vi) f(x)=4x2+5√2x−3
=4x2+6√2x−√2x−3 [by splitting the middle term]
=2√2x(√2x+3)−1(√2x+3)
=(√2x+3)(2√2x−1)
So, the value of 4x2+5√2x−3 is zero when √2x+3=0or2√2x−1=0
i.e., When x=−3√2 and 12√2
so, the zeroes of 4x2+5x−3 are −3√2 or x=12√2
∴ sum of zeroes =−3√2+12√2
=−52√2=−5√24
=−(Coefficientofx)(Coefficientofx2)
And product of zeroes =(−3√2)(12√2)=−34
=constant termcoefficient of x2
Hence, verified the relations between the zeroes and the coefficient of the polynomial