We have
f(x)=4x2−4x−3
f(x)=4x2−(6x−2x)−3
f(x)=4x2−6x+2x−3
f(x)=2x(2x−3)+1(2x−3)
f(x)=(2x+1)(2x−3)
∴f(x)=0
=>(2x+1)(2x−3)=0
=>2x+1=0 or 2x−3=0
=>x=−1/2 or x=3/2
So, the zeros of f(x) are −1/2 and 3/2.
Sumofthezeros=−1/2+3/2=(−1+3)/2=22=1=−(coefficientofx)/(coefficientofx2)
Productofthezeros=−12×32=−34=constantterm/(coefficientofx2)