URGENT! Polynomial Equations and Inequalities...

X

xo-hikari

New member
Joined
Sep 18, 2008
Messages
14
Hello! I would like someone to help me by giving me the answers to these various homework questions. I'm not sure if i did them correctly, but i only need the answers to check if im right. Thanks for your time!

Here are the questions *AND my answers (in bold):

1) Solve 2x^ 3 + 5x^2 + 14x + 6 = 0
x= -0.5, -1 +- square root of 5

2) Solve x^3 - 5x^2 + 3x + 9 > 0
x=-1, 3

3) Solve the inequality (x-5)/(x+4) < -2x
x<-5 and -4<x<0.5

4) determine the domain, intercepts, and equation of any vertical, horizontal, or slant asymptotes of

(i) f(x)= x^2/[(x+2)^2]
D={ x E R | x cannot equal -2}
vertical asymptote x = -2
horizontal asymptote y = 1
y-int =0
x-int = 0

(ii) f(x)= x^3/x^2 -1
D= {x E R | x cannot equal +- 1}
vertical asymptote = +- 1
horizontal asymptote = none
slant asymptote y=1x
y-int =0
x-int = 0

5) Solve x-2/x-1 < 6/x+3
x<-3 and 1<x<2

6) Factor 8x^3 - 125
? im still confused?

7) The graph of f(x)=3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts of -4 and 2. Determine the function.


Thanks again! I appreciate any help i can get!
 
xo-hikari said:
Hello! I would like someone to help me by giving me the answers to these various homework questions. I'm not sure if i did them correctly, but i only need the answers to check if im right. Thanks for your time!

Here are the questions:

1) Solve 2x^ 3 + 5x^2 + 14x + 6 = 0

2) Solve x^3 - 5x^2 + 3x + 9 > 0

3) Solve the inequality (x-5)/(x+4) < -2x

4) determine the domain, intercepts, and equation of any vertical, horizontal, or slant asymptotes of

(i) f(x)= x^2/[(x+2)^2]

(ii) f(x)= x^3/x^2 -1

5) Solve x-2/x-1 < 6/x+3

6) Factor 8x^3 - 125

7) The graph of f(x)=3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts of -4 and 2. Determine the function.


Once again, to save you time, you just have to give me the final answers. Thanks again! I appreciate any help i can get!
Click to expand...

We will be happy to check your answers for each of these problems.

Please show us the work you've done on each, and the answer you got for each. When we can see your work, we can determine how best to help you.

If you "just want the final answers," please be aware that though some of us may have been born at night, it wasn't LAST NIGHT.
 
lol its not like i need the answers just because i didnt do the work, otherwise i'd ask for the full steps so i could learn how to do it. At my level, im pretty sure the teachers don't care if i do the homework, as long as i can show my work when it comes to tests and stuff. I'm pretty sure i just need these to finalize with myself that the way i do it is correct, thats all lol.

im not going to type out all the process work i did for each question, because it'd be a waste.
If you'd like to help me out by just giving me the final answers to each question, i'd appreciate it greatly.
 
xo-hikari said:
im not going to type out all the process work i did for each question, because it'd be a waste.
Click to expand...
This is your homework from your instructor for your class to prepare you for your tests. If it's "a waste" for you to bother doing this, why would it somehow be "worthwhile" for us to do it? :shock:

Eliz.
 
lol im sorry but i think you misunderstood what i meant by it being a waste lol
i dont mean doing the work is a waste of time, obviously its not! I was referring to typing out all the steps i did for each question.
I'll post MY ANSWERS if that makes you guys happy. You just need to understand that i just want verification of the answers, i dont need explanations of how to get them (unless im wayyy off in my answer, of course).

sorry for any misunderstandings, im just an idiot.

i'll post the answers i got if someones willing to check if their correct.
 
You need to learn how to check your answers by substituting the various answers back into the original equations.

For instance, on the first one, if you got x=-2, then check by doing the following.

2x^3 + 5x^2 + 14x + 6 = 0
2(-2)^ 3 + 5(-2)^2 + 14(-2) + 6 =
-16 +20 -28 + 6 = -18. Since it should equal 0, this doesn't check. c=-2 is not the answer.

Suppose you got -1/2. The check would be

2(-1/2)^3 + 5(-1/2)^2 + 14(-1/2) + 6 =
2(-1/8) +5(1/4) +14(-1/2) + 6 =
-1/4 +5/4 -7 + 6 = 4/4 -1 = 0. It checks. x=-1/2 is one of the solutions.
 
Please pay attention ...

xo-hikari said:
... I'll post MY ANSWERS if that makes you guys happy ... im just an idiot ... i'll post the answers i got ...
Click to expand...

Hello:

At your prior thread to "just" confirm your results, I asked you to post your exercises and answers together in the future when you want people at this site to confirm your results.

I see your exercises in this new thread. I see that your motivation for posting them is the same. Where are the answers you want somebody to check?

~ Mark :|
 
for some reason the site wasnt letting me post (?), i think my internet connection is to blame,
but i fixed my 1st post by adding the answers
 
Your Internet connection deleted your answers ?!?

xo-hikari said:
1) Solve 2x^ 3 + 5x^2 + 14x + 6 = 0
x= -0.5, -1 +- square root of 5 ? You forgot the imaginary unit on the complex numbers

2) Solve x^3 - 5x^2 + 3x + 9 > 0
x=-1, 3 ? Incorrect. This exercise requires two intervals as a solution

3) Solve the inequality (x-5)/(x+4) < -2x
x<-5 and -4<x<0.5 ? THIS IS CORRECT

4) determine the domain, intercepts, and equation of any vertical, horizontal, or slant asymptotes ? BOTH (4i) and (4ii) ARE CORRECT

(i) f(x)= x^2/[(x+2)^2]
D={ x E R | x cannot equal -2}
vertical asymptote x = -2
horizontal asymptote y = 1
y-int =0
x-int = 0

(ii) f(x)= x^3/x^2 -1
D= {x E R | x cannot equal +- 1}
vertical asymptote = +- 1
horizontal asymptote = none
slant asymptote y=1x ? (Don't write coefficients of 1; write y = x, instead)
y-int =0
x-int = 0

5) Solve x-2/x-1 < 6/x+3
x<-3 and 1<x<2 ? This is not correct. (Try x = -4, -1, and 3, for examples)

6) Factor 8x^3 - 125
? im still confused? ? Think, "Difference of Cubes"

7) The graph of f(x)=3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts of -4 and 2. Determine the function.

I don't see your answers for p and q; did your connection fail again?
Click to expand...
 
why are you being rude? :(
i never blamed my connection for "deleting my answers"
i simply said that after i decided to post the answers i had, the site wasnt letting me post, thats why i ended up posting a bit later than i had planned to.

and #7 i didnt put up because i didn't finish doing it.

but yea, thanks for checking over the answers, i do appreciate it. :wink: :D
 
lol i realize i forgot the 2nd step in #2. The final answer would be x>3 and -1<x<3.

for #6, i just dont know how to do it, i've spent so much time on it and i just cant get it lol.

and for #7, i got p=59 and q=210
 
xo-hikari said:
why are you being rude?
Click to expand...

My sarcasm is an attempt to get you to pay attention.

You did not pay attention the first time. You did not pay attention the second time. (Your excuse for this is lame.)

Hopefully, you will pay attention the next time.

~ Mark :|

PS: I regret if your skin is not thick enough to shield you from sarcasm; there is nothing that I can do about that.
 
xo-hikari said:
lol i realize i forgot the 2nd step in #2. The final answer would be x>3 and -1<x<3. ?This is not correct

for #6, i just dont know how to do it, i've spent so much time on it ... ? Did you look up the factorization for a difference of cubes?

(Do you have any work to show for all of this time? If I could see what you're doing, then I probably could help you.)

and for #7, i got p=59 and q=210 ? This is incorrect; these values give an x-intercept of roughly -1/10 instead of 2

(Please show your work if you would like more help on this.)

Did you correct #5?
Click to expand...

~ Mark :)
 
xo-hikari said:
6) Factor 8x^3 - 125

use

a[sup:2aztba5x]3[/sup:2aztba5x] - b[sup:2aztba5x]3[/sup:2aztba5x] = (a - b)(a[sup:2aztba5x]2[/sup:2aztba5x] + ab + b[sup:2aztba5x]2[/sup:2aztba5x])

? im still confused?

Thanks again! I appreciate any help i can get!
Click to expand...
 
My Answers...

The way i did #2, makes sense to me...

Solve x^3 - 5x^2 + 3x + 9 > 0
the factors of 9 are +-1,+- 3, +-9
so i figured out that f(3) = 0

then i divided x^3 - 5x^2 + 3x + 9 by 3 using the remainder theorem, which yielded the answer (1x^2 - 2x -3)
i factored the answer and so all the factors of the original question are:

(x-3)(x-3)(x+1) > 0
(x-3)^2(x+1) > 0 <-------- i graphed the polynomial by using these as my x-int's


x-int = -1, 3
and the graph is cubic and positive, going from the 3rd quad and continuing forever in the 1st quadrant.
because (x-3) is squared, it tells me that the graph only touches -3, but does not pass through.
so because i am seeing where the function is above zero, the final answer would be when x>3 and when -1<x<3.
thats how i got the answer, so i'm not sure where i went wrong? it seems correct to me lol
I figured out that x= -1, 3
then i used those as the x-int for the graph.

----------------------------------------------------------------------------

for #5, i dont know how to do it, i thought my answer was correct, so can you just go through the steps of how you did it with me?

----------------------------------------------------------------------------
then for #6, the answer i think is possible would be
(x - 5/2) [x - (-5 +- 10 square root of 3 i)/ 4]
if this is not correct, can you show me the steps you did to get the answer? :wink:

----------------------------------------------------------------------------
finally #7, this is how i got my answer (note i re-tried #7, and i have no idea what i did the 1st time, so hopefully this time its correct):
The graph of f(x)=3x^4 + 14x^3 + px^2 + qx + 24 has x-intercepts of -4 and 2. Determine the function.


okay so first i subbed in -4 into the equation.= 768 - 896 + 16p - 4q + 24
= 16p - 4q - 104
4q = 16p -104 .... isolated q
q= 4p - 26 <--------- "Equation #1"

then i subbed in 2 into the equation.
= 48 + 112 + 4p + 2q +24
= 4p + 2q + 184
-2q = 4p + 184 .... isolated q
q= -2p - 92 <--------- "Equation #2"

now i put in equation #2 into equation #1
-2p - 92 = 4p - 26
-2p - 4p = -26 + 92
-6p = 66 ..... divide by -6
p = 11


Now i subbed in p=11 into equation #1 to get q

q= 4p - 26
q= 18



SOOO the answer is p=11 and q=18 ? HOPEFULLY thats right :p

----------------------------------------------------------------------------
 
Re: My Answers...

xo-hikari said:
#2... yielded the answer (1x^2 - 2x -3) ? Do not write coefficients of 1. Write x^2-2x-3, instead

... it seems correct to me ... YES, YOU ARE RIGHT. I COPIED A WRONG SYMBOL THE SECOND TIME AROUND ...

... then for #6, the answer i think is possible would be
(x - 5/2) [x - (-5 +- 10 square root of 3 i)/ 4]

8x^3 - 125 is a difference of cubes.

(2x)^3 - (5)^3

Use the factorization for a difference of cubes a^3 - b^3 posted by Subhotosh earlier.

a = 2x

b = 5

... #7, this is how i got my answer ...

-6p = 66 ..... divide by -6
p = 11

66 divided by -6 is not 11
Click to expand...


I'll work through #5 in a few minutes ...

~ Mark :)
 
for #6, i think i understand it

if i use this formula a[sup:3qv82uje]3[/sup:3qv82uje] - b[sup:3qv82uje]3[/sup:3qv82uje] = (a - b)(a[sup:3qv82uje]2[/sup:3qv82uje] + ab + b[sup:3qv82uje]2[/sup:3qv82uje])
then the answer would be
(2x-5)(4x[sup:3qv82uje]2[/sup:3qv82uje] -10x + 25)

see i was never taught that formula... but i think i understand now!

_______________________________________________________________________

for #7

p= -11
and
therefore q= -70
 
xo-hikari said:
#6 ...

if i use this formula a[sup:2hwg7mw0]3[/sup:2hwg7mw0] - b[sup:2hwg7mw0]3[/sup:2hwg7mw0] = (a - b)(a[sup:2hwg7mw0]2[/sup:2hwg7mw0] + ab + b[sup:2hwg7mw0]2[/sup:2hwg7mw0])
then the answer would be (2x-5)(4x[sup:2hwg7mw0]2[/sup:2hwg7mw0] -10x + 25)

Be careful with your signs. 2x * 5 is not -10x

______________________________________________________________________

#7 ...

p= -11
and
therefore q= -70

THIS IS CORRECT

(I note that the graph of this polynomial also crosses the x-axis at -3 and 1/3, for a total of four x-intercepts.)
Click to expand...

There are different methods for solving #5; here's one way.

\(\displaystyle \frac{x - 2}{x - 1} \;<\; \frac{6}{x + 3}\)

\(\displaystyle \frac{x - 2}{x - 1} \;-\; \frac{6}{x + 3} \;<\; 0\)

\(\displaystyle \frac{x \cdot (x - 5)}{(x + 3) \cdot (x - 1)} \;<\; 0\)

From this last expression, we see that the critical values of x are -3, 0, 1, and 5.

These values divide up the real number line into five intervals.

Pick a test value from within each interval and use it to evaluate the inequality.

You should find that there are two intervals where the lefthand side is less than 0; these two intervals make up the solution set.

Thank you for showing your work.

Cheers,

~ Mark :)
 
You must log in or register to reply here.
Top