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Question
When the polynomial x3 + 2x2 – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x3 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
Solution
It is given that when the polynomial x3 + 2x2 – 5ax – 7 is 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 + ax2 – 12x + 16 is divided by (x + 2), the remainder is B.
∴ x3 + ax2 – 12x + 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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