In Exercises 37–38, use Descartes’s Rule of Signs to determine th...

Question

In Exercises 37–38, use Descartes’s Rule of Signs to determine the possible number of positive and negative real zeros for each given function. f(x) = 3x^4 - 2x^3 - 8x + 5

Accepted Answer

Hi there. So this problem tells us to find the possible number of positive and negative real zeros for the given function by using the cards rule of sites. So the cars full of signs states that for the number of sign changes we have in our polynomial we will have that number of real positive or negative zeros potentially. So let's go ahead and take a look at our polynomial. Some we have the polynomial X to the fourth plus three is the third minus four X plus one. We're gonna check the positive or negative sign in front of each of these coefficients. So two X to the fourth plus three X to the third -4 X. Plus one. Now we don't have a sign in front of twist the fourth but it's not negative. So we're going to assume that this is positive. Let's see if we have any sign changes over each of our signs. So from this plus to this plus we do not change sign power. We do change from this plus this negative. So we have a change and again from negative positive. We have another change diving is that we have to possible positive roots. Now I say possible posit Bruce. Because if we would do the rational zero later we could we would be able to tell if there's actually a positive route there. So the account for that I'm also gonna say or zero positive difference. Just the account for the event that we might not actually have positive roots. Now let's check our negative. So check our negative zeros with Carcillo signs. We're gonna substitute negative X everywhere where there's an X. So we would say then two times negative X. To the fourth plus three negative X. The third minus four times negative X. Plus one. We're gonna simplify this. So we get native X. The fourth that gives us positive X. The fourth negative X. The third that gets us negative three X. The third made it four times negative X. Is positive four X. And we just have plus one. So once again we'll imagine there's a plus in front of that. Check our sons we got from plus to minus that is a change minus the plus, that's another change and plus the plus, which is not a change To sign changes means we have two possible negative fruits. And by the same logic as the positive roots, we have to check by the rational 00 to see if there actually are two negative roots. So I'm going to say or zero possible negative roots. So we look at our answers. Now this answer is going to be answer A We have two or zero positive rail routes, two or zero negative rail routes. Alright, I hope to help you solve the problem. Thank you for watching. Goodbye

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