Question

y^4-4y^3+8y^2-my+n is divided by y+1 and y-1 . The remainders are 10 and 16 respectively . Find the values of m and n

Answer 1

Answer:

if y-1 =0

y=1

substitute value of y from equations (i)

1^4-4(1)^3+8(1)^2-m(1)+n =16

1-4+8-m+n=16

11-m+n=16

-m+n=16-11

-m+n=5................................................................(ii)

now if. y+1=0

y= -1......................................................................(iii)

substitute the value of y from equation (iii)

(-1)^4-4(-1)^3+8(-1)^2-m(-1)+n=10

1-(-4)+8-(-m)+n=10

1+4+8+m+n=10

13+m+n=10

m+n=10-13

m+n= -3 ................................................(iv)

from equation (ii) and (iv) we get :

-m+n=5

m+n= -3

2n=2

n=1

now substitute value of 'n' in equation (iv)

m+(1)= -3

m= -3-1

m=-4

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HOPE IT HELPS U,