Question

did I do the evaluation correctly? pic attached

Answer 1

so after the division, we get
(x^3-x^2+5x+2) = (x^2-3x+2)(x+2) + (9x-2)
gott his ?

Answer 2

*got this ?

Answer 3

so i always have to multiply the original denominator by the answer on top of the division sign and then add the remainder?

Answer 4

DIVIDEND = DIVISOR * QUOTIENT +REMAINDER

Answer 5

thanks hartnn. Our lecturer kinda sucks at teaching us this stuff, i'll re -attempt the question and post a picture

Answer 6

good! note that we will get something like
x+2+(9x-2)/(x^2-3x+2)
apply partial fractions on only last term

Answer 7

okay you lost me there. what i did was have (Ax+B/x^2-3x+2)+(C/x+2)+(D/9x-2)

Answer 8

no...
you didn't get how i got
x+2+(9x-2)/(x^2-3x+2) ??

Answer 9

no i don't, it looks like you took the remainder and quotient and divided them by the divisor. Is the original equation = that equation?

Answer 10

\(\Large \dfrac{ (x^2-3x+2)(x+2) + (9x-2)}{(x^2-3x+2)} \\ \Large =\dfrac{(x^2-3x+2)(x+2)}{(x^2-3x+2)}+\dfrac{9x-2}{(x^2-3x+2)} \\ \Large = x+2 +\dfrac{9x-2}{(x^2-3x+2)} \)

now got it ?

Answer 11

yes I see it now, thanks

Answer 12

so in essence the original numerator = the dividend?

Answer 13

yep, in x/y
x is dividend , y is divisor

Answer 14

i am assuming you are trying partial fraction for last term :)

Answer 15

I'm doing the partial fraction from the mid year exam. In fact, i'm working through the entire paper

Answer 16

i think that is correct! :)
good work!

Answer 17

YEAH! thanks again for the help

Answer 18

welcome ^_^