# Further Maths - Roots of Polynomials

Watch
Announcements

Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 - 2i

I then go to substitute 3 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 2i -> 3 - 2i.

I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.

My workings:

Any help is appreciated, thanks

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 - 2i

I then go to substitute 3 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 2i -> 3 - 2i.

I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.

My workings:

Any help is appreciated, thanks

Last edited by BrandonS15; 6 months ago

1

reply

Report

#2

Have you tried substituting 3 2i into f(x) to check if it’s actually a root? If it isn’t, you know the mistake is in the algebra before that. If it is, you can assume 3-2i will also work and use those to find the other.

Edit: just realised this will be awkward with m and n, but I would still do it for the and - and compare answers.

Edit: just realised this will be awkward with m and n, but I would still do it for the and - and compare answers.

Last edited by Dancer2001; 6 months ago

0

reply

Report

#3

(Original post by

Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 +- 2i

I then go to substitute 3 + 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 + 2i -> 3 - 2i.

I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.

My workings:

Any help is appreciated, thanks

**BrandonS15**)Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 +- 2i

I then go to substitute 3 + 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 + 2i -> 3 - 2i.

I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.

My workings:

Any help is appreciated, thanks

0

reply

Report

#4

**BrandonS15**)

Hello, I have this question:

I have used the product of all 3 roots equalling -d/a relationship to then further solve for alpha. I end up solving a quadratic to which alpha is 3 +- 2i

I then go to substitute 3 + 2i into the 3 expressions of the roots of the cubic and I don’t get the correct answers because one of them should be the complex conjugate of 3 + 2i -> 3 - 2i.

I have checked and made sure the answers to alpha are solved correct. I don’t understand why substitution into the expressions for solved alpha doesn’t yield me the other two correct roots if given those are correct expressions for the 3 roots.

My workings:

Any help is appreciated, thanks

0

reply

Report

#5

Hey, I know it's not very helpful, but I just wanted to say:

"Well done for maintaining such a neat handwriting".

Keep it up :P

"Well done for maintaining such a neat handwriting".

Keep it up :P

0

reply

0

reply

Report

#7

0

reply

(Original post by

Roots of polynomials only come in conjugate pairs if all of the coefficents are real. I haven't done the second part of the question, but I expect you will find that at least one of m and n are not real.

**Pangol**)Roots of polynomials only come in conjugate pairs if all of the coefficents are real. I haven't done the second part of the question, but I expect you will find that at least one of m and n are not real.

0

reply

Report

#11

(Original post by

I checked the algebra and its correct in yielding alpha as 3 +- 2i however I understand that the complex roots only come as conjugate pairs if the coefficients of the polynomial are real. Therefore instead of assuming they should be conjugate pairs, I directly substituted alpha as 3 + 2i into 1/alpha and the other expression to yield the other two given roots to the cubic however, these do not match the correct answers of 3 + 2i, 3 -2i and 4. I’m not sure how to obtain the other two roots of the cubic given the expression of the roots and a correctly calculated value for alpha.

**BrandonS15**)I checked the algebra and its correct in yielding alpha as 3 +- 2i however I understand that the complex roots only come as conjugate pairs if the coefficients of the polynomial are real. Therefore instead of assuming they should be conjugate pairs, I directly substituted alpha as 3 + 2i into 1/alpha and the other expression to yield the other two given roots to the cubic however, these do not match the correct answers of 3 + 2i, 3 -2i and 4. I’m not sure how to obtain the other two roots of the cubic given the expression of the roots and a correctly calculated value for alpha.

0

reply

Report

#12

(Original post by

Are you sure those are the correct answers?

**Pangol**)Are you sure those are the correct answers?

0

reply

Report

#13

(Original post by

Are you sure those are the correct answers? Where does the question come from?

**Pangol**)Are you sure those are the correct answers? Where does the question come from?

It looks to me as if you have done the right thing based on the information given to you, but the way the question is phrased is wrong. I'm trying to see what their error is, but can't come up with anything that puts it quite right.

0

reply

(Original post by

Oh hang on, I've found it. This is one of those Edexcel end of unit tests that they put out when the A Levels were reformed. There are quite a few errors in them, and I think this question is one of them.

It looks to me as if you have done the right thing based on the information given to you, but the way the question is phrased is wrong. I'm trying to see what their error is, but can't come up with anything that puts it quite right.

**Pangol**)Oh hang on, I've found it. This is one of those Edexcel end of unit tests that they put out when the A Levels were reformed. There are quite a few errors in them, and I think this question is one of them.

It looks to me as if you have done the right thing based on the information given to you, but the way the question is phrased is wrong. I'm trying to see what their error is, but can't come up with anything that puts it quite right.

0

reply

(Original post by

I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.

**old_engineer**)I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.

0

reply

Report

#16

(Original post by

I can see them all okay

**Sir Cumference**)I can see them all okay

Perhaps this is a sign that I should be getting used to it

0

reply

Report

#17

**old_engineer**)

I think there may be typos in the question, and suspect (tentatively) that the roots are supposed to be a, 13/a and a + 13/a - 2.

0

reply

Report

#18

(Original post by

It works now that I switched to the 'Beta' version of the new page layout (enabled all the way at the bottom of the page).

Perhaps this is a sign that I should be getting used to it

**RDKGames**)It works now that I switched to the 'Beta' version of the new page layout (enabled all the way at the bottom of the page).

Perhaps this is a sign that I should be getting used to it

1

reply

Report

#19

(Original post by

I'm surprised that they haven't got rid of the old page yet - I'm sure they said they would last year!

**Sir Cumference**)I'm surprised that they haven't got rid of the old page yet - I'm sure they said they would last year!

0

reply

Report

#20

Hi, can anyone share Edexcel AS Further Maths Core Pure end of course test paper?

0

reply

X

### Quick Reply

Back

to top

to top