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It is given that if cubic function $f(x)$ is divided by $(x^2+4)$ and $(x+3)$, the remainders will be $7x-20$ and $-2$ respectively. Also it is said show that $f(x)$ is $x^3 + 6x^2 + 11x +4$. I tried but can i find the $a,b,c$, and $d$ without knowing them in from the start?

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  • $\begingroup$ Consider using Chinese Reaminder Theorem in $\mathbb{Z}[x]$. $\endgroup$ – Zongxiang Yi Jun 19 at 7:03
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$f(x) = (x^2 + 4)(ax +b) +7x - 20$

$f(-3)=-2 = 13(-3a+b)-41 $

therefore $b= 3a+3$

so, $f(x)= (x^2+4)(ax +3a+3)+ 7x-20 $ (non-zero $a$)

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A simple method (based on the chinese remainder theorem) goes as follows:

To solve $f \operatorname{mod} p = u$ and $f \operatorname{mod} q = v$ (where p and q are coprime polynomials):

put $\ \bar{p} = p^{-1}\operatorname{mod} q$ and $\ \bar{q} = q^{-1}\operatorname{mod} p$ then $f = vp\bar{p}+uq\bar{q}$. Indeed we have $ f = upp^{-1}\operatorname{mod} q = u$ and $ f = vqq^{-1}\operatorname{mod} p = v$.

In this case we have for $p = x^2+4$, $q = x+3$, $u = 7x-20$ and $v=-2$ that $\bar{p} = -\frac{x}{39}$ and $\bar{q} = -\frac{x}{13}+\frac{3}{13}$ so that $f = -\frac{19}{39}x^3+\frac{20}{13}x^2+\frac{197}{39}x-\frac{180}{13}$. But this only one of the many solutions, the all differ by a multiple of $pq$. In this case $f + \frac{58}{39}\ pq\ $ yields the given polynomial.

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Solving in pari/gp:

? chinese(Mod(7*x-20,x^2+4),Mod(-2,x+3))
%1 = Mod(3*x^2 + 7*x - 8, x^3 + 3*x^2 + 4*x + 12)
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? Mod(a*x^3+b*x^2+c*x+d,(x^2+4)*(x+3))
%2 = Mod((-3*a + b)*x^2 + (-4*a + c)*x + (-12*a + d), x^3 + 3*x^2 + 4*x + 12)

I.e. $3=-3a+b$, $7=-4a+c$, $-8=-12a+d$.

And $\quad a=1\quad\Longrightarrow\quad b=6,c=11,d=4$.

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