Quadratic factors of a cubic polynomial

**Detanon** · December 8th 2009, 12:27 PM

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?

**Grandad** · December 9th 2009, 08:11 AM

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 $(rx+s)$ . Then:

$(x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p$

Comparing coefficients:

$x^3: r = 1$

So:

$(x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p$

Comparing coefficients:

$x^2: (s+b) = (2b-1) \Rightarrow s = b-1$ (1)

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

So, from (1) and (2):

$b(b-1)=2$

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

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

$\Rightarrow b = 2, -1$

$\Rightarrow s = 1, -2$

Constant term:

$p = 2s$

$\Rightarrow p = 2, -4$

Grandad

**Detanon** · December 9th 2009, 09:26 AM

Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
and what does $<br /> x^3: r <br />$
mean?

Is there a different method to do it?

**Grandad** · December 9th 2009, 10:20 AM

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

and what does $<br /> x^3: r <br />$ 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 $x^3$ . Later on, I look at the coefficients of $x^2, x$ 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.

Grandad

**Detanon** · December 9th 2009, 10:41 AM

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

**Grandad** · December 9th 2009, 01:38 PM

Hello Detanon

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

$(x^2+bx-2)(rx+s)$

and then note that the result has to be identically equal to

$x^3 +(2b-1)x^2 -p$

(Can you do that?)

In this way you get the equations for $r, s, b$ and $p$ 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

**mr fantastic** · December 9th 2009, 04:21 PM

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

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

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

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

**Detanon** · December 10th 2009, 11:49 AM

Where do you get

Constant term:

p = 2s

from??

**Grandad** · December 10th 2009, 12:01 PM

Hello Detanon

Originally Posted by Detanon

Where do you get

Constant term:

p = 2s

from??

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

Hence $p = 2s$

Grandad

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