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When $P(x+4)$ is divided by $P(x)$, the remainder is $3x+m$. When $P(x)$ is divided by $P(x+4)$, the remainder is $nx-6$. Compute $m-n$

Here I wrote down the equations as follows

$$P(x+4) = P(x)Q_1(x)+3x+m$$

$$P(x) = P(x+4)Q_2(x) + nx-6$$

However, these will not give me anything useful. Could you assist me with what I'm missing?

Regards

Bobtrollsten
  • 161

2 Answers2

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Since $P(x+4)$ and $P(x)$ are of the same degree, the quotients $Q_{1,2}(x)$ must be constant. In fact, from the identical leading terms, we see that $Q_{1,2}(x)=1$. So $$3x+m=P(x)-P(x+4) $$ $$nx-6=P(x+4)-P(x) $$ and we conclude $m=6$ and $n=-3$.

Hagen von Eitzen
  • 374,180
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firstly, we write :

$m=P(x+4)-P(x)Q_1(x)-3x,$

$n=[P(x)-P(x+4)Q_2(x)+6]/x=(-3)[P(x)-P(x+4)Q_2(x)+6]/[m-P(x+4)+P(x)Q_1(x)]$

pick $n=\pm{3}$ by irreducible and improve above formula to :

$P(x)(Q_1(x)-1)+P(x+4)(Q_2(x)-1)=6-m,$$P(x)(Q_1(x)+1)-P(x+4)(Q_2(x)+1)=-6-m$

also note that:

$P(x)Q_1(x)-P(x)=P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6,$$P(x)Q_1(x)+P(x)=P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6$

then we can get:

$P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6+P(x+4)(Q_2(x)-1)=6-m$

$P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6-P(x+4)(Q_2(x)+1)=-6-m$

therefore, $m=\pm{6},n=\mp{3}$