Question

Factorise x³ - 23x² + 142x -120....​

Answer 1

Answer:

(x-1)(x-12)(x-10)

Answer 2

Answer:-

(x-1)(x-10)(x-12)

Explanation:-

Let p(x) = x³-23x²+142x-120

Some of the factors of -120 are:-

±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15,

±20, ±24, ±30, ±60

By trial, we found 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)

[Taking (x-1) common]

Now, x²-22x+120 can be factorized either by splitting the middle term or by using Factor theorem. By splitting middle term,we get:-

= x²-22x+120

= x²-12x-10x+120

= x(x-12)-10(x-12)

= (x-10)(x-12)

∴ x³-23x²+142x-120 = (x-1)(x-10)(x-12)