We have $P (x)$ $P(x)$ an polynomial that meets the conditions:rest of dividing $P (x)$ $P(x)$ with $x - 1$ $x-1$ is 3 and $(x - 1) P (x) + x P (x + 2) = 1$ $(x-1)P(x)+xP(x+2)=1$ .How to find rest of dividing $P (x)$ $P(x)$ with $g = x^{2} - 4 x + 3$ $g=x^2-4x+3$ ?

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Answer 1 · 2017-06-12T09:56:23

$- x + 4$ $-x+4$

$P (x) = Q (x) g (x) + r (x) = Q (x) g (x) + a x + b$ $P(x)=Q(x)g(x)+r(x) = Q(x)g(x) + ax+b$

and $g (x) = (x - 1) (x - 3)$ $g(x)=(x-1)(x-3)$

$P (1) = a + b = 3$ $P(1)=a+b= 3$ and
$P (3) = 3 a + b$ $P(3)=3a+b$

but

$(1 - 1) P (1) + 1 P (3) = 1 \Rightarrow P (3) = 1$ $(1-1)P(1)+1P(3)=1 rArr P(3) = 1$ then

${(a+b=3),(3a+b=1):}$

Solving

$a=-1,b=4$ and the division remainder is

$r(x)=-x+4$

We have P(x)P(x)P(x) an polynomial that meets the conditions:rest of dividing P(x)P(x)P(x) with x-1x−1x-1 is 3 and (x-1)P(x)+xP(x+2)=1(x−1)P(x)+xP(x+2)=1(x-1)P(x)+xP(x+2)=1.How to find rest of dividing P(x)P(x)P(x) with g=x^2-4x+3g=x2−4x+3g=x^2-4x+3?