What are all the zeroes of #f(x)= x^3 +2x^2 -3x +20 #?

1 Answer
George C.
Nov 28, 2015

#x = -4#

#x = 1 + 2i#

#x = 1 - 2i#

Explanation:

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# in lowest terms, where #p, q in ZZ#, #p# a divisor of the constant term #20# and #q# a divisor of the coefficient #1# of the leading term.

So the possible rational roots are the divisors of #20#:

#+-1#, #+-2#, #+-4#, #+-5#, #+-10#, #+-20#

After a bit of trial and error, find:

#f(-4) = -64+32+12+20 = 0#

so #x=-4# is a zero and #(x+4)# is a factor of #f(x)#.

#x^3+2x^2-3x+20=(x+4)(x^2-2x+5)#

Find the Complex zeros of the remaining quadratic factor using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (2+-sqrt(4-20))/2 = (2+-sqrt(-16))/2#

#= (2+-4i)/2 = 1 +-2i#

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