I am stuck on this one problem and think it might be the teachers typo but don't want to point it out without being sure. You have to find the zeros (we use long division) of x^5+8x^3+16x but because these are all raised to odd powers they either give me negative or positive numbers, regardless of the rational root I plug in..
Maybe I'm just being clueless..thank you tho!
No worries! Let's figure out the problem together. Finding the zeros of a polynomial can sometimes be tricky, but there are a few methods we can use to approach it.
Since the given polynomial, x^5 + 8x^3 + 16x, has all odd powers, we can't make any immediate conclusions about the rational roots just by observing the signs.
One approach we can use is the Rational Root Theorem. This theorem tells us that if a polynomial has a rational root, it must be of the form p/q, where p is a factor of the constant term (in this case, 16) and q is a factor of the leading coefficient (in this case, 1). By testing these possible rational roots, we may be able to find some that work.
Let's list out the factors of 16: ±1, ±2, ±4, ±8, ±16. These are the possible values of p.
Now, let's list out the factors of 1: ±1. These are the possible values of q.
Combining each possible p with each possible q, we get the following list of potential rational roots: ±1, ±2, ±4, ±8, ±16.
To find if any of these values are zeros of the polynomial, we can perform long division or synthetic division. Start by selecting one of the potential rational roots and divide the polynomial by (x - r), where r is the root we are testing.
Repeat this process for each of the potential rational roots, and check if any of them result in a remainder of 0. If a remainder is zero, then the polynomial is divisible by (x - r), and r is indeed a zero of the polynomial.
If none of the potential rational roots work, it may indicate that there are no rational roots for this polynomial. In that case, the zeros could be irrational or complex numbers.
I hope this explanation helps you in tackling the problem effectively.