How do you use the rational roots theorem to find all possible zeros of P(x)=x5+3x3+2x−6?

1 Answer
George C.
Jul 4, 2018

The "possible" rational zeros are ±1,±2,±3,±6.

The only real zero is x=1

Explanation:

Given:

P(x)=x5+3x3+2x−6

By the rational roots theorem any rational zeros of P(x) are expressible in the form 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

In addition, note that the sum of the coefficients is zero, i.e.:

1+3+2−6=0

Hence we can deduce that x=1 is a zero and (x−1) a factor:

x5+3x3+2x−6=(x−1)(x4+x3+4x2+4x+6)

The remaining quartic has all positive coefficients, so any real zeros must be negative.

Trying −1, −2, −3 and −6 we find that none is a zero.

In fact it has only non-real complex zeros.

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