Math Help Forum - View Single Post

Moo · #3 May 2nd, 2008, 02:42 PM

Hello,

Here is a method, it's the one I apply, it's not necessarily the one you have to use.

$p(x)=12x^3-40x^2+11x+39$
$d(x)=2x-5$

I see what the highest degree in p(x) is, and I find how to multiply x powered to the highest degree to get the first term of p(x).

$In \ p(x) \ : \ 12x^3$
$In \ d(x) \ : \ 2x$

$12x^3={\color{red}6x^2} (2x)$

So I'll write :

$p(x)={\color{red}6x^2} (2x-{\color{magenta}5}) {\color{magenta}+5*6x^2} \ \ -40x^2+11x+39$

$p(x)=6x^2 (2x-5)+30x^2-40x^2+11x+39=6x^2(2x-5)-10x^2+11x+39$

Then I'll do it again, with -10x² :

$-10x^2=-5x (2x)$

---> $p(x)=6x^2 (2x-5)-5x(2x-{\color{magenta}5}){\color{magenta}-25x} \ \ +11x+39$

And continue, again and again...