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The H.C.F of 3(x2−9)(x+4) and 12(x−3)23(x2−9)(x+4) and 12(x−3)2 is |
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Answer» x2−3x+2=x2−2x−x+2x(x−2)−1(x−2)=(x−2)(x−1)Nowx2−4x+3=x2−3x−x+3x(x−3)−1(x−3)=(x−3)(x−1)THUS, the only COMMON FACTOR is (x-1 |
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