factorise:-
x^3 - 23x^2 +142x - 120
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Step-by-step explanation:
refer to the attachment
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Answer:
We know that if the sum of the coefficients is equal to 0 then (x-1) is one of the factors of given polynomial.
x-1 = 0
x = 1
Put x=1,
(1)³-23(1)²+142(1)-120
→ 1-23+142-120
→ 120-120
→ 0
Therefore,(x-1) is a factor of the given polynomial.
Now the factors are (x-1) and (x²-22x+120) {see pic for knowing how (x²-22x+120) is a factor}
Now factorise x²-22x+120
x²-22x+120
→ x²-12x-10x+120
→ x(x-12)-10(x-12)
→ (x-12)(x-10)
Therefore, x²-22x+120 = (x-12)(x-10)
The factors of x³-23x²+142x-120 = (x-1)(x-12)(x-10)
Hope it helps....…
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