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^5-6x^4+14x^3-20x^2+24x-16

Accepted Answer

Welcome back. I am so glad you're here. We're 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 fifth minus three, X to the fourth plus nine X cubed minus X squared plus 18, X minus six. Our answer choices are answer choice. A one with a multiplicity of three negative square at six I and square at six. I answer choice B negative 11 with a multiplicity of two negative square at six I and square at six. I answer choice C one with a multiplicity of four and negative square at six I and answer choice D one with a multiplicity of four and square at six I. All right. When we have got a polynomial function with a degree of five, this one's really long. Our best bet is to start off with the rational zeros theorem. We recall from previous lessons that the rational zeros theorem is going to take P which is all of the possible factors of our constant term here, negative six and divide that by Q which is all of the possible factors of our leading coefficient in this case one. And that's going to give us all of our possible rational zeros. So what are all of the factors of negative six? Our constant term? Well, that's plus or minus six, which would be multiplied by plus or minus one. We could also have plus or minus three which would be multiplied by plus or minus two. How about four Q for the factors of the leading coefficient? Well, that's just plus or minus one. Yes, it's the best when the leading coefficient is one because taking any of our factors of the constant term, any of our P S and dividing them by one or negative one will just give us themselves or their negative version, which are already listed. So we can essentially ignore the Q. Now, what do we do with that? Well, now we utilize the remainder theorem which we recall from previous lessons, use a synthetic um division to test all the possible rational zeros and see if we get a remainder of zero. So we'll take all of the coefficients in the constant term from our polynomial which is one negative 39, negative 1918 and negative six. And we can start testing all of our possible rational zeros. We could start off with six, divided by one and use guess and check to figure it out. However, in this case, we're doing multiple choice and the only rational zeros that they list are one and negative one. So we could start off testing one. They've got that in answer choice. A now let's do the synthetic division. We bring down this one that is the leading coefficient. And now we multiply one multiplied by one is one. Now we add negative three plus one is negative two multiply one multiplied by negative two is negative two. Add nine plus negative two is positive seven multiply one multiplied by seven is positive seven. Add negative 19 plus seven is negative 12. Multiply one multiplied by negative 12 is negative 12, add 18 plus negative 12 is positive six. Multiply one multiplied by six is positive six, add negative six plus six is zero. And yes, thank you. Remainder theorem. We have a remainder of zero which means that one is one of our zeros for this polynomial. Now we've got a new polynomial with new coefficients and constant term. We've got one negative 27, negative 12 and six. And because of that six there, we have the same PS and Qs. And because of our multiple choice solutions, we can just test one again and see what we get. So bringing down the one, we have one multiplied by one which is one, add negative two plus one is negative one multiply one multiplied by negative one is negative one, add seven plus negative one is positive six multiply one multiplied by six is six. Add negative 12 plus six is negative six multiply one multiplied by negative six is negative six, add six plus negative six is zero. Yes, thank you. Again, remainder theorem, we have another zero and again, it is one because we had a remainder of zero. Well, we could go with one or negative one again. What's our new polynomial here? We've got one negative 16, negative six. The constant term is six. Again, they're all the same. Let's test one again. See if that works. Bringing down the one multiply one multiplied by one is one, add negative one plus one is zero. Multiply one multiplied by zero is zero, add six plus zero is six, multiply one multiplied by six is six. Add negative six plus six is zero. Yes, we've got another one. All right. So far we have one with a multiplicity of three. What's our new polynomial here? Well, we've got one X squared. That's just X squared. We have zero X S. So we don't have to put that. And then we have plus six. Our constant term all equal to zero. Well, with this, we can, since it's quadratic, it's got a power of two, we can just solve for X here, we can subtract six from both sides. And we have X squared equals negative six, we can take the square root of both sides and then we have X equals plus or minus the square rut of negative six. Well, we know that the square rut of negative six is the same as the square root of negative one multiplied by six. And the square root of negative one is I. So we can pull an I out. And so we have I, and that's plus or minus plus or minus I square root six. But using this, putting this in our standard form for a complex number, which would be A plus B I, we can put the I after the square at six. But just knowing that that is not inside the radical. So we have plus or minus the square root of six I. And so we have one with a multiplicity of three, a positive square root six I and a negative square at six I, we look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^5-6x^4+14x^3-20x^2+24x-16

Key Concepts

Polynomial Functions

Complex Zeros

Factoring and Synthetic Division