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Let $F(x)$ be a polynomial.

If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$

and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.

What is the remainder if $F(x)$ is divided by $x^2(x-1)$?

My solution : $F(x) = (x-1)^2 P(x) + x+1$, substitute $x=1$ and get $F(1) = 2$.

Differentiate it and get $F'(x) = (x-1)^2 P'(x)+P(x)(2)(x-1)+1$ then $F'(1)=1$.

And do the same to the other equation $( F(0)=3, F'(0) = 2 )$.

Let the remainder of $F(x)$ divided by $x^2(x-1)$ be $R(x)$. $( R(x) = ax^2+bx+c)$

so $F(x) =x^2(x-1)A(x) + R(x)$

$F(0) = R(0) = 3$

$F(1) = R(1) = 2$

$F'(x) = (x)^2(x-1) A'(x)+x^2A(x)+2x(x-1)A(x) + R'(x)$

$F'(0) = R'(0) = 2$

from $R(0) = 3$ then $c=3$

from $R'(0) = 2$ then $R'(x) = 2ax+b$ then $R'(0) = b = 2$

$R(x) = ax^2+2x+3$ from $R(1) = 2$, will get $a+5 = 2 , a=-3$

so the remainder will be $-3x^2+2x+3$

but one said that the answer can be $-3x^2+2x+1$ too, so what's correct answer.

cqfd
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Natnicha Weerasin
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2 Answers2

1

Your answer is correct.

The second answer cannot be also true, as $(-3x^2 + 2x + 3) - (-3x^2 + 2x + 1) =2 $ is not divisible by $(x-1)x^2$.

Here is another elementary solution:

  • $(1):F(x) = (x-1)^2p(x) + x+1$
  • $(2):F(x) = x^2q(x) + 2x+3$
  • Setting $p(x) =x^2f(x)+ax+b$ you get

$$(x-1)^2p(x) + x+1 = x^2q(x) -2bx+ax+b+x+1 $$ $$\stackrel{(2)}{\Rightarrow} b+1 = 3, 1+a-2b = 2\Rightarrow \boxed{b=2, a= 5}$$ $$\Rightarrow p(x) =x^2f(x)+5x+2$$ Plugging this into $(1)$, you need only consider $$(x-1)^2(5x+2) + x+1 = 5x^3-8x^2+2x+3 $$ $$= 5(x^3 - x^2) \boxed{- 3x^2+2x+3}$$

trancelocation
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$\!\!\overbrace{\bmod x\!-\!1\!:}^{\large x\ \equiv\ 1}\,\ \ f\equiv \overbrace{1\!+\!x}^{\Large 2} \equiv \overbrace{2x\!+\!3\! +\! a x^2}^{\Large 2\ +\ 3\ +\ a\ \ \ }\!\iff\! a=-3\!\iff\! f\equiv\overbrace{ 2x\!+\!3\!-\!3x^2}^{\large\rm your\ answer\ \checkmark}\!\!\pmod{\!x^2(x\!-\!1)}$

Update You mention $\!\bmod\!$ is unknown, so we translate the above into more elementary language. $\ f = 2x+3 + ax^2\ $ and $\,1+x\,$ have equal remainders when dividied by $x-1\,$ iff $\,a=-3,\ $ which is easily verified since a polynomial $\, g(x)\, $ has remainder $g(1)$ when divided by $\,x\!-\!1\,$ by the Polynomial Remainder Theorem. Said more simply and explicitly, with full details, we have:

$ f = \color{#0a0}{\,x\!+\!1 + q\cdot (x\!-\!1)^2}\ $ by hypothesis, and also by hypothesis we have

$ f = 2x\!+\!3 + \color{#c00}g\cdot x^2.\, $ Substitute $\,\color{#c00}{g = a + (x\!-\!1)\,h}\ $ (by dividing $\,g\,$ by $x\!-\!1)\,$ to get

$ f = 2x\!+\!3 + \color{#c00}a\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h}\, =\, \color{#0a0}{x\!+\!1 + q\cdot (x\!-\!1)^2} $

Eval at $\,x=1\:\Rightarrow\: 2+3+\color{#c00}{a+0}\: =\: \color{#0a0}{1+1 +0}\ $ so $\ \color{#c00}{a = -3},\, $ which substituted above yields

$ f = 2x\!+\!3 \color{#c00}{-3}\cdot x^2 + x^2\color{#c00}{(x\!-\!1)\,h},\, $ which agrees with your result.

Bill Dubuque
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