P(x) is a polynomial of degree …

{tex}\text {Let when } p(x) \text { is divided by } (x-2)(x-3) \text { it leaves a remainder } (ax-b) \text { quotient } g(x),{/tex}

{tex}\therefore p(x) = (x-2)(x-3)g(x)+(ax+b){/tex}            {tex}...(1){/tex}

{tex}\text {Now, when } p(x) \text {is divided by } (x-2) \text {it leaves a remainder 1,}{/tex}

{tex}\therefore \text {putting value } x=2 \text { in eqn. (1), we get}{/tex}

{tex}p(2) = (2-2)(2-3)g(2)+[a(2)+b]{/tex}

{tex}\Rightarrow 1 = 0+2a+b{/tex}

{tex}\Rightarrow 2a+b=1{/tex}         {tex}...(2){/tex}

{tex}\text {And, when } p(x) \text { is divided by } (x-3) \text { it leaves a remainder 3},{/tex}

{tex}\therefore \text { putting value } x = 3 \text { in eqn. (1), we get}{/tex}

{tex}p(3) = (3-2)(3-3)g(3)+[a(3)+b]{/tex}

{tex}\Rightarrow 3 = 0 + 3a +b{/tex}

{tex}\Rightarrow 3a+b=3{/tex}         {tex}...(3){/tex}

{tex}\text {On solving, eqn. (2) and (3), we get}{/tex}

{tex}a=2 \text { and } b = -3{/tex}

{tex}\text {Putting values of } a \text { and } b \text { in eqn. (1), we get}{/tex}

{tex}p(x) = (x-2)(x-3)+(2x-3){/tex}

{tex}\text {Thus, the required remainder is }(2x-3).{/tex}