Question

Suppose that R(x) is a polynomial of degree 8 whose coefficients are real numbers.

Answer 1

A polynomial can have a maximum no. of roots = its highest degree. in this case it can have a total of 8 roots. Roots can be real or imaginary. but all put together there will be only 8.

If your polynomial has real coefficients then if it has an imaginary root then its conjugate will also be a root. Meaning if a+ib is a root then a-ib will also be a root and vice versa.

Now your are given 4 roots, 3 are real and one imaginary. the imaginary is -2-2i. Its conjugate will also be root. so -2+2i is also a root (Your answer for part one).

Now for real coefficients the imaginary roots will occur only in pairs. meaning a+ib and a-ib are roots.

now you know 5 of them. the other 3 could be all real. But if some of them are imaginary only 2 more could be imaginary since they occur in pairs.

So for you part 2 it can have a maximum of 6 real roots.
For part 3 it can have a max of 4 imaginary roots.

Answer 2

wait a sec im reading rajee

Answer 3

Oh okay i get it! THANK YOU!

Answer 4

so A=-2+2i
B=6
C=4

Answer 5

Not only will it have a maximum number of roots equalling its highest degree, it will have a minimum number of roots equalling its highest degree. Some of the roots may have the same value.

Answer 6

yes to @Taisonmac 's Question

Answer 7

k thanks

Answer 8

Yep, I agree those answers are correct.

Answer 9

thanks whpalmer for the double check!

Answer 10

yeah like @whpalmer4 some of the roots may be duplicated but it will have 8 roots

Answer 11

There are some interesting rules that you can apply when you know the coefficients to further determine how many roots of each type. See Descartes' Rule of Signs...

Answer 12

Also, duplicate roots will influence the shape of the curve. A duplicate root with an even number of roots will cause the curve to dip down to (or bow up to) the x-axis and then retreat. An odd number causes a crossing with an inflection point.

Answer 13

can you guys help me with this?

Answer 14

They have given you the hint . you have to take log on both sides.

Answer 15

oh okay

Answer 16

so like this x-6log6=-5xlog5?

Answer 17

yes

Answer 18

and then I take

Answer 19

(x-6) in parentheses

Answer 20

xlog6-6log6?

Answer 21

Distributing log6 to both x and -6

Answer 22

-5xlog5+xlog6=6log6

Answer 23

+5x

Answer 24

not -5x

Answer 25

oh okay

Answer 26

so 6log6/5log5+xlog6?

Answer 27

sry not xlog6

Answer 28

x( 5log 5 + log 6) = 6log 6

Answer 29

im stuck here 5xlog5+xlog6=6log6

Answer 30

oh okay

Answer 31

Factor out the x, then divide

Answer 32

x = 6log6 / (5log 5 + log6)

Answer 33

so 6log6/5log5+log6?

Answer 34

oh there it is

Answer 35

Pet peeve of mine — 6 log 6/5 log 5 + log 6 != 6 log 6/(5 log 5 + log 6)...

Answer 36

correct me if I'm wrong, but if you take \(\large log_6 \)on the left side, you have to take \(\large log_6\) on the right side as well, otherwise you'd lose the equality

Answer 37

no this is log 6 to the base 10

Answer 38

hmm

Answer 39

start from the beginning @jdoe0001

Answer 40

@jdoe0001 we aren't taking log base 6, we did log base 10 on both sides. the log 6 you are seeing is \(log_{10} 6\)

Answer 41

base of -10

Answer 42

value of x comes out to be about 1.09265296334604...

Answer 43

oh okay

Answer 44

approximately ;-)

Answer 45

wait

Answer 46

so dividing 6log6/5log5+log6 equals 1.092?

Answer 47

They want the answer in log form so no need for value

Answer 48

yeah i was like oh my lol

Answer 49

so what is it in log form?

Answer 50

the one that we found 6log 6 / (5log 5 + log 6)

Answer 51

cause i got log(6^... is that right?

Answer 52

log (6^5 / 5^5)

Answer 53

depending on how you solve it, you might end up with a different expression. For example, WolframAlpha gives \[x = \frac{6 (\ln(2)+\ln(3))}{\ln(2)+\ln(3)+5\ln(5)}\] but remember that adding logs is the same as taking the log of the product, so there's your 6 log 6 on top, and log 6 on the bottom

Answer 54

yeah i got that answer whpalmer4

Answer 55

but im not sure which is the true right answer

Answer 56

What do you mean, the true right answer? What are the candidates?

Answer 57

well it cant be the one you just put up cause they want in log

Answer 58

no?

Answer 59

it doesnt even let me put ln so it cant be that :\

Answer 60

Oh, I see what you're saying. I forgot that they were asking for it in a particular form, and not just the numeric value.

Answer 61

oh okay rajee. so log10(6/5)^5

Answer 62

@rajee_sam that's not correct. loga/logb does not equal log(a/b)

Answer 63

my bad @agent0smith is correct.. brain drain

Answer 64

yeah thats wrong because i cant even put the logwith the little 10 lol

Answer 65

so our earlier answer is the simplest one we can get

Answer 66

\[\Large \frac{ \log6^{6} }{ \log5^{5} +\log 6 } = \frac{ \log6^{6} }{ \log(5^{5} \times 6)}\]

Answer 67

which was 6log6/5log5+log6

Answer 68

yes

Answer 69

yes to who?

Answer 70

yes to you and @agent0smith

Answer 71

You can simplify to what I gave above, or \[\Large \frac{ \log6^{6} }{ \log(5^{5} \times 6)} = \frac{ 6\log6 }{ \log(18750)}\]

Answer 72

so which one should i enter?

Answer 73

lol

Answer 74

The most simplified is the very last one.

Answer 75

whichever one it will let you.. both are right answers @agent0smith 'sis more simplified version

Answer 76

i can put both so

Answer 77

put @agent0smith 's answer

Answer 78

i cant put parenthesis though which i dont think it will make a dif

Answer 79

should i enter it or wait for your reply?

Answer 80

You don't need parentheses. You can also bring the power of 6 back to the log
\[\Large \frac{ \log6^{6} }{ \log(5^{5} \times 6)} = \frac{ \log 46656 }{ \log18750}\]

Answer 81

okay

Answer 82

how you getting those numbers my calc isnt showing those numbers

Answer 83

6^6, and 5^5*6...

Answer 84

k thanks

Answer 85

How did you not get them correct on a calculator??

Answer 86

yet the q is not asking to reduce fully so i wonder if its necesarry to make them numbers or just leave it as 6log6/5log5+log6

Answer 87

and i re entered it and it worked

Answer 88

well thanks anyways!

Answer 89

Either should be acceptable. But if you can't use parentheses, it would not be wise to enter 6log6/5log5+log6... since that's not the same as 6log6/(5log5+log6)

Answer 90

k thanks