In Exercises 49–50, find all the zeros of each polynomial functio...

Question

In Exercises 49–50, find all the zeros of each polynomial function and write the polynomial as a product of linear factors. f(x) = 2x^4 + 3x^3 + 3x - 2

Accepted Answer

this problem does expresses polynomial as a product of its linear and quadratic factors. Do that personally define all zeros of this polynomial. So the final zeros will make use of the rational zeros theorem. So if you recall for your book, the rational zeros theorem states that we have plus or minus factors of some number P Over some number two with those factors. Now P is given to us in our polynomial as well as Q. Where P is coefficient of our final turn. So this case P is equal to four. No Q is very similar. It's just given to us by the coefficient of our first term, which in this case is three. So Q equals three. Maybe when you find the factors of these, the factors of four are just the numbers that multiply the equal four. We have one and four and two and two on the other side with the factors of Q which is three. So we have three which should be just one and three. Now we're gonna take these numbers here into this fraction we have. So there are factors we should have plus or minus And then we have our first soul say is one for the p. one Over one for the queue. Another factor is two for the p and one for the queue and plus or minus or for I P one for you. And now we want to look at cuba's three. So we have plus or minus one for mp three for a Q closer minus 24. RP 34 Q. And finally plus a minus four for RP and +34 RQ. So are possible zero. So we simplify all this. You have a plus or minus one. Plus or minus two plus or minus four Plus or - 3 Plus or -2/3 or plus or -4/3. So now we can go in and check to see which one of these are actually zeroes. So I'm gonna take all these numbers and plug them into our original equation and see if we get zero out of any of these. So So we have three Times 1 to the 4th plus one to the third plus four times one squared Plus two times 1 -44 f of positive one apply that in we get three times one plus one plus four times one plus two minus four where set up equaling six which in this case that is not one of our zeros. Let's try another one. Let's try negative one. Hey the one It's three times negative one to the 4th plus native ones. The third plus four times negative one squared plus two minus four. We go ahead and simplify this we get three minus one plus four Plus 2 -4 which that actually simplifies to being zero. So negative one is one of our zeros. So we would do this until we get through all of our zeros. But I've actually checked this beforehand and our other zero that we find about doing all this is going to be X equals two thirds. So X equals two thirds and X equals negative one. Are the two zeros that we have Now. Let's think about this real quick. We have a polynomial with degree X. The four or degree four. We should have four zeros for this polynomial because we have a three or four but we only gave us out to zero. So we did all of our plugging in. So we have something going on here. We most likely have an imaginary set of zeros somewhere. So I'm gonna go and check our factors. Or rather always check for factors. We want to check our synthetic division. Don't go and use some tag division for this division helps us find the other factors by dividing with one of our current factors. So tech division our division symbol. Now the numerator of this. Division is our big polynomial up here. The three X. The four polynomial. We want every coefficient to go in the top of that fraction. So let's say we have three one Or two and 4. Now we also want to put outside one of our factors. We have extra two thirds. X equals two thirds and X equals negative one. I must say we'll use negative one first. Instead division this number is gonna go down first which is just three. Now we're going to multiply the number. You have outside times this number down here and put it underneath the one. So we have negative one times three is negative three. And that goes right under our one. Then we add those two numbers together. So you have one plus negative three or one minus three. That is negative two. Now we're gonna go ahead and do it again negative one times negative two should be a positive two. So we should get 4-plus 2 which is six. Do it again -1 times six is -6. So we get two plus negative six which is negative four. Finally negative one times negative four is positive four. That gives us zero. So the final numbers are remainder. We got a zero which is good. It's not look at what time we had originally originally we were at X the fourth X the third X the second and exit the first. We do our synthetic division. The number down here turns into X. The third, this is X squared X. The first and the rest is just the coefficients in our remainder at the end. So we got out X. The third senator that is three X. The third -2 x squared plus six X minus four. So now we're gonna do another synthetic division. This time we're going to divide by R two desserts. So I do another bracket here We have 3 -2, 6, -4. And we have 2/3. So once again we bring down our three and we look at the rest of our things here. So we have 2/3 times three. That should just be too. So you multiply those together. Two plus two. Should give us zero. Then we have two thirds of zero and zero which is six. Then we have six times 2/3 Which gives us 12 or three which is four. That gives us zero. So no we have X the third, right here, X to the second and X the first and our coefficient, that means that this is the bottom turns X squared and X the first and our coefficient. So what's left over is now three X squared plus six. Okay, so we actually have our three factors here. However let's just say it's three X three, X squared plus six equals zero powers to take out of three out of that. Get three X squared plus two equals zero. Or in the simplest form of this X squared plus two equals zero. So one of the factors actually X squared plus two. So if we go back We look at our x equals 2/3 And or x equals negative one. If we go ahead and set this equal to zero we gotta move our two thirds over. Well actually we don't want any fractions here. So we're actually not gonna do that. What I'm actually gonna do is just solve by multiplying three sets three x equals two. I have three X subtracting 23, X minus two equals zero. Similar for the icicles Native one, you add one to both sides, you have X plus one equals zero. Think of that are factors are now X plus one and three x minus two. And finally we have what we found by state division X squared plus two that will give us an imaginary number. So those are imaginary solutions. So finally our answer is this X +13 x minus two and X squared plus two. Which that is answer. C. Alright. I hope to help you solve the problem. Thank you for watching. Goodbye.

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