Question

Does the zeroes exist for the polynomial p(x) =x²+16? Justify your answer​

Answer 1

Answer:

The equation is in the form of (a^2 + b ^2)

= (x + 4) (x - 4)

x = - 4 and x= 4

So,this equation is having 2 distinct real roots

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Answer 2

Answer:

Solution:-

$\mathtt \blue{ = {x}^{2} + 16}$

$\mathtt \blue{ = (x + 4)(x - 4)}$

Now find zero of the polynomial

$\mathtt\purple{x + 4 = 0}$

$\mathtt \purple{x = 0 - 4 }$

$\fbox{x = - 4}$

$\mathtt \green{x - 4 = 0}$

$\mathtt \green{x = 0 + 4}$

$\fbox{x = 4}$

Yes zeros are exist for this polynomial.

Step-by-step explanation:

$\fbox\blue{I hope this helps you}$