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Remainder theorem for polynomials (JUEE 1990)

Theorem for Dividing PolynomialsFinding two unknowns in two quadratic polynomials with only knowing the divisorsRemainder of the polynomialWhat is the remainder useful for when dividing a polynomial?A certain polynomial P(x) , $xin R$ when divided by $x-a, x-b,x-c$ leaves the remainders a,b,c respectively…Application of the Chinese Remainder Theorem for polynomialsLet $a in Bbb Z$ such that $gcd(9a^25+10:280)=35$. Find the remainder of $a$ when divided by 70.chinese remainder theorem for polynomialsRemainder of polynomial by product of 2 polynomials

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7

1

$begingroup$

Suppose the polynomial $P(x)$ with integer coefficients satisfies the following conditions:

(A) If $P(x)$ is divided by $x^2 − 4x + 3$, the remainder is $65x − 68$.

(B) If $P(x)$ is divided by $x^2 + 6x − 7$, the remainder is $−5x + a$.

Then we know that $a =$?

I am struggling with this first question from the 1990 Japanese University Entrance Examination. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the polynomial remainder theorem but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?

So for some $g(x)$ and $h(x)$ we have:
$$P(x) = g(x)(x^2-4x+3) + (65x-68),\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$
which looks to have more unknowns than equations. How should I proceed from here?

polynomials chinese-remainder-theorem

share|cite|improve this question

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x-1$ divides both moduli so evaluating both equations at $x = 1$ yields $, -3 = P(1) = -5 + a.$ $ $
  $endgroup$
  – Bill Dubuque
  Apr 17 at 18:16
  
  

  
  
  
  

add a comment
 | 

7

1

$begingroup$

Suppose the polynomial $P(x)$ with integer coefficients satisfies the following conditions:

(A) If $P(x)$ is divided by $x^2 − 4x + 3$, the remainder is $65x − 68$.

(B) If $P(x)$ is divided by $x^2 + 6x − 7$, the remainder is $−5x + a$.

Then we know that $a =$?

I am struggling with this first question from the 1990 Japanese University Entrance Examination. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the polynomial remainder theorem but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?

So for some $g(x)$ and $h(x)$ we have:
$$P(x) = g(x)(x^2-4x+3) + (65x-68),\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$
which looks to have more unknowns than equations. How should I proceed from here?

polynomials chinese-remainder-theorem

share|cite|improve this question

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: $x-1$ divides both moduli so evaluating both equations at $x = 1$ yields $, -3 = P(1) = -5 + a.$ $ $
  $endgroup$
  – Bill Dubuque
  Apr 17 at 18:16
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Suppose the polynomial $P(x)$ with integer coefficients satisfies the following conditions:

(A) If $P(x)$ is divided by $x^2 − 4x + 3$, the remainder is $65x − 68$.

(B) If $P(x)$ is divided by $x^2 + 6x − 7$, the remainder is $−5x + a$.

Then we know that $a =$?

I am struggling with this first question from the 1990 Japanese University Entrance Examination. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the polynomial remainder theorem but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?

So for some $g(x)$ and $h(x)$ we have:
$$P(x) = g(x)(x^2-4x+3) + (65x-68),\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$
which looks to have more unknowns than equations. How should I proceed from here?

polynomials chinese-remainder-theorem

share|cite|improve this question

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

$endgroup$

share|cite|improve this question

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

share|cite|improve this question

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

edited Apr 17 at 8:00

Saad

21.2k99 gold badges2626 silver badges5555 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

asked Apr 17 at 7:52

peekaypeekay

13844 bronze badges

5 Answers
5

active

oldest

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5

$begingroup$

Hint:

If you know that the remainder of the division of some polynomial $Q$ by, say, $x^2-5x+4$ is $7x-8$ you can find some values of $Q$ by substituting $x$ by the zeros of the divisor.

Indeed, the zeros of $x^2-5x+4$ are $x=1$ and $x=4$. So you can find what is $Q(4)$.

$$Q(x)=h(x)(x^2-5x+4)+7x-8$$
$$Q(4)=h(4)(4^2-5cdot 4+4)+7cdot 4-8=h(4)cdot 0+28-8=20$$

share|cite|improve this answer

edited Apr 17 at 8:00

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

7

$begingroup$

You need only $P(1)$, since

• 
  $x^2-4x+3 = (x-1)(x-3)$ and


• $x^2+6x-7 = (x-1)(x+7)$

Hence,

• $P(1) = 65-68 = -3$


• $P(1) = -5+a Rightarrow a=2$

share|cite|improve this answer

answered Apr 17 at 7:58

trancelocationtrancelocation

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$endgroup$

add a comment
 | 

2

$begingroup$

We notied that: $x^2-4x+3=(x-1)(x-3)$ and $x^2+6x-7=(x-1)(x+7)$.

So, we have: $P(1)=65*1-68=-5*(1)+a$.
$implies -3=-5+aiff a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Since $P(x)=g(x)(x^2-4x+3)+(65x-68)$, for $x=1$ you get $P(1)=-3$. This imply that
$P(1)=h(1)cdot 0+(-5+a)$, i.e. $a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number. $x^2-4x+3=x-(-x^2+5x-3)$.

Easier to note the second divisor is $10x-10$ more than the first so $(10x-10)y-5x+a= 65x-68$ so y = 7 produces $70x-70-5x+a=65x-(70-a)$ so $a=2$ works.

share|cite|improve this answer

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

$endgroup$

add a comment
 | 

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5

$begingroup$

Hint:

If you know that the remainder of the division of some polynomial $Q$ by, say, $x^2-5x+4$ is $7x-8$ you can find some values of $Q$ by substituting $x$ by the zeros of the divisor.

Indeed, the zeros of $x^2-5x+4$ are $x=1$ and $x=4$. So you can find what is $Q(4)$.

$$Q(x)=h(x)(x^2-5x+4)+7x-8$$
$$Q(4)=h(4)(4^2-5cdot 4+4)+7cdot 4-8=h(4)cdot 0+28-8=20$$

share|cite|improve this answer

edited Apr 17 at 8:00

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Hint:

If you know that the remainder of the division of some polynomial $Q$ by, say, $x^2-5x+4$ is $7x-8$ you can find some values of $Q$ by substituting $x$ by the zeros of the divisor.

Indeed, the zeros of $x^2-5x+4$ are $x=1$ and $x=4$. So you can find what is $Q(4)$.

$$Q(x)=h(x)(x^2-5x+4)+7x-8$$
$$Q(4)=h(4)(4^2-5cdot 4+4)+7cdot 4-8=h(4)cdot 0+28-8=20$$

share|cite|improve this answer

edited Apr 17 at 8:00

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Hint:

If you know that the remainder of the division of some polynomial $Q$ by, say, $x^2-5x+4$ is $7x-8$ you can find some values of $Q$ by substituting $x$ by the zeros of the divisor.

Indeed, the zeros of $x^2-5x+4$ are $x=1$ and $x=4$. So you can find what is $Q(4)$.

$$Q(x)=h(x)(x^2-5x+4)+7x-8$$
$$Q(4)=h(4)(4^2-5cdot 4+4)+7cdot 4-8=h(4)cdot 0+28-8=20$$

share|cite|improve this answer

edited Apr 17 at 8:00

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

$endgroup$

share|cite|improve this answer

edited Apr 17 at 8:00

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

answered Apr 17 at 7:55

ajotatxeajotatxe

57.7k22 gold badges4545 silver badges9393 bronze badges

7

$begingroup$

You need only $P(1)$, since

• 
  $x^2-4x+3 = (x-1)(x-3)$ and


• $x^2+6x-7 = (x-1)(x+7)$

Hence,

• $P(1) = 65-68 = -3$


• $P(1) = -5+a Rightarrow a=2$

share|cite|improve this answer

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

7

$begingroup$

You need only $P(1)$, since

• 
  $x^2-4x+3 = (x-1)(x-3)$ and


• $x^2+6x-7 = (x-1)(x+7)$

Hence,

• $P(1) = 65-68 = -3$


• $P(1) = -5+a Rightarrow a=2$

share|cite|improve this answer

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You need only $P(1)$, since

• 
  $x^2-4x+3 = (x-1)(x-3)$ and


• $x^2+6x-7 = (x-1)(x+7)$

Hence,

• $P(1) = 65-68 = -3$


• $P(1) = -5+a Rightarrow a=2$

share|cite|improve this answer

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

answered Apr 17 at 7:58

trancelocationtrancelocation

17.3k11 gold badge1111 silver badges3030 bronze badges

2

$begingroup$

We notied that: $x^2-4x+3=(x-1)(x-3)$ and $x^2+6x-7=(x-1)(x+7)$.

So, we have: $P(1)=65*1-68=-5*(1)+a$.
$implies -3=-5+aiff a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

We notied that: $x^2-4x+3=(x-1)(x-3)$ and $x^2+6x-7=(x-1)(x+7)$.

So, we have: $P(1)=65*1-68=-5*(1)+a$.
$implies -3=-5+aiff a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

We notied that: $x^2-4x+3=(x-1)(x-3)$ and $x^2+6x-7=(x-1)(x+7)$.

So, we have: $P(1)=65*1-68=-5*(1)+a$.
$implies -3=-5+aiff a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

answered Apr 17 at 8:00

know dontknow dont

14711 silver badge1313 bronze badges

2

$begingroup$

Since $P(x)=g(x)(x^2-4x+3)+(65x-68)$, for $x=1$ you get $P(1)=-3$. This imply that
$P(1)=h(1)cdot 0+(-5+a)$, i.e. $a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Since $P(x)=g(x)(x^2-4x+3)+(65x-68)$, for $x=1$ you get $P(1)=-3$. This imply that
$P(1)=h(1)cdot 0+(-5+a)$, i.e. $a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Since $P(x)=g(x)(x^2-4x+3)+(65x-68)$, for $x=1$ you get $P(1)=-3$. This imply that
$P(1)=h(1)cdot 0+(-5+a)$, i.e. $a=2$.

share|cite|improve this answer

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

answered Apr 17 at 8:00

Viki 183Viki 183

6633 bronze badges

1

$begingroup$

You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number. $x^2-4x+3=x-(-x^2+5x-3)$.

Easier to note the second divisor is $10x-10$ more than the first so $(10x-10)y-5x+a= 65x-68$ so y = 7 produces $70x-70-5x+a=65x-(70-a)$ so $a=2$ works.

share|cite|improve this answer

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number. $x^2-4x+3=x-(-x^2+5x-3)$.

Easier to note the second divisor is $10x-10$ more than the first so $(10x-10)y-5x+a= 65x-68$ so y = 7 produces $70x-70-5x+a=65x-(70-a)$ so $a=2$ works.

share|cite|improve this answer

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number. $x^2-4x+3=x-(-x^2+5x-3)$.

Easier to note the second divisor is $10x-10$ more than the first so $(10x-10)y-5x+a= 65x-68$ so y = 7 produces $70x-70-5x+a=65x-(70-a)$ so $a=2$ works.

share|cite|improve this answer

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges

answered Apr 17 at 16:27

Roddy MacPheeRoddy MacPhee

75222 gold badges22 silver badges2525 bronze badges