This is the basic and conceptual question from polynomials in which we have given a cubic equation and we have to verify its zeroes are 3,-2,1 and we also have to verify that relationship between its zeroes and coefficient.
Kindly solve the above problem
RS Aggarwal, Class 10, chapter 2B, question no 1
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