Polynomial Questions (1 Viewer)

[Fortian09]

Hi all,

I have a slight problem with my hw and I need a bit of help.
Obviously its got to do with Polynomials.

The question is as follows

Solve the equation x^3 - 14x^2 + 56x - 64=0 if the roots are in geometric progression.

And can somebody tell me how to use superscript on this forum?

Thanks to whoever

 

[lyounamu]

Hi all,

I have a slight problem with my hw and I need a bit of help.
Obviously its got to do with Polynomials.

The question is as follows

Solve the equation x^3 - 14x^2 + 56x - 64=0 if the roots are in geometric progression.

And can somebody tell me how to use superscript on this forum?

Let the roots be e/f, e and ef

Therefore, e/f . e . ef = -d/a
e^3 = 64/1
e = 4
e + e/f + ef = -b/a
4 +4/f+ 4f = 14/1
4f^2 + 4f + 4 = 14f
4f^2 - 10f +4 = 0
2f^2 - 10f + 2 = 0
(2f-1)(f-2) = 0
So f=1/2 or 2.

Therefore, the roots are 8, 4 and 2 or 2, 4 and 8

 

[Fortian09]

Hmm I used roots as
a, ak, ak^2
what is wrong with using roots in that format?

 

[tommykins]

the k^2 does alot of things and makes it messy. you're better off with a/k, a and ak.

 

[lsam]

Hmm I used roots as
a, ak, ak^2
what is wrong with using roots in that format?

Well, you can use that. But namu's format is better. It just makes it easier for him to work out.

Did you get the right answer anyway?

Let the roots be e/f, e and ef

Therefore, e/f . e . ef = -d/a
e^3 = 64/1
e = 4
e + e/f + ef = -b/a
4 +4/f+ 4f = 14/1
4f^2 + 4f + 4 = 14f
4f^2 - 10f +4 = 0
2f^2 - 10f + 2 = 0
(2f-1)(f-2) = 0
So f=1/2 or 2.

Therefore, the roots are 8, 4 and 2 or 2, 4 and 8

Good work, Namu.

 

[shaon0]

Hi all,

I have a slight problem with my hw and I need a bit of help.
Obviously its got to do with Polynomials.

The question is as follows

Solve the equation x^3 - 14x^2 + 56x - 64=0 if the roots are in geometric progression.

And can somebody tell me how to use superscript on this forum?

Thanks to whoever

Is this question from Yr 12 Cambridge 3unit?

 

[Fortian09]

I dont think so...
I'm in yr 11

 

[Fortian09]

I have another question....

find the remainder when x^3+5x^2-7x-6 is divided by (x-1). What number must be added to x^3+5x^2-7x-6 to make the result divisible by (x-1)?

 

[Aerath]

Using long division, you should get a remainder of -7. Therefore, add 7 to make it divisible by one.
Edit: The more I read the question, the more I've convinced myself I've misunderstood the question. (Well, the second bit anyway).

 

[vds700]

I have another question....

find the remainder when x^3+5x^2-7x-6 is divided by (x-1). What number must be added to x^3+5x^2-7x-6 to make the result divisible by (x-1)?

Using the remainder theorem,

R = P(1)
=1 + 5 - 7 -6
=-7

For P(x) to be divisible by (x - 1), R must be 0, so add 7 to P(x)

And you asked about superscripts, this is how u do it:

For subscript x {sub}x{/sub}

For superscript x, {sup}x{/sup} NOTE : Replace {} with []

 

[Aerath]

Using the remainder theorem,

R = P(1)
=1 + 5 - 7 -6
=-7

For P(x) to be divisible by (x - 1), R must be 0, so add 7 to P(x)

Haha, that works too. I totally forgot about Remainder theorem.

 

[lyounamu]

Using the remainder theorem,

R = P(1)
=1 + 5 - 7 -6
=-7

For P(x) to be divisible by (x - 1), R must be 0, so add 7 to P(x)

And you asked about superscripts, this is how u do it:

For subscript x {sub}x{/sub}

For superscript x, {sup}x{/sup} NOTE : Replace {} with []

Great & efficient and hence impressive!

 

[Fortian09]

ahh tyty vds700
and thanks for the help all

 

[conics2008]

Hey buddy for these type of roots just use these

In GM Progression let alpha = a/b >> beta = a and gamma = ab
In AT Progression Series let alpha = a beta = a + d gamma = a +2d and so on.

Just use your basic knowledge of series to solve question relating to poly =}

 

[Fortian09]

Hey so where do we add the seven?

 

[lyounamu]

Hey so where do we add the seven?

x^3+5x^2-7x-6

So it can become divisible by (x-1)

 

[Fortian09]

as in because we have to make the equations remainder zero right?
so in the equation do we add the 7 to the constant?

 

[lyounamu]

as in because we have to make the equations remainder zero right?
so in the equation do we add the 7 to the constant?

Yes. And hence this makes the equation divisible to (x-1)

Check it through Remainder Theorem, you should get P(1) = 0

 

[Fortian09]

ahh ic so the equation becomes x^3+5x^2-7x+1 after adding the seven
hmm does that make the equation unbalanced in anyway?

 

[lyounamu]

ahh ic so the equation becomes x^3+5x^2-7x+1 after adding the seven
hmm does that make the equation unbalanced in anyway?

P(1) = 1^3 + 5 . 1^2 - 7 . 1 + 1 = 1 + 5 - 7 + 1 = 0
Therefore, P(x) is divisible by (x-1)