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Find the H.C.F. of the following :(i) x2 – 4 and x2 + 4x + 4(ii) 4x4 – 16x3 + 12x2 and 6x3 + 6x2 – 72x |
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Answer» (i) x2 – 4 = (x + 2)(x – 2) x2 + 4x + 4 = (x + 2)2 H.C.F. of coefficient = 1 H.C.F. of other factors = (x + 2)1 = x + 2 H.C.F. = x + 2 (ii) 4x4 – 16x3 + 12x2 = 4x2(x2 – 4x + 3) = 4x2(x – 1)(x – 3) 6x3 + 6x2 – 72x = 6x(x2 + x – 12) = 6x(x + 4)(x – 3) = 2x(x – 3) [because H.C.F. of coefficients is 2] = 2x2 – 6x |
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