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Improper algebric expression by remainder theorem

Hi
need help
I try to understand myself but I can't
F (x)=q(x) X divisor + remsinder
In one question they are like that (Ax^2+bx+c)(x-2)+Cx+Do
In another question it is like that (Ax+b)(x+1) +C

I can't figure out how we know that or how to arrange it

Reply 1

Qer

Can anyone explain me with this example

Reply 2

RDKGames

Study Forum Helper

Original post by Qer

Can anyone explain me with this example

There is a theorem (or something to do with Newton - too tired so I cant remember) that states if

f_n(x)

is a polynomial of degree

n

,and

g_m(x)

is a polynomial of degree

m

, AND if

n \geq m

, then

\frac{f_n(x)}{g_m(x)}

can be rewritten in the form

\displaystyle h_{n-m}(x)+\frac{i_{m'}(x)}{g_m(x)}

where

h_{n-m}(x)

is a polynomial of degree

n-m

and

i_{m-1}

is a polynomial of degree

m' < m

So with your example, you have

\displaystyle \frac{x^4+3x^2-4}{x^2+1}=Ax^2+Bx+C+\frac{Dx+E}{x^2+1}

Notice how this agreed with what I said above; since the numerator is of degree 4 and denom is of degree 2, the

Ax^2+Bx+C

is a poly of degree 2, which is 4-2 as expected, and

Dx+E

is a poly of order less than

x^2+1

So multiplying both sides by

x^2+1

gives

x^4+3x^2-4=(Ax^2+Bx+C)(x^2+1)+Dx+E

Anyway, you don't need to know where the form comes from for the exam, but that's where it comes from.

(edited 6 years ago)

Reply 3

Muttley79

Original post by Qer

Can anyone explain me with this example

These are two different methods for the same question.

If you divide x^4 by x^2 then you get a quadratic expression, yes?

So the remainder must be at most linear, yes?

Are you OK so far?

Reply 4

Qer

Original post by Muttley79

These are two different methods for the same question.

If you divide x^4 by x^2 then you get a quadratic expression, yes?

So the remainder must be at most linear, yes?

Are you OK so far?

Got it
Thanks

Reply 5

Qer