Let f(x)=x3−2x2−5x+6
3,−2 and 1 are the zeroes of the polynomial ( given )
Therefore,
f(3)=(3)3−2(3)2−5(−2)+6
=27−18−15+6
=0
f(−2)=(−2)3−2(−2)2−5(−2)+6
−8−8+10+6
=0
f(1)=(1)3−2(1)2−5(1)+6
=1−2−5+6
=0
Verify relations :
General form of cubic equation :ax3+bx2+cx+d
now ,
Consider α=3,β=−2 and y=1
α+β+y=3−2+1=2=−b/a
αβ+βy+αy=3(−2)+(−2)(1)+1(3)=−5=c/a
and αβy=3(−2)(1)=−6=−d/a