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Question

If x3+ax2+bx+6 has (x-2) as a factors and leaves a remainder 3 when divided(x-3). Find a and b

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Solution

Let the given polynomial be p(x) = x3 + ax2 + bx + 6
Given p(x) is divisible by (x – 2), hence p(2) = 0
Put x = 2 in p(x)
p(2) = 23 + a(2)2 + b(2) + 6
⇒ 0 = 8 + 4a + 2b + 6
⇒ 4a + 2b = – 14
⇒ 2a + b = – 7 → (1)
It is also given that when p(x) is divided by (x – 3) the remainder is 3
That p(3) = 3
Put x = 3 in p(x)
p(3) = 33 + a(3)2 + b(3) + 6
⇒ 3 = 27 + 9a + 3b + 6
⇒ 9a + 3b = – 30
⇒ 3a + b = – 10 → (2)
Subtract (2) from (1)
2a + b = – 7
3a + b = – 10
-------------------
– a = 3
∴ a = – 3
Substitute a = – 3 in (1), we get
2(– 3) + b = –7
∴ b = –1

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