# First, find the maximum number of real zeros. Then, use Decartes' Rule of Signs to determine the possible numbers of positive and negative roots.`f(x)=3x^5-2x^3-4x-7`

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First, let's find the possible number of real roots. The maximum possible number of roots *without considering other facts* will always be the **highest degree** of in your polynomial. Here, our x-term with the greatest power is: `3x^5` so, the degre of the polynomial will be 5.

Therefore, **the maximal number of real roots before we proceed will be 5 roots**.

Side note, but important: Because this is a polynomial of odd degree with real coefficients, we can say for certain there is** at least 1 real root**! This is based on the theorem of complex conjugates.

Now, we can find the possible numbers of positive roots with the equation as it is, rewritten here:

`f(x) = 3x^5 - 2x^3-4x-7`

So, to quantify the number of positive roots, **we find the number of sign changes between adjacent terms (we ignore where coefficients are zero)**. So, I only see one sign change: between `3x^5` and `-2x^3` (+3 to -2). There are no other sign changes because the rest of the terms are negative!

Therefore there is 1 positive real root.

Now to find the number of negative roots, **we flip each sign that's associated with an odd power of** `x`. Here, we flip each sign except for that in front of the 7 (it's a constant). This gives us a new f(x):

`f(x) = -3x^5 + 2x^3 + 4x - 7`

Now, we have 2 sign changes! The first between `-3x^5` and `2x^3` (-3 to +2), and the second between `4x` and `-7` (+4 to -7). This indicates that we may have 2 roots.

But we can't stop there! Decartes' Rule of Signs says that the **number of roots will be the number of changes of signs, or it can be that number minus a multiple of 2**. In our first case, we knew there was 1 positive root, but we can't subtract 2 from that to say we have -1 positive roots! However, in this case, we can say there are either 2 negative roots (the number of sign changes) or 0 negative roots (number of sign changes - 2).

So there's our answer for negative roots: 2 or 0.

After this analysis, we can now **change the maximum number of real roots**. We have shown that there must be 1 positive root and either 2 or 0 negative roots. Based on this, there are a **maximum of 3 possible real roots**. This would also indicate that there are at least 2 complex roots that are not real.

Of course, we only know this after our analysis, so it really depends on the order the question asks you about these things to say which is the right answer!

I hope that helps!