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Algebra/Complex Zeros

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Expert: - 4/4/2010


Question
Find all the complex zeros of the polynomial:
p(x)=x^3-7x+36

Answer
Note p(-1) = 42, p(-2) = 42, p(-3) = 30, p(-4) = 0 - aha! I found it.
SO, p(x) = (x+4)(x²-4x+9)- go ahead, multiply it out.

Now in the quadratic formula, we have √(b²-4ac), where a=1, b=-4, and c=9.
That would be √(16-36), which can't be done in real numbers.

If complex numbers are allowed, the roots of x²-4x+36 are (4±√-20)/2 = 2±√-5 = 2±i√5.

So all of the roots are -4, 2+i√5, and 2-i√5.

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