What Are Some Tips for Factorising Cubic Functions?

[Matty R]

Hello again.

I'm having more trouble with my homewrork, but this one really isn't fair. We haven't done the work in class, and as far as I can tell it isn't on the syllabus, but we're still expected to do it.

Homework Statement

http://img17.imageshack.us/img17/4379/questions7.jpg

Homework Equations

http://img23.imageshack.us/img23/122/equation7.jpg

The Attempt at a Solution

http://img219.imageshack.us/img219/9325/attempt7a.jpg http://img23.imageshack.us/img23/4422/attempt7b.jpg http://img23.imageshack.us/img23/7175/attempt7c.jpg



For part c, I have no idea where to go. From the googling I've done, I'm supposed to find a value of x where f(x)=0. I would have thought that the point of part b was to get a value of of x to make f(x)=0, so the trial-and-error bit would be removed.

I used some software called Mathematica to plot the graph of this function between x=-4 and 4, and the only point where it crosses the x-axis is somewhere between x=2.3 and 2.4.

I plotted between those points because later on I have to sketch the graph and plot the points when x=0 to x=3, and only those points.

I originally made a mistake where I have -250 - 16. I added 16 and got -234, then came up with this :

http://img23.imageshack.us/img23/5506/wrongh.jpg

This is my third day on this question, and it's driving me nuts.

Does anyone have any hints/tips or know what I'm supposed to do?

 

[elect_eng]

This is my third day on this question, and it's driving me nuts.

Does anyone have any hints/tips or know what I'm supposed to do?

There is an easier way to approach the problem

16=54*(x-5/16)^3

8/27=(x-5/16)^3

the cubed root of 8/27 has three possible answers: 2/3, 2(e^(-j2*pi/3)/3, 2e^(j2/3)/3

You can take if from there.

 

[Matty R]

There is an easier way to approach the problem

16=54*(x-5/16)^3

8/27=(x-5/16)^3

the cubed root of 8/27 has three possible answers: 2/3, 2(e^(-j2*pi/3)/3, 2e^(j2/3)/3

You can take if from there.

Thanks for the reply.

I really appreciate it, but I haven't seen anything like that before. I don't even know how the maths of it works. I can't see us being expected to do that.

I'm going to email my teacher and see if it's meant to be +16. There have been typing errors before.

I'm going to keep a note of your reply though, just in case it does pop up on the course at some point.

Thanks again.

 

[tiny-tim]

the cubed root of 8/27 has three possible answers: 2/3, 2(e^(-j2*pi/3)/3, 2e^(j2/3)/3

Thanks for the reply.

I really appreciate it, but I haven't seen anything like that before. I don't even know how the maths of it works. I can't see us being expected to do that.

Hi Matty R!

I take it you're confused by the "three possible answers"?

If you haven't done complex numbers, then just ignore that …

follow elect_eng's hint, and get to 8/27=(x-5/16)³ …

then x - 5/16 must be the cube root of 8/27, which is … ?

 

[Matty R]

Hi Matty R!

I take it you're confused by the "three possible answers"?

If you haven't done complex numbers, then just ignore that …

follow elect_eng's hint, and get to 8/27=(x-5/16)³ …

then x - 5/16 must be the cube root of 8/27, which is … ?

Hello again Tim. I'm sorry I didn't reply sooner. There's been a family emergency.

I've got the cube root of 8/27 as 2/3, so does that mean x=47/48?

I don't understand how to get (x-5/16), and according to my graph, the line doesn't cross the x-axis at 47/48.

I'm so confused.

Complex Numbers is on the syllabus, but we haven't done it yet.

I haven't had a reply from my teacher either.

 

[tiny-tim]

Hello again Tim. I'm sorry I didn't reply sooner. There's been a family emergency.

I've got the cube root of 8/27 as 2/3, so does that mean x=47/48?

I don't understand how to get (x-5/16), and according to my graph, the line doesn't cross the x-axis at 47/48.

Hello again Matty R!

on loooking at the original equation, I think elect_eng copied wrong, and meant 5/3, not 5/16 (which you should have spotted … you should check everything you use here) …

does that put it right?

I'm so confused.

Complex Numbers is on the syllabus, but we haven't done it yet.

I haven't had a reply from my teacher either.

Forget complex numbers.

 

[elect_eng]

I think elect_eng copied wrong, and meant 5/3, not 5/16

Yes, I'm very sorry about that. I'd like to say it was just a "slip of the pen", but since we don't use pens anymore, I've got no excuse.

 

[tiny-tim]

Yes, I'm very sorry about that. I'd like to say it was just a "slip of the pen", but since we don't use pens anymore, I've got no excuse.

we goldfish have to use pens that write underwater …

we can't get underwater computers …

i can only access the internet with the help of of my mouse ​

 

[Matty R]

Hello again Matty R!

on loooking at the original equation, I think elect_eng copied wrong, and meant 5/3, not 5/16 (which you should have spotted … you should check everything you use here) …

does that put it right?

That looks great, and fits with my graph. x=7/3. So I only have one value for x because the line crosses the x-axis only once. I'm getting there. It's slow, but I'm getting there.

elect and tim, thank you so much. I was stuck on that question for over a week, and the answer you've helped me work out has also cleared up the next question.

Yes, I'm very sorry about that. I'd like to say it was just a "slip of the pen", but since we don't use pens anymore, I've got no excuse.

we goldfish have to use pens that write underwater …

we can't get underwater computers …

i can only access the internet with the help of of my mouse ​

:rofl: :rofl: :rofl: