OpenStudy (anonymous):

Real zero questions! Picture below! Need some help on the process to get the answer!

3 years ago
OpenStudy (anonymous):

@jim_thompson5910 hey! do you recall what to do for these types of questions?

3 years ago
OpenStudy (jim_thompson5910):

since the degree is so high (16), you have to use a graphing calculator to find the roots

3 years ago
OpenStudy (jim_thompson5910):

you could factor out the GCF x^10, but you'd still be left with a very messy polynomial that you cannot solve by hand

3 years ago
OpenStudy (anonymous):

ugh yikes. i cant remember how to do these!

3 years ago
OpenStudy (jim_thompson5910):

do you have a graphing calculator?

3 years ago
OpenStudy (anonymous):

yup

3 years ago
OpenStudy (anonymous):

how do i find the roots though?

3 years ago
OpenStudy (jim_thompson5910):

the roots are the points where the graph crosses or touches the x-axis

3 years ago
OpenStudy (jim_thompson5910):

roots = x-intercepts

3 years ago
OpenStudy (jim_thompson5910):

also, the expression factors to x^(10) * ( x^6-2x^5-x^4+4x^3-x^2-2x+1 )

3 years ago
OpenStudy (jim_thompson5910):

so it might be easier to graph x^6-2x^5-x^4+4x^3-x^2-2x+1

3 years ago
OpenStudy (anonymous):

okay! so the roots are 2 and -2.

3 years ago
OpenStudy (jim_thompson5910):

did you graph x^6-2x^5-x^4+4x^3-x^2-2x+1 ?

3 years ago
OpenStudy (anonymous):

i think so! that cant be right though can it. hold on.

3 years ago
OpenStudy (anonymous):

oh i had made a mistake! the roots are 1 and -1

3 years ago
OpenStudy (jim_thompson5910):

of x^6-2x^5-x^4+4x^3-x^2-2x+1, yes don't forget to list the multiplicities

3 years ago
OpenStudy (anonymous):

hmm i dont understand. how is x=0 with any multiplicity?

3 years ago
OpenStudy (jim_thompson5910):

well you need to solve x^(10) = 0 the other factor

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

how do you find the multiplicity?

3 years ago
OpenStudy (jim_thompson5910):

well notice how the root at x = -1 is touching and not crossing the x axis

3 years ago
OpenStudy (jim_thompson5910):

so the multiplicity has to be an even number

3 years ago
OpenStudy (jim_thompson5910):

same with x = 1

3 years ago
OpenStudy (jim_thompson5910):

so that means we can eliminate choice A. That choice has x = -1 with multiplicity 3 (odd multiplicity)

3 years ago
OpenStudy (anonymous):

ah! i totally forgot about that. we did that in a previous unit. okay so that meant the answer is either B or C

3 years ago
OpenStudy (jim_thompson5910):

yes

3 years ago
OpenStudy (jim_thompson5910):

notice how the root at x = -1 is more parabolic (symmetric) while the root at x = 1 isn't so parabolic. That makes me think the multiplicity at x = -1 is 2

3 years ago
OpenStudy (anonymous):

hmm okay so the larger the multiplicty, the less parabolic?(typically)

3 years ago
OpenStudy (jim_thompson5910):

yeah the more flatter it looks (I think of it as it is touching more of the x axis, sorta)

3 years ago
OpenStudy (anonymous):

ahh okay. that makes more sense. i feel like this type of stuff is made up of little tricks like that. I actually have some more, would you mind helping out? lol like usual! sorry!

3 years ago
OpenStudy (jim_thompson5910):

3 years ago
OpenStudy (anonymous):

i am thinking the answer is A.

3 years ago
OpenStudy (jim_thompson5910):

choice A is correct

3 years ago
OpenStudy (jim_thompson5910):

nice work

3 years ago
OpenStudy (anonymous):

awesome! thanks! if i have issues with any other problems i will attach a picture!

3 years ago
OpenStudy (jim_thompson5910):

alright

3 years ago
OpenStudy (jim_thompson5910):

did you have another question?

3 years ago
OpenStudy (anonymous):

i did, but i figured out the answer! i actually have a new one though. this one i am stumped on.

3 years ago
OpenStudy (jim_thompson5910):

were you able to graph?

3 years ago
OpenStudy (anonymous):

yes! and all i see is that it hits the graph at one point..which is why it doesnt make sense.

3 years ago
OpenStudy (jim_thompson5910):

3 years ago
OpenStudy (anonymous):

on the x axis i mean^

3 years ago
OpenStudy (anonymous):

oh my bad! i had something wrong in the equation. okay let me see if i can figure it out now.

3 years ago
OpenStudy (anonymous):

is it b?

3 years ago
OpenStudy (jim_thompson5910):

no, it is not

3 years ago
OpenStudy (jim_thompson5910):

look for the root that has a cubic piece going through it

3 years ago
OpenStudy (jim_thompson5910):

that root has a multiplicity of 3

3 years ago
OpenStudy (anonymous):

ah, it must be d then?

3 years ago
OpenStudy (jim_thompson5910):

correct

3 years ago
OpenStudy (anonymous):

do i have the correct answer on this one?

3 years ago
OpenStudy (jim_thompson5910):

incorrect

3 years ago
OpenStudy (jim_thompson5910):

let p(x) = x^3 - 1 this is the numerator

3 years ago
OpenStudy (jim_thompson5910):

dividing by x-4 will give a remainder of p(4) = ??

3 years ago
OpenStudy (anonymous):

hmm im not sure, i was doing it by synthetic division

3 years ago
OpenStudy (jim_thompson5910):

I'm using the remainder theorem, so you can plug 4 into the x of the numerator

3 years ago
OpenStudy (jim_thompson5910):

and evaluate

3 years ago
OpenStudy (anonymous):

so 16? haah im confused! where am i plugging 4 into?

3 years ago
OpenStudy (jim_thompson5910):

x^3 - 1

3 years ago
OpenStudy (anonymous):

oh okay, so then that would be 63.

3 years ago
OpenStudy (jim_thompson5910):

correct

3 years ago
OpenStudy (anonymous):

oh okay! thanks!

3 years ago
OpenStudy (jim_thompson5910):

np

3 years ago