Question

an unknown polynomial of degree greater than two given remainders 2 and 1 when it is divided by 1 and 2 respectively. Find the remainder when it is divided by x^2-3x+2.​

any human to ans this question!?​

Answer 1

Here a polynomial, say $p(x),$ of degree greater than 2 leaves remainder 2 on division by $(x-1).$ This implies,

$\longrightarrow p(1)=2$

Also the polynomial $p(x)$ leaves remainder 1 on division by $(x-2).$ This implies,

$\longrightarrow p(2)=1$

We're asked to find the remainder obtained on dividing $p(x)$ by $x^2-3x+2=(x-1)(x-2).$

So let,

$\longrightarrow p(x)=(x-1)(x-2)\,q(x)+r(x)\quad\quad\dots(i)$

where $q(x)$ is the quotient and $r(x)$ is the remainder obtained on dividing $p(x)$ by $(x-1)(x-2).$

Since $p(x)$ has a degree greater than 2, the degree of the remainder $r(x)$ should be at most 2 for every possible $p(x).$ We take $r(x)$ with maximum possible degree, 2.

Let,

$\longrightarrow r(x)=ax^2+bx+c\quad\quad\dots(ii)$

such that (i) becomes,

$\longrightarrow p(x)=(x-1)(x-2)\,q(x)+ax^2+bx+c$

For $x=1,$

$\longrightarrow p(1)=2$

$\longrightarrow (1-1)(1-2)\,q(1)+a(1)^2+b(1)+c=2$

$\longrightarrow a+b+c=2\quad\quad\dots(1)$

For $x=2,$

$\longrightarrow p(2)=1$

$\longrightarrow (2-1)(2-2)\,q(2)+a(2)^2+b(2)+c=1$

$\longrightarrow 4a+2b+c=1\quad\quad\dots(2)$

From (1) and (2) we obtain the following (3) and (4):

$\longrightarrow3a+b=-1\quad\quad\dots(3)$

$\longrightarrow 2a-c=-3\quad\quad\dots(4)$

But from (ii),

$\longrightarrow r(x)=ax^2+bx+c$

$\longrightarrow r(x)=ax^2-3ax+2a+3ax-2a+bx+c$

$\longrightarrow r(x)=a\left(x^2-3x+2\right)+(3a+b)x-(2a-c)$

From (3) and (4),

$\longrightarrow r(x)=a\left(x^2-3x+2\right)-x+3$

Hence (i) becomes,

$\longrightarrow p(x)=(x-1)(x-2)\,q(x)+a\left(x^2-3x+2\right)-x+3$

$\longrightarrow p(x)=\left(x^2-3x+2\right)\,q(x)+a\left(x^2-3x+2\right)-x+3$

$\longrightarrow p(x)=\left(x^2-3x+2\right)\big(q(x)+a\big)+3-x$

This implies our polynomial $p(x)$ leaves remainder $\bf{3-x}$ on division by $x^2-3x+2.$

Hence $\bf{3-x}$ is the answer.