13. Factorise x³ - 23x² + 142x - 120
Answers
=x3 - 23x2 +142x -120
=x3 -x2 - 22x2 +22zx +120x -120
=x2 (x-1) -22x (x-1) +120 (x-1)
=(x-1) (x2 -22x +120)
=(x -1) [x2 -10x-12x+120
=(x-1)[x(x-10)- 12(x-10
=(x-2)(x-10)(x-12)
Step-by-step explanation:
Given :-
x³ - 23x² + 142x - 120
To find :-
Factorise x³ - 23x² + 142x - 120 ?
Solution :-
General method :-
Given expression is x³ - 23x² + 142x - 120
=> x³-x²-22x²+22x+120x-120
=> (x³-x²)-(22x²-22x)+(120x-120)
=> x²(x-1)-22x(x-1)+120(x-1)
=> (x-1)(x²-22x+120)
=> (x-1)(x²-10x-12x+120)
=>(x-1)[x(x-10)-12(x-10)]
=> (x-1)(x-10)(x-12)
x³ - 23x² + 142x - 120 = (x-1)(x-10)(x-12)
Trial and error method :-
Given expression is x³ - 23x² + 142x - 120
Put x = 1 then
=> 1³-23(1)²+142(1)-120
=> 1-23+142-120
=> 143-143
=> 0
Since the value of expression at x=0 is zero
x-1 is a factor of the given expression.
To get another factors we divide given expression by (x-1)
x-1)x³ - 23x² + 142x - 120(x²-22x+120
x³-x²
(-)
________________
0 -22x²+142x
-22x² +22x
(-)
________________
0 +120x -120
120x -120
(-)
_________________
0
___________________
We get
x³ - 23x² + 142x - 120 = (x-1)(x²-22x+120)
=> (x-1)(x²-10x-12x+120)
=>(x-1)[x(x-10)-12(x-10)]
=> (x-1)(x-10)(x-12)
x³ - 23x² + 142x - 120 = (x-1)(x-10)(x-12)
Answer:-
The factorization of the given expression
x³ - 23x² + 142x - 120 is (x-1)(x-10)(x-12)