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no.33
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Answer:
f(x)=x³-23x²+142x-120
We know that if the sum of the coefficients is equal to 0 then (x-1) is one of the factors of given polynomial.
f(1)=1³-23×1²+142(1)-120
= 1-23+142-120
=0
therefore (x-1) is a factor of x³-23x²+142x-120.
deviding x³-23x²+142x-120 by (x-1) we get (x²-22x+120)
finding factors of (x²-22x+120)
x²-22x+120
= x²-10x-12x+120
=x(x-10)-12(x-10)
=(x-12)(x-10)
factors of x²-22x+120 are (x-12)(x-10)
hence factors of x³-23x²+142x-120 are (x-1) (x-12)(x-10)
refer→→ https://brainly.in/question/1227722
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