Question

How many real solutions does the equation x⁷ + 14x⁵ + 16x³ + 30x − 560 = 0 have?

(A) 7

(B) 1

(C) 3

(D) 5

Answer 1

Answer is (B) 1

x⁷ + 14x⁵ + 16x³ + 30z - 560 = 0

Let f(x) = x⁷ + 14x⁵ + 16x³ + 30x. Then

f'(x) = 7x⁶ + 70x⁴ + 48x² + 30 > 0 ∀ x

Therefore, f(x) is a strictly increasing function for all x. 

So, it can have at the most one solution.