Find the zeros: f(x)= x^5+2x^4+18x^3+36x^2-19x-38

Answer 1

Rational Root theorem says:

P = +or- { 1,2,19,38}

Q = +or-{1}

so P/Q = P

x=1 is a solution since f(1) = 1+2+18+36-19-38 = 57-57 = 0

x=-1 is a solution since f(-1) = -1 + 2 - 18 + 36 + 19 - 38

= 1-18+36+19-38

= -17 + 36 + 19-38

= 19+19-39

= 0

x=2 is a solution since f(-2) = -32 + 32 - 144 + 144 + 38-38

= 0

Synthetic division:

1 | 1 2 18 36 -19 -38

1 3 21 57 38

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1 3 21 57 38 0

-1 | 1 3 21 57 38

-1 -2 -19 -38

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1 2 19 38 0

x^3 + 2x^2 + 19x + 38 =

x^2(x+2) + 19(x+2) =

(x+2) (x^2 + 19) =

the solution set is x = { 1, -1, -2, i*sqrt(19), -i*sqrt(19) }