factorise x2-23x2+142-120
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x3- 23x2+142x-120
let p(x)= x3- 23x2+142x-120
g(x)=x-1
0=x-1
x=1
p(1)=(1)3-23(1)2+142(1)-120
=1-23+142-120
=143-143
=0
Hence x-1 is a factor of x3- 23x2+142x-120
by long division
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+12)(x+10)
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let p(x)= x3- 23x2+142x-120
g(x)=x-1
0=x-1
x=1
p(1)=(1)3-23(1)2+142(1)-120
=1-23+142-120
=143-143
=0
Hence x-1 is a factor of x3- 23x2+142x-120
by long division
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+12)(x+10)
PLZ MARK ME AS BRAINLIEST
sheelu1277tu:
sorry wrong answer
see replace 3 with 2 and then proceed u will get answer
PLZ MARK ME AS BRAINLIEST
(x-1)(x-10)(x-12) this is wrong answer
sorry
this is wright answer
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