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

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Answer 1 · 2017-11-21T17:38:08

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

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

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 $CC, because (x^2+3)=(x+isqrt3)(x-isqrt3),$

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

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