Factorise x³-23x²+142x-120

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Factorise x³-23x²+142x-120

Posted by Akshay Mundhe 6 years, 9 months ago

CBSE > Class 09 > Mathematics

2 answers

Parth Khandelwal 2 years, 9 months ago

1) Let p(x) = x3 – 23x2 + 142x – 120 The factors of –120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60. By hit and trial method, we find that p(1) = 0. So x – 1 is a factor of p(x). Now, x3 – 23x2 + 142x – 120 = x3 – x2 – 22x2 + 22x + 120x – 120 = x2(x –1) – 22x(x – 1) + 120(x – 1) = (x – 1) (x2 – 22x + 120) [ On taking (x – 1) common] Now x2 – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we get x2 – 22x + 120 = x2 – 12x – 10x + 120 = x(x – 12) – 10(x – 12) = (x – 12) (x – 10) So, x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)

1Thank You

Parth Khandelwal 2 years, 9 months ago

Solution x3 - 23x? + 142x - 120 =x3 - x? - 22x? + 22x + 120x - 120 =x?(x - 1) - 22x(x- 1) + 120(x - 1) =(x - 1)(x? - 22x + 120) =(x- 1) [x2 - 10x - 12x + 120] =(x - 1) [x(x - 10) – 12(x - 10)] =(x - 1)(x - 10)(x - 12)

1Thank You

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CBSE > Class 09 > Mathematics