For Exercises 40–46, (a) List all possible rational roots or r...

Question

For Exercises 40–46, (a) List all possible rational roots or rational zeros. (b) Use Descartes's Rule of Signs to determine the possible number of positive and negative real roots or real zeros. (c) Use synthetic division to test the possible rational roots or zeros and find an actual root or zero. (d) Use the quotient from part (c) to find all the remaining roots or zeros. $f (x) = x^{3} + 3 x^{2} - 4$

Accepted Answer

were they given polynomial function? Use the carts full of signs to find possible rational zeros and actionable rational zeros. So let's go and look at this. The cards full of science tells us that if we have a specific number of science changes in our polynomial, that means we can figure out if there are real roots or not. So let's go and look at our polynomial on them all. We have to execute minus four X squared minus one. So we will see how many sign changes there are in the polynomial kind of imagine that there's a plus right here for to execute. So if we look at this we are from plus to minus and the - minus. There's only one sign change. So you have one sign change here. So We noticed that that measure is one positive real good. We don't know where it is yet. But just by that little signs we know there's one positive real root. Now you can check our negative the check our negative, let's plug in the negative X into everyone where there's an X. So you remember how to do this for our book. But we have we can have two Now instead of X. One negative X. The third minus four negative X squared minus one. This is going to help us figure out on how many negative railroads we have. So go and simplify that. We're gonna turn this -2. Sit there minus four X squared and minus one. So you notice here we're going from negative to negative two negative. There are no sign changes which means we have no negative real roots. You have one positive, no negative. Let's go ahead and check our possible rational zeros. So possible rational zeros are given by the rational 00. So rational zero theorem states that we have possible minus factors of P over factors of Q. Where p is the number, that is our trailing coefficient. In other words it is our final number and the polynomial which in this case is one. Similarly for Q Q is the coefficient of our first term, which determined the beginning or two. So go ahead and plus in two our question back here. So we need to find the factors of one or RP Well for one that's just one and one similar for two, that's just one and two. So we're going to play this into our fraction. We have plus or minus are factors of pr just one. So you have one on the top Divided by one on the bottom And for our other Q We have plus or -1 on the top and two on the bottom. So possible rational zeros Are plus or - and plus or -1 half. I'm gonna check which of these are actually the zeros from their actual zeros. So to find our actual zero, I'm gonna plug in our numbers we got into our equation. So let's say we take our first one f of positive one. We have two times one cubes minus four times one squared minus one. Let's see if this equals zero. So we solve this we get to -4-plus 1 that is negative one, that is not equal to zero. So we're gonna move on to our next one. Let's say f of -1. I use a different color here But f of negative one three times eight of one to the third one is four times eight of one squared minus one. That will give us 2001 is -2. Native four times positive one is 4 -1, that is negative seven. That is also not a zero rbl zero that we can use. Let's go ahead and check um positive one half. So you get two times one half cubed minus four times one half squared minus one. That is equal to one half to the third which is 1/8, 2 times 1/8 minus 41 half squared is 1/4 And we have -1, So we're gonna get 1 4th here. Then we have a minus one minus 1, 1/4 minus two Which is so not equal to zero but I'm gonna check neither one half, So native one half, We're gonna have two times 8-1 half to the third minus four times half squared minus one where you get two times negative 1/8 -4 times native 1/ squared Which is native fun and not even negative anymore. It's positive 1/4 minus one Which is gonna be equal to negative 1 4th - which is still not equal to zero. So we've checked all of them. None of them are equal to zero. So we actually have no rational zeros many of that. Our answer is actually answer D. A. Possible rational zeros are plus or minus one And plus or - half. But we actually have no rational Zeros. After we check. Okay I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Rational Root Theorem

Descartes's Rule of Signs

Synthetic Division