Let f(x) = x2 + 7x + 10
By splitting the middle term, we get
f(x) = x2 + 5x + 2x + 10 [∵ 7 = 2 + 5 and 2×5 = 10]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
On putting f(x) = 0 , we get
(x + 2) (x + 5) = 0
⇒ x + 2 = 0 or x + 5 = 0
⇒x = – 2 or x = – 5
Thus, the zeroes of the given polynomial x2 + 7x + 10 are – 2 and – 5
Verification
So, the relationship between the zeroes and the coefficients is verified.