If a cubic polynomial $p(x)$ with leading co-efficient 1 is divided by $x-1$, $x-2$ and $x-3$ respectively it leaves $r(x)$ $1$, $4$ and $9$. What will be the remainder when it is divided by $x-4$? I've searched for this everywhere but I couldn't find an answer. Help would be much appreciated.
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Sounds like a bog standard polynomial interpolation problem, or a standard chinese remainder theorem problem. Surely your textbook talks about these in the very chapter it asks this problem? – Jun 18 '17 at 09:02
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Actually, it isn't in any of the textbooks or guides that I own. That's why I asked the question here. – Prakhar Nagpal Jun 18 '17 at 09:03
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Are you sure you gave us all hypothesis? There are infinitely many answers, not just one. – José Carlos Santos Jun 18 '17 at 09:04
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Do you have the leading coefficient? – CY Aries Jun 18 '17 at 09:04
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yeah, I'm pretty sure that this is all the information that my teacher gave. – Prakhar Nagpal Jun 18 '17 at 09:05
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I'm sorry whats a leading coefficient? – Prakhar Nagpal Jun 18 '17 at 09:05
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The coefficient of $x^3$ in this case. If you follow my solution, you will find that $p(x)=A(x-1)(x-2)(x-3)+x^2$. $A$ can be any constant. So, there are infinitely many answers. – CY Aries Jun 18 '17 at 09:08
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oh okay, I'll check and clarify. – Prakhar Nagpal Jun 18 '17 at 09:09
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Can you please tell us what is it that you do not understand within the answer provided by CY Aries? It seems very clear to me. And please do not write in all caps. That's the equivalent of shouting, and I don't think that we deserve that. – José Carlos Santos Jun 18 '17 at 09:31
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Sorry. It's just so frustrating. – Prakhar Nagpal Jun 18 '17 at 09:41
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Anyway, I don't understand how we can write fx as px - x2? – Prakhar Nagpal Jun 18 '17 at 09:43
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@PrakharNagpal What CY Aries did was to define $f(x)$ as $p(x)-x^2$, which is a natural choice, since $p(k)=k^2$ for each $k\in{1,2,3}$. – José Carlos Santos Jun 18 '17 at 11:04
2 Answers
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Let $f(x)=p(x)-x^2$. Then $f(x)$ is a cubic polynomial.
Note that $f(1)=p(1)-1^2=1-1=0$, $f(2)=p(2)-2^2=4-4=0$ and $f(3)=p(3)-3^2=9-9=0$
$f(x)$ has roots $1$, $2$ and $3$. So,
$$f(x)=A(x-1)(x-2)(x-3)$$
for some constant $A$.
$$p(x)=A(x-1)(x-2)(x-3)+x^2$$
When $p(x)$ is divided by $x-4$, then remainder is
$$p(4)=6A+16$$
Without other information on $p(x)$, we cannot proceed further.
CY Aries
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HINT:
Let $$f(x)=(x-1)(x-2)(x-3)(x-4)\left(\dfrac A{x-1}+\dfrac B{x-2}+\dfrac C{x-3}+\dfrac D{x-4}\right)$$
lab bhattacharjee
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