Question

y+1/4y=2 then 16y³+1/4y³=
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Answer 1

ANSWER

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21So put y=−21 in expression we get

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21So put y=−21 in expression we getp(−21)=4(−21)3+4(−21)2−(−21)−1

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21So put y=−21 in expression we getp(−21)=4(−21)3+4(−21)2−(−21)−1⇒p(−21)=4×−81+4×41−21−1

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21So put y=−21 in expression we getp(−21)=4(−21)3+4(−21)2−(−21)−1⇒p(−21)=4×−81+4×41−21−1⇒p(−21)=21+1−21−1

ANSWERGiven  p(y)=4y3+4y2−y−1   is a multiply of (2y+1)To check that 2y+1 is the factor the value of 2y+1=0 so y=−21So put y=−21 in expression we getp(−21)=4(−21)3+4(−21)2−(−21)−1⇒p(−21)=4×−81+4×41−21−1⇒p(−21)=21+1−21−1⇒p(−−21)=0

−21)=0So remainder is  zero when 2y-1 divided polynomial  4y3+4y2−y−1 

−21)=0So remainder is  zero when 2y-1 divided polynomial  4y3+4y2−y−1 Then 2y+1  is  a multiple or factor  of polynomial  4y3+4y2−y−1