1 is a root by rational root theorem.
-1|  1  0  -4  -4  -1
         -1   1   3    1
     1  -1  -3  -1   0
Using synthetic division we see that (x4+0x3-4x2-4x-1)/(x-(-1))=x3-x2-3x-1
Again -1 is a root by the rational roots theorem
Again we use synthetic division
-1| 1  -1  -3  -1
         -1    2   1
      1 -2   -1   0
(x3-x2-3x-1)/(x+1)=x2-2x-1 whose roots are, by the quadratic formula
(2±√8)/2 or 1+√2 and 1-√2
x^4-4x^2-4x-1=x^4-4x(x+1)-1=x^4-1-4x(x+1)
(x^4-1)-4x(x+1)=(x^2-1)(x^2+1)-4x(x+1)
(x-1)(x+1)(x^2+1)-4x(x+1)=(x+1)[(x-1)(x^2+1)-4x]
(x+1)(x^3-x^2+x-1-4x)=(x+1)(x^3-x^2-3x-1)
we know -1 is a root
I thought that I could factor x^3-x^2-3x-1 by grouping again like in the beginning but not so-
look at the rational roots theorem: +factors of -1/factors of coefficient of x^3; it appears -1 is a root again
(x^3-x^2-3x-1)/(x+1)=x^2-2x-1(I divided on paper to get my answer)
as you can see in the other solution(s) that synthetic division is much easier
for example using x^3-x^2-3x-1 and synthetic division we have:
-1     1  -1   -3   -1
            -1    2     1
         1  -2   -1    0
so now we have to work with x^2-2x-1 using the quadratic equation
[-b+sqrt(b^2-4ac)]/2a=[2+sqrt(4+4)]/2=[2+sqrt(8)]/2=[2+2sqrt(2)]/2=1+sqrt(2)
 
           
 
 
Hey Efqeffqe -- the roots we may test are (+/- last term factors) / (+/- 1st term factors). 
 
Try +/- 1 ... -1 works => divide by x+1 ==> x^3 -x^2 -3x -1 ...
-1 works again ... divide by x+1 ==> x^2 -2x -1 => (x^2 -2x +1) -2 ... (x-1)(x-1) - 2 ...
(x-1) = +/- sqrt 2 ==> x = 1 +/- sqrt 2
 
Four roots: -1, -1, 1+ sqrt 2, 1- sqrt 2 ... Best regards :)