Question

Aglebra 2 Help!! If P(x)=x^3-4x^2+cx+6 , for what value of c is P(3)=0?

Answer 1

do you have any idea how to solve this?

Answer 2

Plug in the 3 getting:
3^3 - 4 (3^2) + 3c + 6 = 0
\[3^{3} - 4(3^{2}) + 3c + 6=0\]

Answer 3

Can you now solve for c ?

Answer 4

so c=1 ?

Answer 5

No. working it out 27 - 36 + 3c + 6 = 0 combine and solve.

Answer 6

so what is the answer?

Answer 7

i thought that it would be 1

Answer 8

combine 27-36+6 can you add signed numbers?

Answer 9

yes i can it equals -3

Answer 10

LOL I goofed so -3 + 3c=0 and 3c=3 and c does indeed = 1 Sorry I was wrong.

Answer 11

its okay! can you please help me with another problem?

Answer 12

Yes, if I can learn how to add signed numbers.

Answer 13

LOL thanks! IM going to type it right now.

Answer 14

If a number is selected at random from {-4, (2+2i)^2, (3-i)^2, (2i)^6, 0, (negative sqrt 3 -isqrt3)^2} what is the probability that the number is real?

Answer 15

the i is an imaginary number

Answer 16

Wow that course is something else, they are testing you on probability and the ability to check out for real numbers from complex ones. First would you agree that they have given you a total of 6 numbers (including 0)

Answer 17

yup!

Answer 18

i dont know if this answer is right but I got 1/2. Is that what you think it is?

Answer 19

OK, 4 is real
2i^6 is real
0 is real
Now for the (2+2i)^2 =4 + 8i +4i^2 = 8i imaginary
(3-i)^2 =9 - 6i -1 =8 - 6i a complex number with imaginary component
\[(-\sqrt{3}-i \sqrt{3})^{2}\] =6i which is imaginary

Answer 20

okay so then would you set up the fraction of 3/6 which simplifies to 1/2?

Answer 21

Looks like half are real and half are complex 50% probability wouldn't you say, in fact you did say. Do you have a choice for 50% or 1/2 or is this a multiple choice problem?

Answer 22

the 3 because there are 3 real roots and then 6 because there are 6 numbers that you can choose from.

Answer 23

it is an open ended question, but this makes sense to me.

Answer 24

Well it looks like a good solution. What kind of school are you going to?

Answer 25

im a sophomore in high school taking honors algebra 2.

Answer 26

i dont want to use up all of your time but can you help me with 2 more problems? I solved one of them so far but am not sure if i have the correct answer.

Answer 27

Very good, good luck in your studies, and in softball. My granddaughter got hurt in soft ball, but is all healed up now, she was playing second base, and a runner slid into second and hit her leg with cleats, took 50 stitches!

Answer 28

OK, maybe I can

Answer 29

wow! 50 stitches I hope she recuperates well and does good this season!

Answer 30

the problem is to solve ix^2+5x+24i+0 for x. The i is imaginary numbers again.

Answer 31

Is their an equal sign somewhere in there?

Answer 32

yes, ix^2+5x+24=0

SOrry about that. I solved the problem and got -3i and 8i but I am not sure if there are supposed to be two answers.

Answer 33

ix^2+5x+24i=0

Answer 34

sorry again, I wrote the wrong equation down

Answer 35

Is the first term \[ix ^{2}........or (ix)^{2}\]

Answer 36

it is ix^2 where the x is raised to the second power

Answer 37

That factors to (ix + 8) (x + 3i)
is that how you worked it?

Answer 38

ix + 8 = 0
ix = -8 multiply both sides by i
-x = -8i
x = 8i
and the other root is x = -3i

Answer 39

Thats what I got as well! I used the quadratic formula though. Thank you sooo much for your help!!! You've really helped me a lot. Are you up for one more? LOL

Answer 40

One more, and then I got to get off this computer lol

Answer 41

HAHAH!! Thank you!! Here is the problem: Let P(x)=ax^5+bx^3+cx, where a, b, and c are negative. Explain why P cannot have any negative zeros.

Answer 42

I tried to use Descartes Rule of Signs, but didnt know if I was going on the right path because when using that rule, I got that there would be negative zeros. So I am a little confused.

Answer 43

I don't know why! Whats your idea. I can see that when x is positive that the terms would have all negative coefficients there fore negative values,
If x is negative since they are all odd powers the value would be negative and with negative coefficients become positive and could result in negative roots ???????sorry about this, but I am also confused.

Answer 44

I guess I am missing something here.

Answer 45

Thats what I was thinking. because when I used Descartes Rule of Signs, I resulted in 2 negative roots. Oh well. LOL. Thank you so much for everything that you have helped me with!! I feel confident about my math homework now! Thanks :)

Answer 46

I have one more question though....Just kidding! LOL :) Have a good night.

Answer 47

You too, I don't mind trying to help when I see that the student is trying also. Again good luck.

Answer 48

I realize you have closed this question, still I have been thinking about it.
"Here is the problem: Let P(x)=ax^5+bx^3+cx, where a, b, and c are negative. Explain why P cannot have any negative zeros."
Factoring getting P(x)= x(ax^4 + bx^2 +c) Note that one zero is x=0. Now the other factor has all even powers and with negative coefficients would all be negative resulting in positive roots. I think!?

Answer 49

Oh! I understand what you are saying! I've been trying to figure this out. When you factor the remaining quadratic, the roots are positive! I am going to plug it in right now. Thanks! You're awesome!

Answer 50

What you said makes sense to me, so I am trying to come up with an equation to test this.