1. ## algebra

Polynomial p(x) and a divisor d(x) are given. Find the quotient q(x) and the remainder r(x) when p(x) is divided by d(x).

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

b,
p(x)=x^4+x^2+2
d(x)=x^2+x+1

2. Originally Posted by gumi
Polynomial p(x) and a divisor d(x) are given. Find the quotient q(x) and the remainder r(x) when p(x) is divided by d(x).

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

b,
p(x)=x^4+x^2+2
d(x)=x^2+x+1
This is just long division of polynomials. I'll walk you through the first one.
We need to find the correct multiple of $2x - 5$ to get to the level of $12x^3$. That multiple is $6x^2$.

So we multiply $6x^2 \cdot (2x-5) = 12x^2 - 30x^2$ and subtract that from $12x^3 - 40x^2 + 11x + 39$, which leaves us with $-10x^2 + 11x + 39$. Now, how many times will $2x - 5$ go into this? $-5x$.

So $-5x \cdot (2x - 5) = -10x^2 + 25x$ and we subtract out $-10x^2 + 25x$, leaving us with $-14x + 39$. Finally, we see that we need a multiple of -7 for the last step:

$-7(2x - 5) = -14x + 35$ and we subtract this from $-14x + 39$ to yield 4.

$q(x) = 6x^2 - 5x - 7$

$r(x) = \frac{4}{2x - 5}$

3. 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...

4. now i get it thanks