How do you find all zeros with multiplicities of #f(x)=2x^5+3x^4-18x-27#?

1 Answer
Nov 21, 2017

Answer:

(1) The Set of zeroes#={-3/2,+-sqrt3} sub RR.#

(2) The Set of zeroes#={-3/2,+-sqrt3,+-isqrt3} sub CC.#

Explanation:

Let #Z(f)# be the set of zeroes of #f.#

We have,

#f(x)=ul(2x^5+3x^4)-ul(18x-27),#

#=x^4(2x+3)-9(2x+3),#

#=(2x+3)(x^4-9),#

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

#=(2x+3)(x^2+3)(x+sqrt3)(x-sqrt3).#

# :. Z(f)={-3/2,+-sqrt3} sub RR.#

But, in #CC, because (x^2+3)=(x+isqrt3)(x-isqrt3),#

#:. Z(f)={-3/2,+-sqrt3,+-isqrt3} sub CC.#