Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary.* ƒ(x)=x^4+x^3-9x^2+11x-4

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Welcome back. I am so glad you're here. We're told for the following polynomial function to determine all the complex zeros, right? The answer in exact form our function is F of X equals X to the fourth minus three, X cubed minus six X squared plus X minus 24. Our answer choices are answer choice A three and negative two with the multiplicity of three answer choice B negative three and two with the multiplicity of three answer choice C negative three with a multiplicity of three and two and answer choice D three with a multiplicity of three and negative two. All right, when we have a polynomial with a degree of four, our best bet here is to start off with the rational zeros theorem which we recall from previous lessons is going to take P which is all of the factors of our constant term here negative divide that by Q which is all of the factors of our leading coefficient in this case one. So what are all of the factors of negative 24? Well, we'll have plus or minus 24 which would be multiplied by plus or minus one. We have a plus or minus 12 which would be multiplied by plus or minus two. We have a plus or minus eight which would be multiplied by a plus or minus three. And we have a plus or minus six which would be multiplied by a plus or minus four. Just a couple. Now, how about all of the possible factors of Q, the leading coefficient there? Well, that's just one we've got plus or minus one. Yay, it is the best when our leading coefficient is one because taking any of our PS and dividing them by a positive or a negative one is just going to get us itself for the negative, which is already included here. So this is great. Now, we use all of these possible rational zeros and we utilize the remainder theorem. We recall from previous lessons that for the remainder theorem, we're going to take the possible rational zeros and use synthetic division to see if we get a remainder of zero. And that confirms that it is a rational zero. So in this case, we take to set up synthetic division, we take all of our coefficients in constant term from the polynomial function. So we'll have one negative three, negative 6, 28 negative 24. And then for the possible rational zero, we've got so many options. We've got 24 divided by one. We could start with 24 use guess and check. However this is multiple choice and that narrows it down considerably. They've got threes, negative threes, negative twos and positive twos. So we might as well start with what answer choice A has, which is a positive three. So let's try dividing with a positive three. We'll bring down this one and then we will have the three multiplied by one which is three. Then we add negative three plus three. That's zero, multiply three multiplied by zero is zero. Add negative six plus zero is negative six. Multiply three multiplied by negative six is negative 18. Add 28 plus negative 18 is positive 10. Multiply three multiplied by 10 is 30. Add negative 24 plus 30 is not zero. So three doesn't work. So what's the next thing? Let's set this up again. We have one negative three negative 6 negative 24. So we know answer choice A is incorrect. We've eliminated that answer choice B they have a negative three. Let's test out negative three again, bringing down the one now negative three multiplied by one is negative three. Add negative three plus negative three is negative six. Multiply negative three multiplied by negative six is a positive 18. Add negative six plus 18 is positive multiply negative three multiplied by 12 is a negative 36. Add 28 plus negative 36 is a negative eight multiply negative three multiplied by negative eight is a positive 24. Add negative 24 plus 24 is zero. Yes, thank you. Remainder theorem. We've got a +01 of our zeros is negative three which we determined using the remainder theorem. So now what? Well, now we've got new PS and Qs, we have our constant term is now negative eight and so that eliminates some of those possible PS um 24 is out, one is still in 12 is out two is still in eight is still in three plus or minus three is out plus or minus six is out and plus or minus four is still in. So looking at our answer choices that have negative three as a possibility which is answer choices B and C, they are paired with twos. We know we're not gonna have more negative threes because we eliminated that with our new constant term of negative eight. So let's test out two. So going to a different part of our screen, we've got new coefficients in constant terms of one negative 6, 12, negative eight. And we're testing out two, we can bring down the one. Now we multiply two multiplied by one is two. Add negative six plus two is negative four, multiply two, multiplied by negative four is negative eight, add 12 plus negative eight is four, multiply two multiplied by four is eight, add negative eight plus eight is zero. Yes, thank you. Remainder theorem. We've got a zero of two. All right. Now what do we have for? Our polynomial, we have one X squared, that's minus four, X plus the constant term four, all equal to zero. Now that we've got a quadratic, we can see if we can factor this and we can, if we set up two open sets of parentheses, set it equal to zero. What multiplied by what gives us X squared that's X multiplied by X and those are our firsts and what two factors of positive four add up to negative four. And that's going to be a negative two and a negative two. Those are our lasts. So setting those factors X minus two equal to zero and solving for X by adding two to both sides, we get X equals two, but there's two of them. So we have two more twos. So for all of our zeros, we've got negative 322 and two. So that's two with a multiplicity of three, we look at our answer choices and this matches with answer choice. B 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^4+x^3-9x^2+11x-4

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Key Concepts

Complex Zeros

Polynomial Functions

Factoring Polynomials