Question

Show that the polynomial x^4 +4x^2 +6 has no zero

Answer 1

Step-by-step explanation:'

let x²= u

polynomial becomes

u² + 4u+6 =0

finding discriminant

= b² -4ac

=16-24

= -8

means if discriminant is - ve(negative)

no real root exist .

thus this polynomial has no zeros .

TO UNDERSTAND CONCEPTUALLY :

for a polynomial to have zeros the value of polynomial must becomes zero at any value of x .

now look at the question  v carefully ,

it has three terms

x⁴

4 x²

6

now all three of them are in addition .

x⁴ can never be negative as it is a perfect square of x²  ,

means (x²) * (x²) = x⁴

you may put any value of x here it will never be negative for ex if we put x=  -2 , x⁴  would be 16

same is the case for 4 x², it will never be negative.

6 is always positive.

that means you are adding anything positive to 6 , means  polynomial x^4 +4x^2 +6  can never be equal to zero .

note: the minimum value of  polynomial (x^4 +4x^2 +6  )occurs when x=0 , that would be 6