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Question

If the polynomials 2x3+ax2+3x-5 and x3+x2-4x+a leave the same remainder when divided by x-2, Find the value of a.


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Solution

Step 1: Determine the value of x

Take 2x3+ax2+3x-5=g(x)

x3+x2-4x+a=f(x).

On dividing g(x) and f(x) by x-2 gives the same remainder.

x-2 is a factor of g(x) and f(x).

x-2=0[x-2isafactor]x=2

Step 2: Equate the functions

By the remainder theorem, the remainder of f(x) when divided by x-a is f(a)

So, the remainder of f(x)when divided by x-2 is f(2)

and the remainder of g(x)when divided by x-2 is g(2)

Now, since the remainder of f(x) and g(x) when divided by x-2 is equal

,g(2)=f(2)2(2)3+a(2)2+3(2)-5=(2)3+(2)2-4(2)+a16+4a+6-5=8+4-8+a4a+17=a+43a=-13a=-133

Hence, the value of a is -133.


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