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Remainder theorem Polynomial

  • Thread starter terryds
  • Start date
  • #1
392
13

Homework Statement


[/B]
Polynomial f(x) is divisible by ##x^2-1##. If f(x) is divided by ##x^3-x##, then the remainder is...

A. ##(x^2-x)f(-1)##
B. ##(x-x^2)f(-1)##
C. ##(x^2-1)f(0)##
D. ##(1-x^2)f(0)##
E. ##(x^2+x)f(1)##

Homework Equations


Remainder theorem

The Attempt at a Solution


[/B]
f(x) is divisible by ##x^2-1## which means

##f(x) = (x^2-1) H(x)+0 \\
f(x) = (x+1)(x-1) H(x) + 0 \\
f(1) = 0 \\
f(-1) = 0##

f(x) is divided by ##x^3-x## which means

##f(x) = (x^3-x) H(x) + (px+q) \\
f(x) = x (x^2-1) H(x) + px + q \\
f(x) = x(x+1)(x-1) H(x)+ px + q \\
\\
f(1) = p + q = 0 \\
f(-1) = -p+q = 0 \\
f(0) = q##

And, I got p = 0, and q = 0 which means no remainder for the division.
But, the options is very confusing.
Please help
 

Answers and Replies

  • #2
33,305
4,998

Homework Statement


[/B]
Polynomial f(x) is divisible by ##x^2-1##. If f(x) is divided by ##x^3-x##, then the remainder is...

A. ##(x^2-x)f(-1)##
B. ##(x-x^2)f(-1)##
C. ##(x^2-1)f(0)##
D. ##(1-x^2)f(0)##
E. ##(x^2+x)f(1)##

Homework Equations


Remainder theorem

The Attempt at a Solution


[/B]
f(x) is divisible by ##x^2-1## which means

##f(x) = (x^2-1) H(x)+0 \\
f(x) = (x+1)(x-1) H(x) + 0 \\
f(1) = 0 \\
f(-1) = 0##
Since it is given that f(x) is divisible by ##x^2 - 1##, then you can say that ##f(x) = (x^2 - 1)H(x)## for some polynomial H. You don't need to include the "+ 0".
terryds said:
f(x) is divided by ##x^3-x## which means

##f(x) = (x^3-x) H(x) + (px+q) \\
f(x) = x (x^2-1) H(x) + px + q \\
f(x) = x(x+1)(x-1) H(x)+ px + q \\
\\
f(1) = p + q = 0 \\
f(-1) = -p+q = 0 \\
f(0) = q##
No.
From the earlier work, you have that ##f(x) = (x^2 - 1)H(x)##, so you wouldn't have the equation you have above with ##x^3 - x## in it. Besides this, it can't be true that ##f(x) = (x^2 - 1)H(x)## AND ##f(x) = (x^3-x) H(x)## plus some other terms.

Going back to the first equation I wrote, ##f(x) = (x^2 - 1)H(x)##, if f(x) is divided by ##x^3 - x##, which is equal to ##x(x^2 - 1)##, then everything boils down to whether H(x) is divisible by x or not. Note that saying "if f(x) is divided by ##x^3 - x##", that's not the same as saying "f(x) is divisible by ##x^3 - x##.

Investigate these two cases:
1. If H(x) is divisible by x, what can you say about the terms that make up H(x)? Remember, f(x) is a polynomial, so H(x) must also be a polynomial.
2. If H(x) is not divisible by x, what can you say about the terms that make up H(x)?


terryds said:
And, I got p = 0, and q = 0 which means no remainder for the division.
But, the options is very confusing.
Please help
 
  • #3
392
13
Since it is given that f(x) is divisible by ##x^2 - 1##, then you can say that ##f(x) = (x^2 - 1)H(x)## for some polynomial H. You don't need to include the "+ 0".
No.
From the earlier work, you have that ##f(x) = (x^2 - 1)H(x)##, so you wouldn't have the equation you have above with ##x^3 - x## in it. Besides this, it can't be true that ##f(x) = (x^2 - 1)H(x)## AND ##f(x) = (x^3-x) H(x)## plus some other terms.

Going back to the first equation I wrote, ##f(x) = (x^2 - 1)H(x)##, if f(x) is divided by ##x^3 - x##, which is equal to ##x(x^2 - 1)##, then everything boils down to whether H(x) is divisible by x or not. Note that saying "if f(x) is divided by ##x^3 - x##", that's not the same as saying "f(x) is divisible by ##x^3 - x##.
I put it in wrong way...
I mean...
##f(x) = (x^2-1) H(x)
f(x) = (x+1)(x-1) H(x) + 0 \\
f(1) = 0 \\
f(-1) = 0##

##f(x) = (x^3-x) T(x) + (px+q) \\
f(x) = x (x^2-1) T(x) + px + q \\
f(x) = x(x+1)(x-1) T(x)+ px + q \\
\\
f(1) = p + q = 0 \\
f(-1) = -p+q = 0 \\
f(0) = q##

I know I shouldn't have put H(x) in both equations.. So, I make it different now... H(x) for division by ##x^2 - 1## and T(x) for division by ##x^3-x##..
And I know that it is stated that f(x) is divided by ##x^3-x##.
But, by plugging x=0, x=1, and x=-1, the term 'x(x+1)(x-1) T(x)' will cancel out, and I got two equations relating p and q. And, I found out that p and q are zero.


Investigate these two cases:
1. If H(x) is divisible by x, what can you say about the terms that make up H(x)? Remember, f(x) is a polynomial, so H(x) must also be a polynomial.
2. If H(x) is not divisible by x, what can you say about the terms that make up H(x)?
1. If H(x) is divisible by x, it means that there are no constant in H(x).
2. If H(x) is not divisible by x, it means that there is a constant in H(x).
 
  • #4
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4,998
I put it in wrong way...
I mean...
##f(x) = (x^2-1) H(x)
f(x) = (x+1)(x-1) H(x) + 0 \\
f(1) = 0 \\
f(-1) = 0##
Your LaTeX is a little screwy because you omitted a \\ above
The first line above should really be two lines, like so
##f(x) = (x^2-1) H(x) \\
f(x) = (x+1)(x-1) H(x) + 0 \\

Again, there's no point in adding the 0 term in the last line. It's not wrong, but it's not needed.
terryds said:
1. If H(x) is divisible by x, it means that there are no constant in H(x).
2. If H(x) is not divisible by x, it means that there is a constant in H(x).
I'm not sure that writing ##f(x) = (x^3-x) T(x) + (px+q)## is helpful. To my way of thinking, this is more helpful: ##f(x) = (x^2-1) H(x)##. If f(x) is divided by ##x^3 - x##, what you get on the right is ##\frac{H(x)}{x}##. This is the reasoning behind my hint with the two cases. If H(x) is divisible by x, there is no constant term, so f(x) would be divisible by ##x^3 - x##, hence the remainder is zero.

As you have already said, (and related to the fact that ##f(x) = (x^2-1) H(x)##), clearly, f(1) = f(-1) = 0. That would tend to eliminate answers A, B, and E.
 
  • #5
ehild
Homework Helper
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##f(x) = (x^3-x) T(x) + (px+q) ##
I agree with @Mark44 that writing f(x) in this form is not really helpful.
x3-x=x(x2-1).
f(x)=(x2-1)H(x) and H(x) is also a polynomial, that you have to divide by x when you divide the original function by x3-x. You can write H(x)=xQ(x) + q. So the original function is f(x)= (x2-1)(xQ(x) + q ) What is the remainder if you divide it by (x2-1)x?
 
  • #6
392
13
Your LaTeX is a little screwy because you omitted a \\ above
The first line above should really be two lines, like so
##f(x) = (x^2-1) H(x) \\
f(x) = (x+1)(x-1) H(x) + 0 \\##

Again, there's no point in adding the 0 term in the last line. It's not wrong, but it's not needed.


I'm not sure that writing ##f(x) = (x^3-x) T(x) + (px+q)## is helpful. To my way of thinking, this is more helpful: ##f(x) = (x^2-1) H(x)##. If f(x) is divided by ##x^3 - x##, what you get on the right is ##\frac{H(x)}{x}##. This is the reasoning behind my hint with the two cases. If H(x) is divisible by x, there is no constant term, so f(x) would be divisible by ##x^3 - x##, hence the remainder is zero.

As you have already said, (and related to the fact that ##f(x) = (x^2-1) H(x)##), clearly, f(1) = f(-1) = 0. That would tend to eliminate answers A, B, and E.
Alright, so this is it
##f(x) = x(x^2-1) \frac{H(x)}{x}##

But, I don't know if H(x) is divisible or not since f(x) is unknown.

I agree with @Mark44 that writing f(x) in this form is not really helpful.
x3-x=x(x2-1).
f(x)=(x2-1)H(x) and H(x) is also a polynomial, that you have to divide by x when you divide the original function by x3-x. You can write H(x)=xQ(x) + q.
##f(x) = (x^2-1)H(x) \\
H(x) = f(x)/(x^2-1) \\##

Since f(x) is unknown, I can't find out what H(x) is.
So, I don't know if H(x) is divisible by x, or not.
So the original function is f(x)= (x2-1)(xQ(x) + q ) What is the remainder if you divide it by (x2-1)x?
So, it will be

##
f(x)= (x)(x^2-1)(Q(x) + q/x ) \\
f(x) = x(x^2-1)Q(x) + (x^2-1)q##

But, what is q? I can't find out what q is
 
Last edited:
  • #7
ehild
Homework Helper
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##
f(x)= (x)(x^2-1)(Q(x) + q/x ) ##


But, what is q? I can't find out what q is
It will be
##
f(x)= (x^2-1)(xQ(x) + q ) ##
What is the remainder in terms of q? What do you get if x=0?
 
  • #8
392
13
It will be
##
f(x)= (x^2-1)(xQ(x) + q ) ##
What is the remainder in terms of q? What do you get if x=0?
f(0) = -q
So, q = -f(0)

Alright, so, the remainder is D. ##(1-x^2)f(0)##, right??
 
  • #9
ehild
Homework Helper
15,427
1,827
yes.
 
  • #10
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But, I don't know if H(x) is divisible or not since f(x) is unknown.
But H(x) either is or is not divisible by x. That was my reason for suggesting that you investigate the two cases back in post #2.
 

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