How do you use the rational roots theorem to find all possible zeros of P(x)=x3−4x2+x+6?

1 Answer
George C.
Jul 30, 2016

Zeros: −1,2,3

Explanation:

P(x)=x3−4x2+x+6

By the rational roots theorem, any rational zeros of P(x) are expressible as pq for integers p,q with p a divisor of the constant term 6 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1,±2,±3,±6

We find:

P(−1)=−1−4−1+6=0

So x=−1 is a zero and (x+1) a factor:

x3−4x2+x+6=(x+1)(x2−5x+6)

We could simply evaluate P(x) for the remaining possible zeros, but x2−5x+6 factors fairly easily:

x2−5x+6=(x−2)(x−3)

So the other two zeros are x=2 and x=3.

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