factorise x^3-23x^2+142x-120
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Answer:
x
3
−23x
2
+142x−120
=x
3
−x
2
−22x
2
+22x+120x−120
=x
2
(x−1)−22x(x−1)+120(x−1)
=(x−1)(x
2
−22x+120)
=(x−1)[x
2
−10x−12x+120]
=(x−1)[x(x−10)−12(x−10)]
=(x−1)(x−10)(x−12)
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