Question

factors of x cube minus 23 x square + 142 X - 120

Answer 1

QUESTION :

P(X) = X³-23X²+142X-120

ANSWER :

P(X) = X³-23X²+142X-120

SO BY TRIAL AN ERROR METHOD

P(1) = (1)³-23(1)²+142(1)-120

=1-23+142-120

=120-120

=0

SO IT IMPLIES THAT (X-1) IS A FACTOR OF P(X)

SO BY SOLVING BY LONG DIVISION METHOD

P(X)=(X-1)(X²-22X+120)

SO BY SPLITTING THE MIDDLE TERM METHOD

P(X)=(X-1)(X²-12X-10X+120)

P(X)=(X-1)(X(X-12)-10(X-12))

P(X)=(X-1)(X-12)(X-10)

SO (X-1),(X-12) AND (X-10) ARE THE FACTORS OF P(X) = X³-23X²+142X-120

SO ZEROES OF P(X) ARE {1,12,10}