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 $\displaystyle 2x - 5$ to get to the level of $\displaystyle 12x^3$. That multiple is $\displaystyle 6x^2$.
So we multiply $\displaystyle 6x^2 \cdot (2x-5) = 12x^2 - 30x^2$ and subtract that from $\displaystyle 12x^3 - 40x^2 + 11x + 39$, which leaves us with $\displaystyle -10x^2 + 11x + 39$. Now, how many times will $\displaystyle 2x - 5$ go into this? $\displaystyle -5x$.
So $\displaystyle -5x \cdot (2x - 5) = -10x^2 + 25x$ and we subtract out $\displaystyle -10x^2 + 25x$, leaving us with $\displaystyle -14x + 39$. Finally, we see that we need a multiple of -7 for the last step:
$\displaystyle -7(2x - 5) = -14x + 35$ and we subtract this from $\displaystyle -14x + 39$ to yield 4.
The answer is therefore:
$\displaystyle q(x) = 6x^2 - 5x - 7$
$\displaystyle r(x) = \frac{4}{2x - 5}$
Hello,
Here is a method, it's the one I apply, it's not necessarily the one you have to use.
$\displaystyle p(x)=12x^3-40x^2+11x+39$
$\displaystyle d(x)=2x-5$
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).
$\displaystyle In \ p(x) \ : \ 12x^3$
$\displaystyle In \ d(x) \ : \ 2x$
$\displaystyle 12x^3={\color{red}6x^2} (2x)$
So I'll write :
$\displaystyle p(x)={\color{red}6x^2} (2x-{\color{magenta}5}) {\color{magenta}+5*6x^2} \ \ -40x^2+11x+39$
$\displaystyle p(x)=6x^2 (2x-5)+30x^2-40x^2+11x+39=6x^2(2x-5)-10x^2+11x+39$
Then I'll do it again, with -10x² :
$\displaystyle -10x^2=-5x (2x)$
---> $\displaystyle p(x)=6x^2 (2x-5)-5x(2x-{\color{magenta}5}){\color{magenta}-25x} \ \ +11x+39$
And continue, again and again...