Find all complex zeros of each polynomial function. Give exact va...

Question

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^4+4x^3+6x^2+4x+1

Accepted Answer

Welcome back. I am so glad you're here. We are asked for the following polynomial function to determine all the complex zeros. Write the answer in exact form. Our polynomial function is F of X equals X to the fourth plus eight X cubed plus 24 X squared plus 32 X plus 16. Our answer choices are answer choice. A negative two with a multiplicity of four answer choice B two with a multiplicity of four answer choice C negative two with a multiplicity of two and two with a multiplicity of two and answer choice D negative two with a multiplicity of three and two with a multiplicity of one. All right. So for a polynomial function, there's a lot of different ways we can go about trying to discern the zeros. But when we have five terms here, we've got a degree of four, we've got quite a bit going on. The easiest way to start is with the rational zeros theorem, which as you recall from previous lessons will give us all the possible rational zeros when we take P which is all of the factors of the constant term in this case 16 and we divide that by Q, which is all of the factors of the leading coefficient in this case, one. So what are all of the possible factors of our constant term? 16? Well, we've got plus or minus 16, which would be multiplied by plus or minus one. We've got plus or minus eight, which would be multiplied by plus or minus two. And we've got plus or minus four, which would be multiplied by itself. Now, how about all of the possible factors for our leading coefficient? Which is one? Well, that's just plus or minus one. And yay, it's the best when our Q is just plus or minus one because then we only have to focus on the P S. Anything divided by one is itself. So what we would do next is we would utilize the remainder theorem and we would do synthetic division. So we write down all of the coefficients and the constant term from our polynomial function. So we've got one 8 32 16. And then we choose one of our P divided by Q S and you could start just at the beginning here. We've got a positive 16 and you could start going through and do the synthetic division. However, and this is multiple choice. And for these multiple choice questions, the only possible zeros are negative two and positive two. So we can just answer choice A has negative two. We can start off with negative two and test that out with synthetic division. So we start off by bringing down the one and then we're going to multiply that negative two multiplied by one which is negative two. And then we add eight plus negative two. That's going to be a positive six, multiply negative two, multiplied by six is negative 12. Add 24 plus negative 12 is positive 12, multiply negative two multiplied by 12 is negative 24. Add 32 plus negative 24 is a positive eight, multiply negative two, multiplied by eight is negative 16, add 16 plus negative 16 is zero. And there with our remainder theorem, we know that negative two is one of our zeros. It's a real zero here. OK? So now we've got a new polynomial with coefficients of 16 12 and a constant term eight. And we've got new PS and Qs our P S. So that's for the constant term are going to be plus or minus eight, which would be multiplied by plus or minus one plus or minus four which are multiplied by plus or minus two. And that's it for our Qs again, the leading coefficient is plus or minus one. Yay. So we just need to focus on those PS and again, we could choose anything guess and check but we know it's just gonna be negative two or two. So we can try negative two again. And so we bring down the one multiply negative two multiplied by one is negative two, add six plus negative two is positive four, multiply negative two, multiplied by four is negative eight, add 12 plus negative eight is a positive four, multiply negative two, multiplied by four is negative eight and eight plus negative eight is zero. Thank you. Remainder theorem, we've got another zero of negative two. So now we've got a new polynomial and we've got a leading coefficient of one. So that's X squared, then plus four X then plus the constant term four, that's all equal to zero. And sometimes at this point, we can factor this or we can use the quadratic formula. And let's see if we can factor this. We set up two open sets of parentheses equal to zero. What multiplied by what gives us X squared that's X multiplied by X for our first. Then what two factors of positive four are going to add up to a positive four. That's going to be a positive two and a positive two for our lasts. Well, we've got X plus two as a factor and we said that equal to zero to solve for X. And we get X equals negative two, but there's two of them. So it's got a multiplicity of two. There, we've got four negative twos. So that is negative two with a multiplicity of four. We look at our answer choices and that matches with answer choice. A well done. We'll catch you on the next One.

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Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^4+4x^3+6x^2+4x+1

Zeros of Polynomial Functions practice set