How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=x32x28x?

1 Answer
Aug 8, 2018

The zeros of f(x) are 0, 2 and 4

Explanation:

Given:

f(x)=x32x28x

Note that the signs of the coefficients of f(x) are in the pattern +. With one change of sign, Descartes' Rule of Signs tells us that f(x) has exactly one positive real zero.

The signs of the coefficients of f(x) are in the pattern +. With one change of sign, Descartes' Rule of Signs tells us that f(x) has exactly one negative real zero.

In order to usefully apply the rational zeros theorem we need a non-zero constant term, but f(x) has none, or in other words it is zero. That's because x=0 is a zero of f(x). Let's separate x out as a factor first:

x32x28x=x(x22x8)

Now applying the rational zeros theorem to x22x8, any rational zeros of this quadratic are expressible in the form pq for integers p,q with p a divisor of the constant term 8 and q a divisor of the coefficient 1 of the leading term.

That means that its only possible rational zeros are:

±1,±2,±4,±8

Trying these, we find:

(2)22(2)8=4+48=0

(4)22(4)8=1688=0

So the zeros of f(x) are 0, 2 and 4