What are the zeroes of #f(x) = x^4 – x^3 + 7x^2 – 9x – 18#?

1 Answer
Nov 28, 2015

Answer:

#x=-1#, #x=3i#, #x=-3i# and #x=2#

Explanation:

First notice that by reversing the signs of the coefficients of the terms with odd degree, the sum is zero. So #x = -1# is a zero:

#f(-1) = 1+1+7+9-18 = 0#

and #(x+1)# is a factor:

#x^4-x^3+7x^2-9x-18#

#=(x+1)(x^3-2x^2+9x-18)#

then factor by grouping...

#=(x+1)((x^3-2x^2)+(9x-18))#

#=(x+1)(x^2(x-2)+9(x-2))#

#=(x+1)(x^2+9)(x-2)#

then take square root of #-9# to find:

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

So the zeros are #x=-1#, #x=3i#, #x=-3i# and #x=2#