if y^2-1 is a factor of y^4+my^3+2y^2-3y+n find value of m and n
Answers
Answered by
69
Step-by-step explanation:
y⁴+my³+2y2-3y+n=0
y²=1
y=±1
put y=1
(1)⁴+m(1)³+2(1)²-3(1)+n=0
1+m+2-3+n=0
m+n=0---1
put y=-1
(-1)⁴+m(-1)³+2(-1)²-3(-1)+n=0
1-m+2+3+n=0
m=6+n
m -n=6---(2)
solve 1 and 2
2m=6
m=3
from eqn(1)
n=-3
so(m,n)=(3,-3)
mkr9426:
Thanks a lot
But I doesn't understand that why 2m=6
Please proove it
by adding eq 1 and eq 2 , we get 2m=6
Answered by
32
Answer:
y⁴+my³+2y2-3y+n=0
y²=1
y=±1
put y=1
(1)⁴+m(1)³+2(1)²-3(1)+n=0
1+m+2-3+n=0
m+n=0---1
put y=-1
(-1)⁴+m(-1)³+2(-1)²-3(-1)+n=0
1-m+2+3+n=0
m=6+n
m -n=6---(2)
solve 1 and 2
2m=6
m=3
from eqn(1)
n=-3
so(m,n)=(3,-3)
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Step-by-step explanation:
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