Question

x³ − 23x² + 142x − 120

Answer 1

Let  `f(x) = x^3 - 23x^2 + 142x -120` be the given polynomial.

Now, putting  x=1,we get

`f(1) = (1)^3 - 23(1)^2 + 142(1) - 120`

        ` = 1 -23 + 142 - 120`

        ` = 143 - 143 = 0`

Therefore, (x-1)is a factor of polynomial f(x).

Now,

`f(x) = x^2(x-1) - 22x(x-1) + 120(x -1)`

`=(x-1){x^2 - 22x + 120}`

` = (x -1) {x^2 + 12x - 10x + 120}`

`=(x - 1)(x - 10)(x-12)`

Hence  (x-1),(x-10) and (x-12) are the factors of polynomial f(x).