Question

determine the number of rational and irrational zeros of the function: f(x) = x^5-3x^4-2x^3+6x^2-3x+9

Answer 1

$x^5 - 3x^4 - 2x^3 + 6x^2 - 3x + 9 = 0$

Fifth degree polynomial; scary stuff.  We know it has five complex roots, counting multiplicities.  How many are real?

The rational root test tells us we only have to try the divisors of 9, so 1, -1, 3, -3, 9, -9.  We find x=3 gives

243 - 243 - 54 + 54 - 9 + 9 = 0

in nice pairs.

So x-3 is a factor and we can divide to get a 4th degree polynomial.  The division is a bit easier than usual.

         x^4           - 2x^2              - 3

x - 3 | x^5 - 3x^4 - 2x^3 + 6x^2 - 3x + 9

         x^5 - 3x^4

                        0 - 2x^3 + 6x^2

                                                0 - 3x + 9

                                                               0

So we get

$(x-3)(x^4 - 2x^2 - 3) = 0$

It's downhill from here.  The new factor is really just a quadratic in x squared, and factors easily:

$(x-3)(x^2 - 3)(x^2 + 1) = 0$

And now if we descend into irrational and complex numbers, we can further factor

$(x-3)(x - \sqrt 3)(x+ \sqrt{3})(x-i)(x+i) = 0$

and we can read off our roots,

$x= 3, \sqrt{3}, -\sqrt{3}, i, -i$

We have one rational root, namely 3, and two irrational roots, the square roots of three, and two purely imaginary roots.

Answer: 1 rational, 2 irrational

Answer 2

Answer:

One rational and 2 irrational zeroes

Step-by-step explanation:

I agree with the other guy.

f(3) = 0 so  x = 3 is a rational root.

There also are 2 irrational roots and 2 complex roots.

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