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