1. ## Polynomial

Find the quotient and remainder when $P(x)=2x^4+3x^3-x^2+5x-6$ is divided by $x^2+x-2$.

Quotient= $2x^2+x+2$
Remainder= $5x-2$

If yes, then...

When $x=-1$,

P(-1)=-13
D(-1)=-2
Q(-1)=3
R(-1)=-7

(-13)/(-2)
=6 remain 1
≠3 remain -7

When $x=0$,

P(0)=-6
D(0)=-2
Q(0)=2
R(0)=-2

(-6)/(-2)
=3 remain 0
≠2 remain -2

When $x= 1$,

P(1)=3
D(1)=0
Q(1)=5
R(1)=3

(3)/(0)
=undefined
≠5 remain 3

When $x=2$,

P(2)=56
D(2)=4
Q(2)=12
R(2)=8

(56)/(4)
=14 remain 0
≠12 remain 8

Why???

2. Find the quotient and remainder when P(x)=2x^4+3x^3-x^2+5x-6 is divided by x^2+x-2.

Quotient=2x^2+x+2
Remainder=5x-2

yes

When x=0,

P(0)=-6
D(0)=-2
Q(0)=2
R(0)=-2

(-6)/(-2)
=3 remain 0
≠2 remain -2
But these are equal. Think about what you mean when you say "2 remain -2". You mean $2+ \frac {-2}{-2}$. This is the same as 3.

=6 remain 1
≠3 remain -7
6 remain 1 is wrong. $6 * (-2) + 1 = -11 \not = -13$.
It should be 6 remain -1 or 7 remain 1, both of which are the same as 3 remain -7.

When x= 1,

P(1)=3
D(1)=0
Q(1)=5
R(1)=3

(3)/(0)
=undefined
≠5 remain 3
You have here $\frac {3}{0} = \frac {3}{0} +5$. It is entirely understandable to find this weird, but when you deal with infinities you get this kind of stuff.

I think the main issue here is understanding what the quotient and remainder actually are. What you discovered at the very start was that
$
\frac {2x^4+3x^3-x^2+5x-6}{x^2+x-2} = 2x^2+x+2+\frac {5x-2}{x^2+x-2}$

This is not contradicted by any substitution for x although things get weird if the divisor is 0.

3. Find the quotient and remainder when $P(x)=2x^4+3x^3-x^2+5x-6$ is divided by $x^2+x-2$.

Quotient= $2x^2+x+2$
Remainder= $5x-2$

If yes, then...

Let,
Quotient=Q(x)= $2x^2+x+2$
Remainder=R(x)= $5x-2$
Divisor=D(x)= $x^2+x-2$

When $x=-1$,

P(-1)=-13
D(-1)=-2
Q(-1)=3
R(-1)=-7

$\frac {P(-1)}{D(-1)}$
= $\frac {-13}{-2}$
=6 $\frac {1}{2}$
=6 remain 1
≠Q(-1) remain R(-1)
≠3 remain -7

When $x=0$,

P(0)=-6
D(0)=-2
Q(0)=2
R(0)=-2

$\frac {P(0)}{D(0)}$
= $\frac {-6}{-2}$
=3
=3 remain 0
≠Q(0) remain R(0)
≠2 remain -2

When $x= 1$,

P(1)=3
D(1)=0
Q(1)=5
R(1)=3

$\frac {P(1)}{D(1)}$
= $\frac {3}{0}$
=undefined
≠Q(1) remain R(1)
≠5 remain 3

When $x=2$,

P(2)=56
D(2)=4
Q(2)=12
R(2)=8

$\frac {P(2)}{D(2)}$
= $\frac {56}{4}$
=14
=14 remain 0
≠Q(2) remain R(2)
≠12 remain 8

Why???

I still don't understand what you talking about...

4. do you know polynomial long division or synthetic division? if this remainder theorem stuff is confusing you, maybe you should try that approach

5. I know this, but the long division like not correct sometimes just like the example I gave upside.

6. Originally Posted by SengNee
Find the quotient and remainder when $P(x)=2x^4+3x^3-x^2+5x-6$ is divided by $x^2+x-2$.

Quotient= $2x^2+x+2$
Remainder= $5x-2$

If yes, then...

Let,
Quotient=Q(x)= $2x^2+x+2$
Remainder=R(x)= $5x-2$
Divisor=D(x)= $x^2+x-2$

When $x=-1$,

P(-1)=-13
D(-1)=-2
Q(-1)=3
R(-1)=-7

$\frac {P(-1)}{D(-1)}$
= $\frac {-13}{-2}$
=6 $\frac {1}{2}$
=6 remain 1
≠Q(-1) remain R(-1)
≠3 remain -7
I think you are missing the main idea. What does the phrase "Q(-1) remain R(-1)" mean?

We need to know that
$P(x) = Q(x)D(x) + R(x)$
From this, as long as $D(x) \neq 0$
$\frac{P(-1)}{D(-1)} = Q(-1) + \frac{R(-1)}{D(-1)}$
Now you have already calculated $\frac{P(-1)}{D(-1)} = \frac{13}2$
so let us do the remaining,
$Q(-1) + \frac{R(-1)}{D(-1)} = 3 + \frac{-7}{-2} = \frac{6+7}{2} = \frac{13}{2}$

I still don't understand what you talking about...
I think none of us here are understanding what you are saying!

I know this, but the long division like not correct sometimes just like the example I gave upside.
This generally does not happen in maths. If a method is proven, then it will work ALWAYS!