Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

ugh yikes. i cant remember how to do these!

Answer 5

do you have a graphing calculator?

Answer 6

yup

Answer 7

how do i find the roots though?

Answer 8

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

Answer 9

roots = x-intercepts

Answer 10

also, the expression factors to

x^(10) * ( x^6-2x^5-x^4+4x^3-x^2-2x+1 )

Answer 11

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

Answer 12

okay! so the roots are 2 and -2.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

of x^6-2x^5-x^4+4x^3-x^2-2x+1, yes

don't forget to list the multiplicities

Answer 17

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

Answer 18

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

the other factor

Answer 19

ohh okay! forgot about that.

Answer 20

how do you find the multiplicity?

Answer 21

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

Answer 22

so the multiplicity has to be an even number

Answer 23

same with x = 1

Answer 24

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

Answer 25

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

Answer 26

yes

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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!

Answer 31

go ahead

Answer 32

i am thinking the answer is A.

Answer 33

choice A is correct

Answer 34

nice work

Answer 35

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

Answer 36

alright

Answer 37

did you have another question?

Answer 38

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

Answer 39

were you able to graph?

Answer 40

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

Answer 41

try readjusting your window

Answer 42

on the x axis i mean^

Answer 43

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

Answer 44

is it b?

Answer 45

no, it is not

Answer 46

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

Answer 47

that root has a multiplicity of 3

Answer 48

ah, it must be d then?

Answer 49

correct

Answer 50

do i have the correct answer on this one?

Answer 51

incorrect

Answer 52

let p(x) = x^3 - 1

this is the numerator

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

and evaluate

Answer 57

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

Answer 58

x^3 - 1

Answer 59

oh okay, so then that would be 63.

Answer 60

correct

Answer 61

oh okay! thanks!

Answer 62

np