What's the remainder?
What is the remainder when x 2 0 0 6 − x 2 0 0 5 + ( x + 1 ) 2 is divided by x 2 − 1 ?
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2 solutions
using remainder theorm, its 2 or 4
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I think remainder theorem may not apply, unless you rewrite the function as f ( x 2 ) , then find f ( 1 ) . Note that the divisor is second degree the remainder can be first degree as in this case ( x + 3 ) and zeroth degree without x just a constant.
I have redone the solution. Hope you like it.
The remainder will have degree 1 since the degree of the remainder is always less than that of the divisor.
Hence it has the form a x + b .
x 2 0 0 6 − x 2 0 0 5 + ( x + 1 ) 2 = ( x 2 − 1 ) . q ( x ) + ( a x + b )
x 2 0 0 6 − x 2 0 0 5 + ( x + 1 ) 2 = ( x + 1 ) ( x − 1 ) . q ( x ) + ( a x + b )
Now substitute x by 1 and we get,
1 − 1 + 4 = a + b
a + b = 4 ....................(1)
Now if we replace x by − 1 ,we get the following,
− a + b = 2 ....................(2)
From equations (1) and (2) we get a= 1 and b= 3
Hence the remainder i.e a x + b = x + 3
x 2 − 1 x 2 0 0 6 − x 2 0 0 5 + ( x + 1 ) 2 = ( x − 1 ) ( x + 1 ) x 2 0 0 5 ( x − 1 ) + ( x + 1 ) 2 = x + 1 x 2 0 0 5 + x − 1 x + 1 = x + 1 x 2 0 0 5 + 1 − 1 + x − 1 x − 1 + 2 = x + 1 ( x + 1 ) ( x 2 0 0 4 − x 2 0 0 3 + x 2 0 0 2 − . . . + 1 ) − 1 + x − 1 x − 1 + 2 = x 2 0 0 4 − x 2 0 0 3 + x 2 0 0 2 − . . . + 1 − x + 1 1 + 1 + x − 1 2 = x 2 0 0 4 − x 2 0 0 3 + x 2 0 0 2 − . . . + 2 + x 2 − 1 − x + 1 + 2 x + 2 = x 2 0 0 4 − x 2 0 0 3 + x 2 0 0 2 − . . . + 2 + x 2 − 1 x + 3
Therefore, the remainder is x + 3 .