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Question

When the polynomial x3+2x25ax7 is divided by (x1), the remainder is A and when the polynomial x3+ax212x+16 is divided by (x+2), the remainder is B. Find the value of 'a' if 2A+B=0.

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Solution

It is given that when the polynomial x3+2x25ax7is divided by (x – 1), the remainder is A.
(1)3+ 2(1)2 – 5a(1) – 7 = A
1 + 2 – 5a – 7 = A
– 5a – 4 = A …(i)
It is also given that when the polynomial x3+ax212x+16 is divided by (x + 2), the remainder is B.
x3+ax212x+16 = B
(2)3​​​​​​​​​​​​​​ + a(2)2​​​​​​​​​​​​​​ ​​​​​​​ – 12(-2) + 16 = B
-8 + 4a + 24 + 16 = B
4a + 32 = B …(ii)
It is also given that 2A + B = 0
Using (i) and (ii), we get,
2(-5a – 4) + 4a + 32 = 0
-10a – 8 + 4a + 32 = 0
-6a + 24 = 0
6a = 24
a = 4


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