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The H.C.F. of (x2 - 4), (x2 - 5x - 6) and (x2 + x - 6) is -
- A1
- B(x - 2)
- C(x + 2)
- D(x2 + x - 6)
Answer not set
Discussion 1
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CCommunity
x2 - 4 = (x + 2)(x - 2)
x2 - 5x - 6 = x2 - 6x + x - 6
= (x - 6)(x + 1)
x2 + x - 6 = x2 + 3x - 2x - 6
= (x + 3)(x - 2)
Clearly there is no common factor
So H.C.F = 1
x2 - 5x - 6 = x2 - 6x + x - 6
= (x - 6)(x + 1)
x2 + x - 6 = x2 + 3x - 2x - 6
= (x + 3)(x - 2)
Clearly there is no common factor
So H.C.F = 1