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Remainder of polynomial division when divisor is square without calculus

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3

2

$begingroup$

The problem is given as follows:

Let $p(x) = x^2004 - x^1901 - 50$. What is the remainder of the
division of $p(x)$ by $(x-1)^2$.

The solution is straightforward when using the derivative of $p(x)$. However, considering that I stumbled upon this problem in a high school textbook, I'm assuming that an elegant solution exists without the use of calculus?

polynomials

share|cite|improve this question

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

$endgroup$

add a comment | 

3

2

$begingroup$

The problem is given as follows:

Let $p(x) = x^2004 - x^1901 - 50$. What is the remainder of the
division of $p(x)$ by $(x-1)^2$.

The solution is straightforward when using the derivative of $p(x)$. However, considering that I stumbled upon this problem in a high school textbook, I'm assuming that an elegant solution exists without the use of calculus?

polynomials

share|cite|improve this question

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

$endgroup$

add a comment | 

$begingroup$

The problem is given as follows:

Let $p(x) = x^2004 - x^1901 - 50$. What is the remainder of the
division of $p(x)$ by $(x-1)^2$.

The solution is straightforward when using the derivative of $p(x)$. However, considering that I stumbled upon this problem in a high school textbook, I'm assuming that an elegant solution exists without the use of calculus?

polynomials

share|cite|improve this question

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

share|cite|improve this question

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

edited Mar 14 at 15:15

Maria Mazur

47.9k1260120

asked Mar 14 at 15:03

koralakraljkoralakralj

163

asked Mar 14 at 15:03

koralakraljkoralakralj

163

3 Answers
3

active

oldest

votes

2

$begingroup$

Let $$
x^2004-x^1901-50=q(x)(x-1)^2+ax+b.
$$ Translating by $-1$, we have
$$beginalign*
q(x+1)x^2+ax+(a+b)&=(x+1)^2004-(x+1)^1901-50\&=sum_j=0^2004binom2004jx^j-sum_j=0^1901binom1901jx^j-50
\&= text(higher order terms)+103x-50.
endalign*$$ This gives $a=103$ and $b=-153$.

share|cite|improve this answer

answered Mar 14 at 15:10

SongSong

18.5k21550

$endgroup$

add a comment | 

2

$begingroup$

Write: $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax+b$$

Setting $x=1$ we get $$-50 = a+bimplies b=-50-a$$

so $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax-a-50$$

so $$x^1901(x^103-1) = (x-1)^2k(x)+a(x-1)$$

so $$ x^1901(x^102+ x^101+...+x+1)= (x-1)k(x)+a$$

and finally putting $x=1$ again we get $$103 =aimplies r(x)=103x-153$$

share|cite|improve this answer

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

$endgroup$

add a comment | 

2

$begingroup$

$x!=!1,Rightarrow, p = -50 + (x!-!1)color#c00g. $ $, g = dfracp+50x-1=x^1901,dfracx^103!-!1x!-!1 = x^1901(x^102+cdots+x+1),$

so $,x!=!1,Rightarrow,color#c00g = 103+(x!-!1)h $ thus $ p = -50+103(x!-!1)+(x!-!1)^2h$

share|cite|improve this answer

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We twice applied the Remainder Theorem $,f(x) = f(1) = (x-1)g(x),$ for some polynomial $,g(x) $ See also the Double Root Test in algebraic form.
  $endgroup$
  – Bill Dubuque
  Mar 14 at 15:38
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Let $$
x^2004-x^1901-50=q(x)(x-1)^2+ax+b.
$$ Translating by $-1$, we have
$$beginalign*
q(x+1)x^2+ax+(a+b)&=(x+1)^2004-(x+1)^1901-50\&=sum_j=0^2004binom2004jx^j-sum_j=0^1901binom1901jx^j-50
\&= text(higher order terms)+103x-50.
endalign*$$ This gives $a=103$ and $b=-153$.

share|cite|improve this answer

answered Mar 14 at 15:10

SongSong

18.5k21550

$endgroup$

add a comment | 

2

$begingroup$

Let $$
x^2004-x^1901-50=q(x)(x-1)^2+ax+b.
$$ Translating by $-1$, we have
$$beginalign*
q(x+1)x^2+ax+(a+b)&=(x+1)^2004-(x+1)^1901-50\&=sum_j=0^2004binom2004jx^j-sum_j=0^1901binom1901jx^j-50
\&= text(higher order terms)+103x-50.
endalign*$$ This gives $a=103$ and $b=-153$.

share|cite|improve this answer

answered Mar 14 at 15:10

SongSong

18.5k21550

$endgroup$

add a comment | 

$begingroup$

Let $$
x^2004-x^1901-50=q(x)(x-1)^2+ax+b.
$$ Translating by $-1$, we have
$$beginalign*
q(x+1)x^2+ax+(a+b)&=(x+1)^2004-(x+1)^1901-50\&=sum_j=0^2004binom2004jx^j-sum_j=0^1901binom1901jx^j-50
\&= text(higher order terms)+103x-50.
endalign*$$ This gives $a=103$ and $b=-153$.

share|cite|improve this answer

answered Mar 14 at 15:10

SongSong

18.5k21550

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:10

SongSong

18.5k21550

answered Mar 14 at 15:10

SongSong

18.5k21550

answered Mar 14 at 15:10

SongSong

18.5k21550

2

$begingroup$

Write: $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax+b$$

Setting $x=1$ we get $$-50 = a+bimplies b=-50-a$$

so $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax-a-50$$

so $$x^1901(x^103-1) = (x-1)^2k(x)+a(x-1)$$

so $$ x^1901(x^102+ x^101+...+x+1)= (x-1)k(x)+a$$

and finally putting $x=1$ again we get $$103 =aimplies r(x)=103x-153$$

share|cite|improve this answer

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

$endgroup$

add a comment | 

2

$begingroup$

Write: $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax+b$$

Setting $x=1$ we get $$-50 = a+bimplies b=-50-a$$

so $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax-a-50$$

so $$x^1901(x^103-1) = (x-1)^2k(x)+a(x-1)$$

so $$ x^1901(x^102+ x^101+...+x+1)= (x-1)k(x)+a$$

and finally putting $x=1$ again we get $$103 =aimplies r(x)=103x-153$$

share|cite|improve this answer

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

$endgroup$

add a comment | 

$begingroup$

Write: $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax+b$$

Setting $x=1$ we get $$-50 = a+bimplies b=-50-a$$

so $$x^2004 - x^1901 - 50 = (x-1)^2k(x)+ax-a-50$$

so $$x^1901(x^103-1) = (x-1)^2k(x)+a(x-1)$$

so $$ x^1901(x^102+ x^101+...+x+1)= (x-1)k(x)+a$$

and finally putting $x=1$ again we get $$103 =aimplies r(x)=103x-153$$

share|cite|improve this answer

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

$endgroup$

share|cite|improve this answer

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

edited Mar 14 at 15:20

J. W. Tanner

3,5831320

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

answered Mar 14 at 15:13

Maria MazurMaria Mazur

47.9k1260120

2

$begingroup$

$x!=!1,Rightarrow, p = -50 + (x!-!1)color#c00g. $ $, g = dfracp+50x-1=x^1901,dfracx^103!-!1x!-!1 = x^1901(x^102+cdots+x+1),$

so $,x!=!1,Rightarrow,color#c00g = 103+(x!-!1)h $ thus $ p = -50+103(x!-!1)+(x!-!1)^2h$

share|cite|improve this answer

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We twice applied the Remainder Theorem $,f(x) = f(1) = (x-1)g(x),$ for some polynomial $,g(x) $ See also the Double Root Test in algebraic form.
  $endgroup$
  – Bill Dubuque
  Mar 14 at 15:38
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$x!=!1,Rightarrow, p = -50 + (x!-!1)color#c00g. $ $, g = dfracp+50x-1=x^1901,dfracx^103!-!1x!-!1 = x^1901(x^102+cdots+x+1),$

so $,x!=!1,Rightarrow,color#c00g = 103+(x!-!1)h $ thus $ p = -50+103(x!-!1)+(x!-!1)^2h$

share|cite|improve this answer

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We twice applied the Remainder Theorem $,f(x) = f(1) = (x-1)g(x),$ for some polynomial $,g(x) $ See also the Double Root Test in algebraic form.
  $endgroup$
  – Bill Dubuque
  Mar 14 at 15:38
  
  

  
  
  
  

add a comment | 

$begingroup$

$x!=!1,Rightarrow, p = -50 + (x!-!1)color#c00g. $ $, g = dfracp+50x-1=x^1901,dfracx^103!-!1x!-!1 = x^1901(x^102+cdots+x+1),$

so $,x!=!1,Rightarrow,color#c00g = 103+(x!-!1)h $ thus $ p = -50+103(x!-!1)+(x!-!1)^2h$

share|cite|improve this answer

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654

answered Mar 14 at 15:36

Bill DubuqueBill Dubuque

212k29195654