Question

20)If sum of remainders obtained by dividing ax^3-3ax^2+7x+5 by(x-1)and(x+1)is-36 the find a​

Answer 1

Answer:

We have, p(x) = ax3 − 3ax2 + 7x + 5Let q(x) = x + 1  

;  r(x) = x + 2By remainder theorem, when p(x) is divided by q(x), then remainder is given by p(−1).

Now,        p(x) = ax3 − 3ax2 + 7x + 5

⇒p(−1) = a(−1)3−3a(−1)2+7(−1)+5

⇒p(−1) = −a − 3a − 7 + 5⇒p(−1) =−4a − 2

By remainder theorem, when p(x) is divided by r(x), then remainder is given by p(−2).

Now,        p(x) = ax3 − 3ax2 + 7x + 5

⇒p(−2) = a(−2)3−3a(−2)2+7(−2)+5

⇒p(−2) =−8a − 12a − 14 + 5

⇒p(−2) =−20a − 9

Now,     p(−1)  + p(−2) = −36

⇒ (−4a − 2) + (−20a − 9) = −36

⇒−4a−2−20a−9=−36

⇒−24a−11 = −36⇒24a + 11 = 36⇒24a = 36 − 11

⇒24a = 25

⇒a = 25/24

Answer 2

the answer is 23/3

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