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प्रश्न
The polynomial p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.
उत्तर
Given, p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7
When we divide p(x) by x + 1, then we get the remainder p(–1).
Now, p(–1) = (–1)4 – 2(–1)3 + 3(–1)2 – a(–1) + 3a – 7
= 1 + 2 + 3 + a + 3a – 7
= 4a – 1
p(–1) = 19
⇒ 4a – 1 = 19
⇒ 4a = 20
∴ a = 5
∴ Required polynomial = x4 – 2x3 + 3x2 – 5x + 3(5) – 7 ...[Put a = 5 on p(x)]
= x4 – 2x3 + 3x2 – 5x + 15 – 7
= x4 – 2x3 + 3x2 – 5x + 8
When we divide p(x) by x + 2, then we get the remainder p(–2)
Now, p(–2) = (–2)4 – 2(–2)3 + 3(–2)2 – 5(–2) + 8
= 16 + 16 + 12 + 10 + 8
= 62
Hence, the value of a is 5 and remainder is 62.
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