Stubborn polynomial

P(x)=(x-1)q+2.........(1)
P(x)=(x-2)w+5.........(2)
P(x)=(x-1)(x-2)t+ax+b......(3). where q,w,t are quotients
Putting x=1 in 1 we get
P(1)=2,
Putting x=2 in 2
P(2)=5
Putting x=1 in 3
P(1)=a+b
Putting x=2 in 3
P(2)=2a+b...
Solving all this we get a=3

From the question , we have: $\frac{p(x)}{x-1}=f(x)+\frac{2}{x-1}$ $(1)$ $\frac{p(x)}{x-2}=h(x)+\frac{5}{x-2}$ $(2)$
Substracting $(1)$ from $(2)$ ,we obtain: $\frac{p(x)}{(x-1)(x-2)}=h(x)-f(x)+\frac{3x-1}{(x-1)(x-2)} p(x)=(h(x)-f(x))(x-1)(x-2)+3x-1$ ,therefore the remainder is $3x-1$ , where we obtain $a=3$

Respectively, let $q(x), r(x)$ and $s(x)$ be the quotient when $p(x)$ is divided by $(x-1)$ , $(x-2)$ , and $(x-1)(x-2)$ . Hence we can rewrite $p(x)$ as below:

$p(x)=(x-1)q(x)+2...(1)$

$p(x)=(x-2)r(x)+5...(2)$

$p(x)=(x-1)(x-2)s(x)+ax+b...(3)$

From (1) we have $p(1)=(1-1)q(x)+2=0+2=2$

Similarly, from (2) we have $p(2)=5$

Thus $p(1)\ne p(2)$ . But from (3) we have $p(1)=p(2)=ax+b$

Indirectly, it is stated that $p(1)\ne p(2)$ and $p(1) = p(2)$ at once

This is impossible unless $q(x)=r(x)=s(x)=0$ , such that it results the new conclution: $p(x)=ax+b$ .

Therefore, now we have $p(1)=a+b=2$ and $p(2)=2a+b=5$ . And $p(2)-p(1)=a=\boxed {3}$

p(x) = (x-1)(x-2) f(x) + (ax+b) where f(x) is quotient and ax+b is the remainder. On applying the first two conditions, i.e., when x=1 p(x)=2 and x=2 p(x)=5; a+b = 2 and 2a+b=5. Solving the above two equations, we get a=3 and b=-1

2 = a + b , 5 = 2a + b, So, with substitution method we have b = -1 and a = 3. Answer : 3

From the given statement, we can express that

P(x) = (x-1) Q(x) + 2 => P(1) = 2 P(x) = (x-2) K(x) + 5 => P(2) = 5

Finally: P(x) = (x-1)(x-2) H(x) + ax + b Replace x by 1 and 2, we have a + b = 2 2a + b = 5 => a = 3

It follows easily from Chinese Remainder Theorem .We get: $ax+b=3x-2$ giving $a=\boxed{3}$ .

p(x)=(x-1)Q1(x)+2 p(x)=(x-2)Q2(x)+5 p(x)=(x-1)(x-2)Q3(x)+ax+b

p(1)=2 p(1)=a+b p(2)=5 p(2)=2a+b

Hence, a+b=2 -a-b=-2 2a-a-b+b=-2+5 a=3, b=-1
2a+b=5 2a+b=5 a+b=2, 3+b=2, b=2-3=-1
Hence, the remainder will be ax+b=3x-1

Ans: 3x-1

by Division Algorithm,

Dividend=Divisor*quotient+remainder

So,p(x)=(x-1)*q(x)+2

putting x=1,

p(1)=0+2=2 -(i)

Also, p(x)=(x-2)*q'(x)+5

so, p(2)=5 -(ii)

Now, p(x)=(x-1)(x-2)*Q(x)+(ax+b)

putting x=1 and x=2

p(1)=a+b and p(2)=2a+b

so, a+b=2 and 2a+b=5 [from (i) and (ii)]

solving the two, we get a =3

We first consder the case in which $(x-1)(x-2)=(x-1)$ or $x=3$ . In this case, we have that $2=3a+b$ . Now, let's consider the case in which $(x-1)(x-2)=2(x-1) \Rightarrow x=4$ . Note that if $n \equiv m \pmod{p}$ , then $n \equiv m \pmod{2p}$ or $n \equiv m+p \pmod{2p}$ . We thus get that $2=4a+b$ or $5=4a+b$ . If we use the first equation, $a$ solves to zero, and that's not the case, so we have to use the second equation. This gives us $\boxed{a=3}$ .

Let p(x)=(x-1)(x-2)Q(x)+ax+b dividend=divisor multiplied by quotient plus remainder. Given p(1)=2=a+b and p(2)=5=2a+b by the remainder theorem. Solving, a=3

Let P(x) be

(x-1)(x-2).Q(x) + r(x) where r(x) is Ax + B

so, P(1) = A+B = 2

and, P(2)= 2A+B = 5

On solving we get, A= 3 and B=-1

Hence the answer is $\boxed{3}$

f(x) = (x-1).g(x) + 2 ; f(1) = 2

f(x) = (x-2).h(x) + 5 ;f(2) = 5

f(x) = (x-1).(x-2).s(x) + ax + b

f(1) = a + b = 2 . ......(1)

f(2) = 2a + b = 5 ........(2)

then, by eliminating equation (1) and (2), we get that a is 3