If the remainders of the polynomial f(x) when divided by x – 1, x – 2 are 2, 5 then the remainder of f(x) when divided by (x-1)(x– 2) is 1) 0 2 1 - * 3) 2x-1 4) 3x-1

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( f(x)=k(x-1)+2 ) we know [ f(x)=K_{2}(x-2)+5 quad f(1)=2 ] ( f(2)=5 ) now et remaindu be ( (a x+b) rightarrow ) remcined wiube ont cess degree than divisor Ans ( 3 x-1 quad(4) ) ( f(x)=k_{3}(x-1)(x-2)+a x+b ) ( left.begin{array}{l}f(1)=a cdot(b+b=2 f(2)=2 a+b=5end{array}right] rightarrow begin{array}{l}text { solving these we get } a=3 b=-1end{array} )

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