If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.

Im not sure what method to use to solve this. do i divide them?

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- Dec 8th 2009, 12:27 PMDetanonQuadratic factors of a cubic polynomial
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.

Im not sure what method to use to solve this. do i divide them? - Dec 9th 2009, 08:11 AMGrandad
- Dec 9th 2009, 09:26 AMDetanon
Sorry, but I really dont get that.

I dont get why you suppose the other facotr is (rx+s)

and what does

mean?

Is there a different method to do it? - Dec 9th 2009, 10:20 AMGrandad
Hello DetanonBecause when one factor of a cubic is a quadratic, the other is linear; i.e. it is of the form for some values of and .

Quote:

and what does mean?

Quote:

Is there a different method to do it?

Read it through again - it's really very simple.

Grandad - Dec 9th 2009, 10:41 AMDetanon
I still dont get it.. Is there another method where dividing is involved? I thnik thats the one I get.

- Dec 9th 2009, 01:38 PMGrandad
Hello Detanon

I really don't know what your problem is here. Multiply out the brackets from

In this way you get the equations for and that I have shown you. I repeat: if you understand the basic rules of algebra, then trust me, this is really very simple. (And certainly much easier than trying to do an algebraic long division.)

Grandad

- Dec 9th 2009, 04:21 PMmr fantastic
If is a factor then it should be obvious, after a little thought, that the other factor has to be . This is the only way that you can get the term and the constant term of .

So . Expand this out. You require the coefficient of to be and the coeffcient of to be zero:

.... (1)

.... (2)

Solve equations (1) and (2) simultaneously for b and p.

You need to go back now and have a good hard think about all that's been said about this question.

For checking purposes:

__Spoiler__: - Dec 10th 2009, 11:49 AMDetanon
Where do you get

Constant term:

p = 2s

from?? - Dec 10th 2009, 12:01 PMGrandad