Getting 10/9 with remainder theorem...is it correct?
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Answer:
10/9 is remainder
Step-by-step explanation:
given = 3y-1 to be divisible by 3ycube + 6ysqure+7y-2
so 3y-1=0 3y=1 y=1/3
substituting this in equations
3(1/3)3+6(1/3)3+7(1/3)-2
3(1/27)+6(1/9)+7(1/3)-2
3/27+6/9+7/3-2/1
cancel out
1/9+1/9+7/3-2/1
(take 1 cm as 9)
1/9+6/9+21/9-18/9
1+6+21-18/9=10/9 not equal to zero
hence y=1/3 is not a factor and remainder is 10/9
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