Can x – 1 be the remainder on division of a polynomial p (x) by 2x + 3? Justify your answer
Answers
Answered by
326
Hi
By division algorithm:
**********************************************
Dividend = divisor × quotient +
remainder
Where,
Remainder = r ( x ) = 0 or
Degree of r ( x ) < degree of divisor
**********************************************
According to the problem,
Dividend = p ( x )
Divisor = 2x + 3
Degree of the divisor = 1 ------(1)
Remainder = r ( x ) = x - 1
Degree of the remainder = 1-------(2)
( 1 ) = ( 2 ) it is possible
Therefore ,
x - 1 is not the Remainder of p ( x )
I hope this will useful to you.
****
By division algorithm:
**********************************************
Dividend = divisor × quotient +
remainder
Where,
Remainder = r ( x ) = 0 or
Degree of r ( x ) < degree of divisor
**********************************************
According to the problem,
Dividend = p ( x )
Divisor = 2x + 3
Degree of the divisor = 1 ------(1)
Remainder = r ( x ) = x - 1
Degree of the remainder = 1-------(2)
( 1 ) = ( 2 ) it is possible
Therefore ,
x - 1 is not the Remainder of p ( x )
I hope this will useful to you.
****
Answered by
18
Answer:
By remainder theorem,
degree of the remainder is less than the degree of divisor
But here
remainder = x -1
its degree is 1
and
divisor2x +3
its degree is 1
it is wrong
Therefore........
n-1 is not remainder of p (x)
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