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(IOp) For the gwven equationFiud the all roots and display them on the coorlinaudler the Liven [unct Jn...

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(IOp) For the gwven equationFiud the all roots and display them on the coorlinaudler the Liven [unct Jn

(IOp) For the gwven equation Fiud the all roots and display them on the coorlina udler the Liven [unct Jn



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An equation is given, followed by one or more roots of the equation. In each case, determine the remaining roots. $$x^{4}-22 x^{3}+140 x^{2}-128 x-416=0 ; x=10+2 i$$

We have equation X minus one times X minus two cubed times X minus three is equal to zero and were asked to list each of the distinct roots and their multiplicity so we can write this as X minus one X minus two X minus two X minus two X minus three and setting. Each of these factors equals zero. We see that we get roots one to to two and three. So one is a root with multiplicity. Three. Excuse me with multiplicity one. Two is a root with multiplicity. Three in three is a roots with multiplicity one.

All right, We're giving this polynomial function. Were asked to think about the number of possible routes, whether they're really or imaginary. So our highest exponents are Degree is 1/7. That means we're gonna have seven complex roots. So seven riel plus imaginary roots. Now, what I like to dio is break this down into a table. So we have are really roots, and we have our imaginary routes. Now, imaginary roots must come in conjugated pairs, so there can only be even numbers in the imaginary column. So if we have seven roots total, we could have zero imaginary. We get up to imaginary, we could have four imaginary, or we could have six. Imaginary must be even numbers. We cannot have eight imaginary because the maximum we can have a seven now, riel plus imaginary gives us the number total complex roots. That's what cr complex roots. So if we have zero imaginary roots, we must have a seven real roots seven plus zero is seven. If we have to. Imaginary roots. We must have five real roots to plus five of seven. We have four imaginary will have three riel on if we have six imaginary we will have one riel. So these are possible combinations of riel and imagine interests there. Four possibilities. Now, the last thing we need to think about is rational roots. So the rational roots just gonna put rational roots. We've think about rational has the route ratio in it. So ratio means some number compared to each other through division. So for the rational route, uh, the rational Ruthie rem says that the factors of our coif, of our constant divided by the factors of our leading coefficient, which is a one there's an imaginary one in front of X to the seventh will give us our possible rational roots. So are factors of three of negative three. We have plus or minus three groups, plus or minus three and plus or minus one, and are factors of one are just plus and minus one. So if we take the ratio here, these would give us a possible rational roots for this equation. Excuse me for this pony

We have equation X Times X plus five to the fourth equals zero and were asked to find the distinct roots and the multiplicity So we can rewrite this as X Times X plus five X plus five times X plus five times X plus five equals zero. So setting each of the factors equal to zero we see that the roots are zero negative. Five negative five negative five native five. So zero is a root with more publicity. One and negative five is a root with multiplicity four.

The number and the type of routes that you're going to get here. So what you're going to do here is you can factor out the X to get X squared plus four X minus 21 then, you know, factor that out. So because you're able the factory completely, um, you're gonna end up with three real solution routes. Now, with that, you are gonna have to positive ones and one negative one as roots for this problem.


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