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प्रश्न
When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is,, divided by (x – 2), the remainder is 7. Find – the remainder when it is divided by (x – 1) (x – 2).
उत्तर
When f(x) is divided by (x – 1),
Remainder = 5
Let x – 1 = 0
⇒ x = 1
∴ f(1) = 5
When divided by (x – 2),
Remainder = 7
Let x – 2 = 0
⇒ x = 2
∴ f(2) = 7
Let f(x) = (x – 1)(x – 2)q(x) + ax + b
Where q(x) is the quotient and ax + b is remainder
Putting x = 1, we get:
f(1) = (1 – 1)(1 – 2)q(1) + a x 1 + b
= 0 + a + b
= a + b
and x = 2, then
f(2) = (2 – 1)(2 – 2)q(2) + a x 2 + b
= 0 + 2a + b
= 2a + b
∴ a + b = 5 ....(i)
2a + b = 7 ....(ii)
Subtracting, we get
–a = – 2
⇒ a = 2
Substituting the value of a in (5)
2 + b = 5
⇒ b = 5 – 2 = 3
∴ a = 2, b = 3
∴ Remainder = ax + b
= 2x + 3.
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