How do you find all the zeros of #f(x)=x^4-x^3-3x^2-x+1#?

1 Answer
George C.
Aug 11, 2016

#f(x)# has zeros:

#x_(1,2)=(1+sqrt(21)+-sqrt(6+2sqrt(21)))/4#

#x_(3,4)=(1-sqrt(21))/4+-sqrt(2sqrt(21)-6)/4i#

Explanation:

#f(x) = x^4-x^3-3x^2-x+1#

Notice the symmetry of the coefficients: #1, -1, -3, -1, 1#

So:

#f(x)/x^2 = x^2-x-3-1/x-1/x^2=(x+1/x)^2-(x+1/x)-5#

Let #t = x+1/x#

Then:

#0 = t^2-t-5#

#= (t-1/2)^2-1/4-5#

#= (t-1/2)^2-(sqrt(21)/2)^2#

#= (t-1/2-sqrt(21)/2)(t-1/2+sqrt(21)/2)#

So:

#x+1/x = t = 1/2+-sqrt(21)/2#

Multiply both ends by #x# and rearrange slightly to get:

#x^2-(1/2+-sqrt(21)/2)x+1 = 0#

Writing the solutions of these two possibilities separately, we have solutions given by the quadratic formula:

#x_(1,2) = (1/2+sqrt(21)/2+-sqrt((1/2+sqrt(21)/2)^2-4))/2#

#=(1+sqrt(21)+-sqrt((1+sqrt(21))^2-16))/4#

#=(1+sqrt(21)+-sqrt(6+2sqrt(21)))/4#

#x_(3,4) = (1/2-sqrt(21)/2+-sqrt((1/2-sqrt(21)/2)^2-4))/2#

#=(1-sqrt(21)+-sqrt((1-sqrt(21))^2-16))/4#

#=(1-sqrt(21)+-sqrt(6-2sqrt(21)))/4#

#=(1-sqrt(21))/4+-sqrt(2sqrt(21)-6)/4i#

Related questions
See all questions in Zeros
Impact of this question
1243 views around the world
You can reuse this answer
Creative Commons License