Arun
Last Activity: 6 Years ago
When a polynomial f ( x ) is divided by x - 1 and x - 2, the remainders are 5 and 7 respectively.
So,
We get f ( x ) = ( x - 1 ) q ( x ) + 5
We substitute x = 1 , we get
f ( 1 ) = ( 1 - 1 ) q ( x ) + 5
f ( 1 ) = 5
And
f ( x ) = ( x - 2 ) Q ( x ) + 5
We substitute x = 2 , we get
f ( 2 ) = ( 2 - 2 ) Q ( x ) + 5
f ( 2 ) = 5
Now Let the remainder Ax + B When f ( x ) divide by ( x - 1 ) ( x - 2 ) , So
f ( x ) = ( x - 1 ) ( x - 2 )p ( x ) + Ax + B ----- ( A )
We substitute x = 1 , we
f ( 1 ) = ( 1 - 1 ) ( 1 - 2 )p ( x ) + A( 1 ) + B
f ( 1 ) = 0 + A + B , Substitute value of f ( 1 ) we get
A + B = 2 ----- ( 1 )
And we substitute x = 2 in equation A , we get
f ( 2 ) = ( 2 - 1 ) ( 2 - 2 )p ( x ) + A( 2 ) + B
f ( 2 ) = 0 + 2A + B , Substitute value of f ( 2 ) we get
2A + B = 5 ----- ( 2 )
Now we subtract equation 1 from equation 2 , we get
A = 3 , Substitute that value in equation 1 , we get
3 + B = 2
B = – 1
So,
Remainder = Ax + B = ( 3 ) x – 1 = 3x – 1
