How do you find all the zeros of x3+x2−4x+6?

1 Answer
Shwetank Mauria
Apr 9, 2016

The only real zero of x3+x2−4x+6 is −3. However, if we extend the domain to Complex numbers 1+2i and 1−2i are other two zeros.

Explanation:

In x3+x2−4x+6, zeros of the function would be factors of 6 i.e. among {1,−1,2,−2,3,−3,6,−6}.

It is apparent that one such zero is −3 as putting x=−3 in x3+x2−4x+6, we get (−3)3+(−3)2−4(−3)+6=−27+9+12+6=0.

Hence (x+3) is a factor of x3+x2−4x+6 and dividing latter by former we get x2−2x+2, for which discriminant is (−2)2−4⋅1⋅2=−4 and hence cannot be factorized with real coefficients.

Hence, the only zero of x3+x2−4x+6 is −3.

However, if domain of x is extended to complex numbers, we will also have 2±√−42 (as they are zeros of x2−2x+2 using quadratic formula) or 1+2i and 1−2i.

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