h(t) =t2−15=(t)2−(1​5)2=(t+1​5)(t−1​5)zeros of the polynomials are −1​5 and 1​5sum of zeroes = 0−1​5+1​5=00 = 0Product of zeroes =(−15)/1−1​5×1​5=−15−15=15∴ Hence the relationship verified.