1. Quadratic 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?

2. Hello Detanon
Originally Posted by Detanon
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?
Suppose 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)

$\displaystyle x: -2+sb = 0 \Rightarrow sb = 2$ (2)
So, from (1) and (2):
$\displaystyle b(b-1)=2$

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

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

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

$\displaystyle \Rightarrow s = 1, -2$
Constant term:
$\displaystyle p = 2s$

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

3. 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?

4. Hello Detanon
Originally Posted by Detanon
Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
Because 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 $\displaystyle s$.
and what does $\displaystyle x^3: r$mean?
It doesn't mean anything. You have taken it out of context. Read the whole thing. I am comparing coefficients. (Do you know what thay means?) First I am looking at the coefficient of $\displaystyle x^3$. Later on, I look at the coefficients of $\displaystyle x^2, x$ and the constant. I have used a colon (':') each time to show which coefficient I am looking at.
Is there a different method to do it?
No, not a method that is as sensible and straightforward as the one I have shown you.

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

5. I still dont get it.. Is there another method where dividing is involved? I thnik thats the one I get.

6. 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.)

7. Originally Posted by Detanon
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?
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.

For checking purposes:

Spoiler:
b = 2, -1 and p = 2, -4

8. Where do you get

Constant term:

p = 2s

from??

9. Hello Detanon
Originally Posted by Detanon
Where do you get

Constant term:

p = 2s

from??
The constant term when we expand:
$\displaystyle (x^2+bx-2)(rx+s)$ is $\displaystyle -2s$, and this is equal to the constant term in $\displaystyle x^3 +(2b-1)x^2 -p$

Hence $\displaystyle p = 2s$