Question

x^3-23x^2+142x-120 factories​

Answer 1

Step-by-step explanation:

p(x)=x³-23x²+142x-120

Factors of 120 are ±1,±2,±3,±4.....….

Let x=1

p(1)=1³-23(1)²+142(1)-120

=1-23+142-120

=-143+143

=0

As x=1 is a zero of p(x)

Therefore,(x-1) would be a factor of p(x)

(x³-23x²+142x-120)/(x-1)

x-1)x³-23x²+142x-120(x²-22x+120

+(-)x³±x².

0-22x². +142x-120

±22x²+(-)22x.

0 +120x-120

+(-) 120x±120

0

x² -22x. +120=0

x²-8x-15x+120=0

x(x-8)-15(x-8)=0

(x-15)(x-8)=0

x-15=0. x-8=0

x=15. x=8

Therefore , factors are (x-1),(x-15) and (x-8).

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