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Answers
1) 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
2)polynomial , x³ +10x² + ax + b is exactly divisible by ( x -1) and (x -2)
it means x = 1 and 2 are the zeors of given polynomial .
so,
put x = 1
(1)³ + 10(1)² +a(1) + b = 0
1 + 10 + a + b = 0
a + b = -11 --------(1)
again put x =2
(2)³ +10(2)² +a(2) + b =0
8 + 40 + 2a + b = 0
2a + b = -48 ----------(2)
solve equations (1) and (2)
a = -37
b = 26
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