What are the zeroes of #p(x) = 2x^4 + x^3 - 2x -1#?

1 Answer
George C.
Jan 1, 2016

Real zeros #x = 1# and #x = -1/2#

Complex zeros #x = -1/2 +-sqrt(3)/2 i#

Explanation:

Factor #p(x)# by grouping then using the difference of cubes identity:

#2x^4+x^3-2x-1#

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

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

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

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

#=(x-1)(x^2+x+1)(2x+1)#

This has Real zeros #x=1# and #x = -1/2#

It has Complex zeros that you can find using the quadratic formula on #x^2+x+1 = 0#, which is in the form #ax^2+bx+c = 0# with #a=b=c=1# ...

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(-3))/2 = -1/2 +-sqrt(3)/2 i#

Note that #omega = -1/2+sqrt(3)/2 i# is called the primitive Complex cube root of #1#. It occurs frequently in the context of solving cubic equations.

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