How do you find all the real and complex roots of #x^4 - 2x^3 - 4x^2 - 8x - 32 = 0#?

1 Answer
George C.
Jan 28, 2016

Use the rational root theorem, then divide by the factors found to find the remaining roots.

#x=-2#, #x=4#, #x=+-2i#

Explanation:

Let #f(x) = x^4-2x^3-4x^2-8x-32#

By the rational root theorem, the only possible rational roots of #f(x) = 0# are expressible in the form #p/q# for some integers #p# and #q# where #p# is a divisor of the constant term #-32# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are factors of #32#:

#+-1#, #+-2#, #+-4#, #+-8#, #+-16#, #+-32#

Trying each of these in turn, we find:

#f(-2) = 16+16-16+16-32 = 0#

#f(4) = 256-128-64-32-32 = 0#

So #x=-2# and #x=4# are roots and #f(x)# has corresponding factors #(x+2)# and #(x-4)#

#x^4-2x^3-4x^2-8x-32#

#= (x+2)(x^3-4x^2+4x-16)#

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

#= (x+2)(x-4)(x-2i)(x+2i)#

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