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Find the LCM of x2 – 4x + 3, 2x2 – 3x + 1, and 2x2 + 7x + 3.1. (4x2 – 1)(x2 – 9)(x – 1)2. (4x2 + 1)(x2 – 9)3. (x2 – 9)(x – 1)(2x + 3)4. (4x2 + 1)(x + 1) |
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Answer» Correct Answer - Option 1 : (4x2 – 1)(x2 – 9)(x – 1) Given: x2 – 4x + 3, 2x2 – 3x + 1, and 2x2 + 7x + 3 Concept: The LCM of two or more given polynomials is the least polynomial which is exactly divisible by each of them. Calculation: x2 – 4x + 3 = x2 – 3x – x + 3 ⇒ x(x – 3) – 1(x – 3) ⇒ (x – 1)(x – 3) 2x2 – 3x + 1 = 2x2 – 2x – x + 1 ⇒ 2x(x – 1) – 1(x – 1) ⇒ (2x – 1)(x – 1) 2x2 + 7x + 3 = 2x2 + 6x + x + 3 ⇒ 2x(x + 3) + 1(x + 3) ⇒ (2x + 1)(x + 3) LCM of x2 – 4x + 3, 2x2 – 3x + 1, and 2x2 + 7x + 3 ⇒ (x – 1)(x – 3) (2x – 1)(x – 1) (2x + 1)(x + 3) ⇒ (2x – 1)(2x + 1) (x – 3)(x + 3) (x – 1) ⇒ (4x2 – 1)(x2 – 9)(x – 1) ∴ The required LCM is (4x2 – 1)(x2 – 9)(x – 1). |
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