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
Hello DetanonSuppose the other factor is $\displaystyle (rx+s)$. Then:

$\displaystyle (x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p$Comparing coefficients:

$\displaystyle x^3: r = 1$So:$\displaystyle (x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p$Comparing coefficients:$\displaystyle x^2: (s+b) = (2b-1) \Rightarrow s = b-1$ (1)So, from (1) and (2):

$\displaystyle x: -2+sb = 0 \Rightarrow sb = 2$ (2)

$\displaystyle b(b-1)=2$Constant term:

$\displaystyle \Rightarrow b^2-b-2=0$

$\displaystyle \Rightarrow (b-2)(b+1)=0$

$\displaystyle \Rightarrow b = 2, -1$

$\displaystyle \Rightarrow s = 1, -2$

$\displaystyle p = 2s$Grandad

$\displaystyle \Rightarrow p = 2, -4$

- 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 $\displaystyle

x^3: r

$

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 $\displaystyle (rx + s)$ for some values of $\displaystyle r$ and $\displaystyle s$.

Quote:

and what does $\displaystyle

x^3: r

$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

$\displaystyle (x^2+bx-2)(rx+s)$and then note that the result has to be identically equal to

$\displaystyle x^3 +(2b-1)x^2 -p$(Can you do that?)

In this way you get the equations for $\displaystyle r, s, b$ and $\displaystyle p$ 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 $\displaystyle x^2 + bx - 2$ is a factor then it should be obvious, after a little thought, that the other factor has to be $\displaystyle x + \frac{p}{2}$. This is the only way that you can get the $\displaystyle x^3$ term and the constant term of $\displaystyle -p$.

So $\displaystyle x^3 + (2b - 1) x^2 - p = (x^2 + bx - 2)\left(x + \frac{p}{2}\right)$. Expand this out. You require the coefficient of $\displaystyle x^2$ to be $\displaystyle 2b - 1$ and the coeffcient of $\displaystyle x$ to be zero:

$\displaystyle \frac{p}{2} + b = 2b - 1 \Rightarrow p = 2b - 2$ .... (1)

$\displaystyle \frac{bp}{2} - 2 = 0 \Rightarrow bp = 4$ .... (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