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Given the polynmial is exactly divided by $x+1$, when it is divided by $3x-1$, the remainder is $4$. The polynomial leaves remainder $hx+k$ when divided by $3x^2+2x-1$. Find $h$ and $k$.

This is the question which is confusing me.. i have done this question like this:

$p(x) = g1(x)(x+1) +0 $

$p(x) = g2(x)(3x-1) +4 $

$p(x) = g3(x)(3x+2x-1)+hx+k $ $=> p(x) = g3(x)(x+1)(3x-1) + hx+k $

now putting the value of $x$ in each $p(x) $

$p(-1) = 0 \tag{1}$

$p(1/3) = 4 \tag{2}$

$p(-1,1/3) = hx+k \tag{3}$

from equation $(2)$ and $(3)$


now putting the value of $x$ in equation $(4)$.

$h(-1)+k=4 => -h+k=4 \tag{5}$

$h(1/3)+k=4 => h+3k=12 \tag{6}$

now adding equations $(5)$ and $(6)$

$ h+3k-h+k = 4+12 =>4k=16 => k=4$

now puting the value of of $k$ in equation $(5)$


therefore my answer is $h=0,k=4$ but the answers are$ h=3, k=3$. please help me to sort out my problem

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What's the meaning of your equation (3), and the following conclusion (4)? (especially the notation $p(1,1/3)$ is rather obscure) Actually, from $p(-1)=0$ you get $(-1)h+k=0$, and from $p(1/3)=4$, $(1/3)h+k=4$. – Jean-Claude Arbaut Jun 23 '14 at 16:10
up vote 3 down vote accepted

The error is that $\,p(-1) = 0\,\Rightarrow\, h(-1) + k = \color{#c00}0,\,$ not $\,\color{#c00}4.\,$ Fixing that yields the given answer.

Remark $\ $ This is a special case of the Chinese Remainder Theorem (CRT) or, equivalently, Lagrange interpolation. Either of these methods can be applied to solve the general case.

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The error in your solution is that p(−1)=0 implies h(−1)+k=0, not 4,Right

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