Can you find all zeros (real or complex) of f(x) = x^4-2x^3-23x^2+90x+150?

Answer 1

Rational Root theorem says p/q = {factors of 150} are candidates for rational solutions.

It turns out that -5 is a rational solution

Synthetic division says:

-5 | 1 -2 -23 90 150

-5 35 -60 -150

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1 -7 12 30 0

(x+5)(x^3 - 7x^2 + 12x + 30) = f(x)

Recursively applying the Rational Root theorem to the resulting cubic polynomial does not

produce any rational solutions. So the real solution must be found numerically, most easily by

examining the graph.

The irrational solution is approximately -1.3105

Synthetic division again gives the approximate cofficients:

-1.3105 1 -7 12 30

-1.3105 10.89091025 -30

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1 -8.3105 22.89091025 0

x^2 - 8.3105x + 22.89091025 = 0

The final two complex solutions , per quadratic formula, are approximately

x = (8.3105 + 4.743335*i)/2 = 4.15525 + 2.371668* i

x = (8.3105 - 4.743335*i)/2 = = 4.15525 - 2.371668* i