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@DariusX @Hero @SolomonZelman @dan815 @jim_thompson5910
The zeroes are where the function crosses or touches the x-axis. (We'll deal with multiplicity in a minute, let's find the zeroes first) So I see that the function touches the x-axis at x=-5. So -5 is one of our zeroes. Can you list the others?
The zeros of the function are -5, -1, 3.5, 4, 4.5, 7. Correct? @zepdrix
The functions line was drawn a little too thick I suppose :) It's crossing at x=4, yes, but not those two points around it, maybe it just looks like it. So we have our zeroes: -5, -1, 4, and 7
If you remember the basic shape of some of your polynomials, it will really help understand what is going on with multiplicity.
Parabola, a second degree polynomial, is a bowl shape, ya? Do you see that shape at any of our zeroes?
I see it at -5. @zepdrix
Good! So at -5, our zero has a multiplicity of 2, because the shape corresponds to a second degree polynomial.
A linear or first degree polynomial is a straight line, so now let's try to find the locations where the curve doesn't do anything fancy at the zero, it just cuts right through.
Oh ok. That makes sense. I see a straight line going through -1, 3.5, 4, 4.5, and 7.
-1 and 7 sound correct. But we have some weird funny business going on at 4. (again you have to let the 3.5 and 4.5 go, we don't care about those, not zeroes)
are they there to throw me off? So, would -1 and 7 have a multiplicity of 1?
Ok great, yes.
At x=3.5, the blue line is just barely below the x-axis at x=4.5, the blue line is just barely above it. It only crosses the x-axis at x=4, not those surrounding points. I don't think they intended to trick you, it's the fact that the line was drawn too thick.
I'll draw an example of a third degree polynomial just to remind you.
|dw:1449793852960:dw|(Sorry, I'm on my laptop, touchpad lol)
This is the kind of shape we're seeing at x=4, ya?
|dw:1449793942853:dw|it's maybe drawn a bit more flattened out in your function, but the same shape, ya?
This is telling us that we have a multiplicity of 3 for our zero located at x=4
Notice that the multiplicities add up to: 2 + 1 + 1 + 3 = 7. This is a seventh degree polynomial, so your multiplicity will never add up to `more than` 7.
It could have ended up being less than 7 :) But they throw that level of complexity at us luckily.
oh okay. that makes sense. thanks for pointing the line thing out.
Do you understand how to do the second part? If x=-5 is a zero, adding 5 to each side gives us x+5=0 This tells us that (x+5) is a `factor` of the 7th degree polynomial. Not only that, but since it had a multiplicity of 2, (x+5)^2 will be a factor of our final polynomial.
Part 1: List the polynomial’s zeroes with possible multiplicities. Zeros: -5, -1, 4, and 7. The zero, -5, has a multiplicity of 2, because the shape corresponds to a second degree polynomial. The zero, 4, has a multiplicity of 3 because the shape corresponds to a third degree polynomial. Does this sound good for part 1?
I'm not sure if the explanation is necessary as part of your answer. I say that because they teach a different goofy method for determining multiplicity usually. Like the curve "bounces" off of the axis, therefore it has even multiplicity there. That specific explanation you gave might not be what they're looking for. Seems good though :)
I almost have the have part B ... One term short. Been try and error for an hour.
@zepdrix would i be better off excluding the explanation?
okay @retirEEd I appreciate it.
Yes probably :) Zeroes and multiplicity: -5 multiplicity of 2 -1 multiplicity of 1 4 multiplicity of 3 7 multiplicity of 1
Oh here is something to keep in mind...
Okay. Thanks so much. Now for part 2... @zepdrix
|dw:1449794958205:dw|This is the shape of a second degree polynomial. But it is also what a 4th degree polynomial can look like. And a 6th degree... and 8th... Any even polynomial can look like this, with both lines going up on the ends.
So when they list "POSSIBLE" multiplicities, they want you to understand that at x=-5, that shape merely tells us that the multiplicity is even. We are able to determine that the multiplicity can't be 4, 6, or any larger even number based on the fact that then we have multiplicities adding up to more than 7. Just something to keep in mind :) That applies to the weird snake shape I drew earlier as well. That not only works for a 3rd degree polynomial, but any odd polynomial greater than degree 1.
wow thanks for the information. i see what you're saying about the parabola you drew and the roots.
so about part two can you explain to me what you drew? @retirEEd
Since x=-5 is a zero, adding 5 to each side gives us (x+5)=0. This tells us that (x+5) is a factor of the polynomial that we want to build. Since it has a multiplicity of 2, we have two of these factors. Multiplying them together gives us (x+5)(x+5) We do the same with all of our factors. x=-1 gives us a factor of (x+1) so only polynomial so far is (x+5)(x+5)(x+1)
His numerator is what we're trying to build with this process. I'm not sure why he has a 5000 or 5500 in the denominator lol that seems unnecessary :)
It is a scaling factor to match the graph.
Oh I see ;) that's clever
Alright. But, I'm confused what about the 7 and 4 and their multiplicities? would we add them to your (x + 5)(x + 5)(x + 1)??
Yes, we need those factors as well.
x=7 gives us a factor of (x-7), yes?
so would it look like this: (x + 5)(x + 5)(x + 1)(x - 4)(x - 4)(x - 4)(x - 7)?
Yes, good. That is a "possible" polynomial for the one that they graphed. They want you to expand out all of the brackets.
Right it's a "possible" one. What does expanding out all of the brackets mean?
(x+5)(x+5) = (x^2 + 10x + 25) "FOIL" out the first pair like this, and then keep repeating that process. ya?
\[\large\rm (x+5)^2(x+1)(x-4)^3(x-7)\]\[\large\rm (x^2+10x+25)(x+1)(x-4)^3(x-7)\]It's going to be a real pain to do the rest :P I don't think I have the energy for it.
You can use an online algebraic calculator to do the multiplication for you if you want. I mean unless of course you need the multiplication practice, then perhaps you shouldn't skip this step :)
OH I misread the question, sorry sorry. It says "factored form", not standard form. Leave it like that, in the sets of brackets. My mistake!
Awesome, so my answer for part two is: (x+5)^2(x+1)(x−4)^3(x−7)? or should i put: (x + 5)(x + 5)(x + 1)(x - 4)(x - 4)(x - 4)(x - 7)?
I'd say it looks better with the exponents :)
Yes me too. Wow, thank you so much for your help. You're explanations make it much easier for me. Cheers!
yay we did it! np :)
Thank you as well @retirEEd
And thank you for teaching us how to teacher, teacher.