Question

prove that 2x^4-6x^3-ax+b is exactly divisible by x^2-x+2 i. by actual division ii. without actual division​

Answer 1

Let f(x) = 2x4 – 6x3 + 3x2 + 3x – 2 and g(x) = x2 – 3x + 2 = x2 – 2x – x + 2 = x(x – 2) – 1 (x – 2) = (x – 2) (x – 1) For f(x) to be exactly divisible by g(x), (x – 1) and (x – 2) should be the factors of f(x), i.e., f(1) = 0 and f(2) = 0. Now, f(1) = 2. (1)4 – 6.(1)3 + 3.(1)2 + 3.1 – 2 = 2 – 6 + 3 + 3 – 2 = 0 f(2) = 2.(2)4 – 6(2)3 + 3(2)2 + 3.2 – 2 = 32 – 48 + 12 + 6 – 2 = 0. ∴ (x – 1) and (x – 2) are factors of f(x) ⇒ f(x) is exactly divisible by g(x).Read more on Sarthaks.com - https://www.sarthaks.com/1000327/without-actual-division-show-that-2x-4-6x-3-3x-2-3x-2-is-exactly-divisible-by-x-2-3x-2

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Answer 2

Answer:

Let f(x) = 2x4 – 6x3 + 3x2 + 3x – 2 and g(x) = x2 – 3x + 2 = x2 – 2x – x + 2 = x(x – 2) – 1 (x – 2) = (x – 2) (x – 1) For f(x) to be exactly divisible by g(x), (x – 1) and (x – 2) should be the factors of f(x), i.e., f(1) = 0 and f(2) = 0. Now, f(1) = 2. (1)4 – 6.(1)3 + 3.(1)2 + 3.1 – 2 = 2 – 6 + 3 + 3 – 2 = 0 f(2) = 2.(2)4 – 6(2)3 + 3(2)2 + 3.2 – 2 = 32 – 48 + 12 + 6 – 2 = 0. ∴ (x – 1) and (x – 2) are factors of f(x) ⇒ f(x) is exactly divisible by g(x).

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