Question

Show that the polynomial x⁴ + 4x² + 5 has no real zero.​

Answer 1

Question:-

• Show that the polynomial x⁴ + 4x² + 5 has no real zero.​

Solution:-

= x⁴ + 4x² + 5

= (x²)² + 4x²+ 5

Now , Putting  x² = a

= x⁴+ 4x² + 6 = (x²)² + 4x² + 5

= a² + 4a + 5

→so , To get the zero of the polynomial

→we are Taking  a²+ 4a + 5 = 0

hence, there is no real zero .

____________________________________

hope it helps :D

Answer 2

Step-by-step explanation:

Given, f(x) = x4 + 4x2 + 6 We can rewrite the given equation as ⇒ f(x)= (x2)2 + 4x2 + 6 Let x2 = a ⇒ f(x) = a2 + 4a + 6 If the discriminant of the equation < 0 then the given polynomial has no zeros. Discriminant = b2- 4ac = 42 - 4(1)(6) =16 - 24 = - 8 < 0 Therefore, there is no zero for given polynomial