Question

Hi!

f(x) is a polynomial such that f(1)=8 and f(3)=16. r(x) is the remainder when f(x) is divided by (x−1)(x−3). What is r(5)?

Answer 1

I'd start by saying g(x) = (x-1)(x-3). I wanna give that expression a name.

Answer 2

Let q be the quotient. We have: f(x) = q * g(x) + r(x)

Answer 3

We at least know that: g(5) = (5-1)(5-3) = (4)(2) = 8

Answer 4

We know that f(1) = 8, and g(1) = (1-1)(1-3) = 0.
f(3) = 16, and g(3) = (3-1)(3-3) = 0

Answer 5

So we know that r(1) = 8 and r(3) = 16 regardless.

Answer 6

I kind of wonder if there really is only one solution to this.

Answer 7

Tampering with: f(x) = q * g(x) + r(x)
We get: f(x) - r(x) = q * g(x).

So the sum of the degrees of q and g(x) is equal to the max of the degrees of f(x) and r(x).

We also know that f(x) and r(x) both must at least be of degree 1, because their points showed they weren't constant.

Answer 8

well you're just on the right track.
do this
let r(x) = Ax + B
solve for A and B.

Answer 9

@shubhamsrg I actually did that, but I kinda got stuck because I'm pondering whether r(x) could be a higher degree polynomial.

Answer 10

since divisor is of 2nd degree, r(x) cant have a degree greater than that,ofcorse..

Answer 11

So we get to assume that q is constant with respect to x?

Answer 12

?

Answer 13

Suppose q(x) = x^5. That means that q(x) times g(x) is degree 7

Answer 14

you mean f(x) = x^5 ?

Answer 15

Assuming that q is constant with respect to x, i.e. degree 0 would give us an upper bound.

Answer 16

q(x) can have any degree whatsoever.

when the divisor is of degree \(r\), the degree of the remainder is \(< r\).

Here. the degree of the divisor is \(2\), hence the remainder can have a degree of at max \(1\), hence the assumption that \(r(x) = Ax + B\)

Answer 17

Why does the remainder have to have a lower degree than the divisor?

Answer 18

The Remainder Theorem:
When you divide a polynomial f(x) by x-c the remainder r will be f(c)

Answer 19

Well, it need not. But if you take the division to as low of a degree as possible, then the degree of the remainder is indeed less than that of the divisor.

Let me illustrate what I mean to say:
Suppose you want to perform \(x^5/(x^2 -1)\).

You can very well write this as \(x^5 = x^3*(x^2 -1) + x^3\)
which indicates that the quotient is \(x^3\) and the remainder is \(x^3\) which has a higher degree than the divisor.
You can very well leave the division here itself. This is not "wrong".

However note that, the remainder is of a degree equal or higher than the divisor, implying that further division is possible.
Continuing further, to write the answer of the complete division, we get
\(x^5 = (x^3+x)*(x^2-1) + x\)

At this point further division is not possible, due to the fact that the degreeof the remainder is less than the degree of the divisor.

With such questions as given, we generally assume that division is taken to the furthest extent possible so that the assumption that the degree of the remainder is less than the divisor is reasonable.

Answer 20

nice explanation!

Answer 21

@FoolAroundMath I understand what you're saying. So my next question then is whether or not there is more than one solution to the problem, supposing you didn't divide further.

Basically, does this assumption eliminate other solutions in the process of leading us to a unique solution?

Answer 22

r(x)=f(p)....where p is 1, 3(as both return a pole)... so r(x) has two values!!....

Answer 23

Further division is generally assumed otherwise, there would probably be infinite many possibilities.

And it defies what we are used to, for example, no one would say that 4 when divided by 2, leaves a remainder of 2, with quotient as 1 which is mathematically correct. (Number theorists might though ;) )

Answer 24

"And it defies what we are used to, for example, no one would say that 4 when divided by 2, leaves a remainder of 2, with quotient as 1 which is mathematically correct."

Well if you give me the dividend and divisor, I can kind of doctor the quotient and remainder to be whatever I want. So I wasn't making this connection, but I see what you are getting at.

Answer 25

Under the principle of "remainder CAN'T be higher than the divisor", I guess you end up with "remainder polynomial's degree CAN'T be higher than the divisor polynomial's degree".

Answer 26

precisely that ^