AMaths (Partial fractions/Polynomials)

Only-Way-4-Destiny! SG

Hi all, can anyone help me with this? Or explain how do I solve this type of questions. It's the first time I came across this type.

Qn:

f(x) = g(x) . (x-1)(x-2)(x-3) . If f(x) is divided by (x-1), (x-2) and (x-3), the remainders are 4, -1 and 2 respectively. Using partial fractions or otherwise, find an expression for f(x) .

I know:

f(1) = 4

f(2) = -1

f(3) = 2

By remainder theorm. But how do I continue on?

Thanks!

wee_ws SG

Hi,

It is an unusual question.

I see that you have not used the suggestion to apply partial fractions?

Why the suggestion? Notice that f(x) / [ (x - 1)(x - 2)(x - 3) ] = g(x).

So we can write g(x) = A / (x - 1) + B / (x - 2) + C / (x - 3).

Then we use remainder theorem to solve for A, B, C.

Try it, thanks!

Cheers,
Wen Shih

Only-Way-4-Destiny! SG

Sorry but I do not quite get it. I kept doing this question for one hour already and I could not get the answer. I need help on this question still.

wee_ws SG

Hi,

I'm not sure what you do not understand of my explanation. Please be more specific. Thanks!

Cheers,
Wen Shih

SBS n SMRT SG

may i know what is the otherwise method.......

Only-Way-4-Destiny! SG

Hello Wen Shih,

" So we can write g(x) = A / (x - 1) + B / (x - 2) + C / (x - 3)."

I do not understand how to move on from this part. First time I am attempting this type of questions so I am having a lot of difficultly continuing from here. How can I apply the remainder theorm ?

Do I equate like this?

f(1) = 4

f(2) = -1

f(3) = 2

g(1) = 4

g(2) = -1

g(3) = 2

wee_ws SG

Hi,

g(x) = A / (x - 1) + B / (x - 2) + C / (x - 3)

=> f(x) / { (x - 1)(x - 2)(x - 3) } = A / (x - 1) + B / (x - 2) + C / (x - 3)

=> f(x) = A (x - 2)(x - 3) + B (x - 1)(x - 3) + C (x -1)(x - 2), by comparing numerators

Now apply the facts that f(1) = 4, f(2) = -1 and f(3) = 2 to solve A, B and C.

Thanks!

Cheers,
Wen Shih

Only-Way-4-Destiny! SG

Wow!!!! Thanks!!!! Now it seems more familiar now

wee_ws SG

Hi,

It's a tough question at sec sch level. Where did it originate from? Thanks!

Cheers,
Wen Shih

Only-Way-4-Destiny! SG

Lol. Good question. I am surprised that you found it tough for sec sch level. Anyway, my amaths teacher gives this kind of hard questions for us to do. Well, a good training before 'O' levels!

wee_ws SG

Hi,

I see. Always use the suggestion provided, to start the ball rolling.

The purpose of unseen or rather hard questions is to sieve out the more able students in the examination.

Often, I see many students struggling to solve unusual questions and neglect doing the other commonly asked questions. It is a misconception that if one is able to solve difficult questions, one will be able to tackle all questions. Do go for a wide variety of questions nevetheless.

Jiayou!

Cheers,
Wen Shih

Only-Way-4-Destiny! SG

Thanks! Good advice.

eagle SG

Nice solutions by Mr Wee :D

But Mr Wee, maybe I mental block or something, is there a way to tell if the power of f(x) is smaller than 3? If not, one can't do partial fractions with it.

thanks.

wee_ws SG

Hi,

Actually, the question should state that the highest degree is less than 3, or the question cannot be solved by the said method of partial fractions. Good observation!

Cheers,
Wen Shih

Alex31584 SG

can anyone help to solve this question?

Express (x^4 + 1) / (x^3 + 2x) in partial fraction? thank you thank you. :)

Doraemonboi SG

Hi,

It is an unusual question.

I see that you have not used the suggestion to apply partial fractions?

Why the suggestion? Notice that f(x) / [ (x - 1)(x - 2)(x - 3) ] = g(x).

So we can write g(x) = A / (x - 1) + B / (x - 2) + C / (x - 3).

Then we use remainder theorem to solve for A, B, C.

Try it, thanks!

Cheers,
Wen Shih

wa chim

wee_ws SG

can anyone help to solve this question?

Express (x^4 + 1) / (x^3 + 2x) in partial fraction? thank you thank you. :)

Hi,

First, check whether the expression is proper.

(x^4 + 1) / (x^3 + 2x) is improper, so we need to simplify further by long division.

So (x^4 + 1) / (x^3 + 2x) = x - 2 + (4x + 1) / x(x^2 + 2).

Now we just need to break (4x + 1) / x(x^2 + 2) into partial fractions

A/x + (Bx + C) / (x^2 + 2)

which is routine.

Try it, thanks!

Cheers,
Wen Shih