Question

factorise this sum please

Answer 1

by hit and trial if we pit x=1 we will get
$( {1)}^{3} - 23 {(1)}^{2} + 142(1) - 120 \\ = 1 - 23 + 142 - 120 \\ = 143 - 143 \\ = 0$
this shows x-1 is a factor of given polynomial.
on dividing with x-1 we will get
$(x - 1)( {x}^{2} - 22x + 120) \\ = (x - 1)( {x}^{2} - 12x - 10x + 120) \\ = (x - 1){x(x - 12) - 10(x - 12)} \\ = (x - 1)(x - 12)(x - 10)$

Answer 2

x^3-23x^2+142x-120

let x=1

==> p(1) = (1)^2-23(1)^2+142(1)-120

==> 1-23+142-120

==> 143-143

==> 0

x =1

==> x-1 = 0

NOW divsible x^3-23x^2+142x-120 by x-1

you get we get quoitent x^2-22x+120

==> factorise x^2-22x+120

==> x^2-10x-12x+120

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

==> (x-10)(x-12)'

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