Let f(x)=3x³−10x²−27x+10

5,−2 and 1/3​ are the zeroes of the polynomial ( given )

Therefore,
f(5)=3(5)³−10(5)²−27(5)+10
=3×125−250−135+10
=0

f(−2)=3(−2)³−10(−2)²−27(−2)+10
−24−40+54+10
=0

f(1/3)=3(1/3)³−10(1/3)²−27(1/)³+10
=1/8−10/9−9+10
=0

Verify relations :
General form of cubic equation :ax³+bx²+cx+d
now ,
Consider α=5,β=−2  and y=31​

α+β+y=5−2+1​/3=10/3​=−b/a​

αβ+βy+αy=5(−2)+(−2)(1​/3)+(5×1/3)=−10+(−2)​/3+5/3​=−9=c/a​

and αβy=5(−2)(1/3​)=−10/3​=−d/a​

General Form of cubic polynomial whose zeroes are a. b and c is :
x³−(a+b+c)x²+(ab+bc+ca)x−abc..........(1)

To find : cubic polynomial whose zeroes are 1/2​,1 and –3

Let us say , a=1/2​,b=1 and c=−3
Putting the values of a,b and c in equation (1) we get :

=x³−(1/2+1−3)x²+(1/2​−3−3/2​)x−(−3/2​)

=x³−(−3)/2​x²−4x+3/2

=2x³+3x²−8x+3
which is required polynomial.

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