You were stumped by that?Originally Posted by phgao
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.
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!
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)