1.

Factorise `x^3-23 x^2+142 x-120`.

Answer» Here, `P(x) = x^3-23x^2+142x-120`
Factors of constant term `120` are
`+-1,+-2,+-3,+-4,+-5,+-6,+-8,+-10,+-12,+-15,+-20,+-24,+-30,+-40,+-60,+-120`
`P(1) = 1-23+142-120 = 0`
From Factor theoram,
`(y-a)` is a factor of `P(y) if P(a) = 0`
Here, as `P(1)` is `0`, so, `(x-1)` is a factor of `P(x)`.
If, we divide `P(x)` by `(x-1)`, we get, `x^2-22x+120`.
Please refer video for the division.So, `P(x) = (x-1)(x^2-22x+120)`
`= (x-1)(x^2-12x-10x+120)`
`=(x-1)(x-10)(x-12)` that are the required factors.


Discussion

No Comment Found

Related InterviewSolutions