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Finding value of expression with roots of a given polynomial.

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2

1

$begingroup$

If $p(x) = x^3-3x^2+2x+5$ and $p(a)=p(b)=p(c)=0$, what is the value of $(2-a)(2-b)(2-c)$?

At first glance, it seemed to me that I needed to find the roots of the given polynomial. But this polynomial cannot be factorized using factors of the constant term. Is there any other method to find the values of a, b and c ?

polynomials

share|cite|improve this question

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Vieta's formulae?
  $endgroup$
  – gt6989b
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know Vieta's formula. What is it?
  $endgroup$
  – SPat04
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Vieta's formulas relate polynomial coefficients to roots
  $endgroup$
  – J. W. Tanner
  8 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

If $p(x) = x^3-3x^2+2x+5$ and $p(a)=p(b)=p(c)=0$, what is the value of $(2-a)(2-b)(2-c)$?

At first glance, it seemed to me that I needed to find the roots of the given polynomial. But this polynomial cannot be factorized using factors of the constant term. Is there any other method to find the values of a, b and c ?

polynomials

share|cite|improve this question

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Vieta's formulae?
  $endgroup$
  – gt6989b
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know Vieta's formula. What is it?
  $endgroup$
  – SPat04
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Vieta's formulas relate polynomial coefficients to roots
  $endgroup$
  – J. W. Tanner
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $p(x) = x^3-3x^2+2x+5$ and $p(a)=p(b)=p(c)=0$, what is the value of $(2-a)(2-b)(2-c)$?

At first glance, it seemed to me that I needed to find the roots of the given polynomial. But this polynomial cannot be factorized using factors of the constant term. Is there any other method to find the values of a, b and c ?

polynomials

share|cite|improve this question

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

gt6989b

37.5k22557

edited 8 hours ago

gt6989b

37.5k22557

asked 8 hours ago

SPat04SPat04

112

New contributor

SPat04 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

SPat04SPat04

112

3 Answers
3

active

oldest

votes

4

$begingroup$

Here is an approach assuming there are no other roots of $p(x)$ except $a,b,c$. By Vieta's formulae, the product of the roots is the constant terms of the polynomial, so you can compute the constant term of $p(2-x)$ and that will be your answer.

---

UPDATE

@ajotatxe's answer is much better.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

gt6989bgt6989b

37.5k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gt6968b I'll delete the comments in order to reduce the "noise"
  $endgroup$
  – PierreCarre
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

$$p(x)=(x-a)(x-b)(x-c)$$

(Assuming that $p$ has no more roots, real or complex).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, much better than mine, very clever, should be the accepted answer.
  $endgroup$
  – gt6989b
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

If $a,b,c$ are distinct, you know that $p(x)=(x-a)(x-b)(x-c)$ and so the value we want to compute is simply $(2-a)(2-b)(2-c) = p(2)$, which we know to be equal to 5. In this case, two of these numbers will be complex conjugates.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I understand the problem now.
  $endgroup$
  – SPat04
  8 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Here is an approach assuming there are no other roots of $p(x)$ except $a,b,c$. By Vieta's formulae, the product of the roots is the constant terms of the polynomial, so you can compute the constant term of $p(2-x)$ and that will be your answer.

---

UPDATE

@ajotatxe's answer is much better.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

gt6989bgt6989b

37.5k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gt6968b I'll delete the comments in order to reduce the "noise"
  $endgroup$
  – PierreCarre
  6 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Here is an approach assuming there are no other roots of $p(x)$ except $a,b,c$. By Vieta's formulae, the product of the roots is the constant terms of the polynomial, so you can compute the constant term of $p(2-x)$ and that will be your answer.

---

UPDATE

@ajotatxe's answer is much better.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

gt6989bgt6989b

37.5k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gt6968b I'll delete the comments in order to reduce the "noise"
  $endgroup$
  – PierreCarre
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Here is an approach assuming there are no other roots of $p(x)$ except $a,b,c$. By Vieta's formulae, the product of the roots is the constant terms of the polynomial, so you can compute the constant term of $p(2-x)$ and that will be your answer.

---

UPDATE

@ajotatxe's answer is much better.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

gt6989bgt6989b

37.5k22557

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

gt6989bgt6989b

37.5k22557

answered 8 hours ago

gt6989bgt6989b

37.5k22557

answered 8 hours ago

gt6989bgt6989b

37.5k22557

2

$begingroup$

Hint:

$$p(x)=(x-a)(x-b)(x-c)$$

(Assuming that $p$ has no more roots, real or complex).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, much better than mine, very clever, should be the accepted answer.
  $endgroup$
  – gt6989b
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

$$p(x)=(x-a)(x-b)(x-c)$$

(Assuming that $p$ has no more roots, real or complex).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1, much better than mine, very clever, should be the accepted answer.
  $endgroup$
  – gt6989b
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

$$p(x)=(x-a)(x-b)(x-c)$$

(Assuming that $p$ has no more roots, real or complex).

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

answered 8 hours ago

ajotatxeajotatxe

55.6k24291

1

$begingroup$

If $a,b,c$ are distinct, you know that $p(x)=(x-a)(x-b)(x-c)$ and so the value we want to compute is simply $(2-a)(2-b)(2-c) = p(2)$, which we know to be equal to 5. In this case, two of these numbers will be complex conjugates.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I understand the problem now.
  $endgroup$
  – SPat04
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

If $a,b,c$ are distinct, you know that $p(x)=(x-a)(x-b)(x-c)$ and so the value we want to compute is simply $(2-a)(2-b)(2-c) = p(2)$, which we know to be equal to 5. In this case, two of these numbers will be complex conjugates.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. I understand the problem now.
  $endgroup$
  – SPat04
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $a,b,c$ are distinct, you know that $p(x)=(x-a)(x-b)(x-c)$ and so the value we want to compute is simply $(2-a)(2-b)(2-c) = p(2)$, which we know to be equal to 5. In this case, two of these numbers will be complex conjugates.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

answered 8 hours ago

PierreCarrePierreCarre

3,3631216

answered 8 hours ago

PierreCarrePierreCarre

3,3631216