# Cubic

• Feb 12th 2006, 02:58 AM
phgao
Cubic
Another that has stumpted me... (sorri about the typos ticbol)

I have

P(x) = x^3 - 3x + 2
So i find that P(2) is a value as it = zero.

So I have:

P(x) = (x-2)(ax^2 + bx + 1) (typo)

Then I divide x-2 / x^3 - 3x + 2

But I get x^2 +2x + 1 Remainder 4!!

So what went wrong, how do you factorise the equation?! Need help!
• Feb 12th 2006, 03:07 AM
ticbol
Quote:

Originally Posted by phgao
Another that has stumpted me...

I have

P(x) = x^2 - 3x + 2
So i find that P(2) is a value as it = zero.

So I have:

P(x) = (x-2)(ax^2 + bx + 1) (typo)

Then I divide x-2 / x^2 - 3x + 2

But I get x^2 +2x + 1 Remainder 4!!

So what went wrong, how do you factorise the equation?! Need help!

You were stumped by that?

If there is somebody who will not be stumped by that as well, he must be a genius.

I thought I knew Math.

You may want to rewrite or correct the data as given.
• Feb 12th 2006, 03:09 AM
phgao
Very very sorry!!!

Yeah its all fixed now. :eek: :eek:
• Feb 12th 2006, 03:10 AM
phgao
Fixed... i think so. Hope you understand it now.

Another that has stumpted me... (sorri about the typos ticbol)

Polynumeral(x) = x^3 - 3x + 2
One value of P = 2 as when you sub it in you get 0.

So I have:

P(x) = (x-2)(ax^2 + bx + 1)

Then I divide x-2 / x^3 - 3x + 2 (by long division)

But I get x^2 +2x + 1 Remainder 4!!

Another Q I had went like this:

P(x) = x^3 + 7x^2 + 14x + 8
P(x) = (x+a)(x+b)(x+c)
abc = 8

So, I found that P(-1) is a possible answer.

Then Eventually after long division I got:

P(x) = (x+1)(x^2+6x+8)
= (x+1)(x+2)(x+4)
• Feb 12th 2006, 03:26 AM
CaptainBlack
Quote:

Originally Posted by phgao
Fixed... i think so. Hope you understand it now.

Another that has stumpted me... (sorri about the typos ticbol)

Polynumeral(x) = x^3 - 3x + 2
One value of P = 2 as when you sub it in you get 0.

If $\displaystyle a$ is a root of a polynomial $\displaystyle P(x)$ (that is $\displaystyle P(a)=0$) then $\displaystyle (x-a)$
is a factor of $\displaystyle P(x)$. Now in your case $\displaystyle P(x)=x^3-3x+2$, and so
$\displaystyle P(2)=4$ so $\displaystyle (x-2)$ is not a factor of $\displaystyle P(x)=x^3-3x+2$.

Now $\displaystyle P(1)=0$, so $\displaystyle (x-1)$ is a factor of $\displaystyle P(x)$.

Factorising gives:

$\displaystyle P(x)=x^3-3x+2=(x-1)(x^2+x-2)$$\displaystyle =(x-1)(x-1)(x+2)$.

RonL
• Feb 12th 2006, 10:05 AM
earboth
Quote:

Originally Posted by phgao
Fixed... i think so. Hope you understand it now.

Another that has stumpted me... (sorri about the typos ticbol)

Polynumeral(x) = x^3 - 3x + 2
One value of P = 2 as when you sub it in you get 0.

Hello,

I'm little puzzled, 'cause when I put 2 instaed of x I don't get zero:
$\displaystyle (2)^3-3\cdot2+2=8-6+2$

Take P(-2):

$\displaystyle (-2)^3-3\cdot(-)2+2=(-8)+6+2$

Bye
• Feb 13th 2006, 02:05 AM
phgao
Thanks!