Question

The number of zeros of polynomial f (x)=(x-2)2+4can have is

Answer 1

Step-by-step explanation:

I think the question is

$f(x)=(x-2)^2+4$

First find the zeroes of the polynomials

Let;   $f(x)=0$

$(x-2)^2+4=0$

Expanding the following equation

$x^2-4x+4+4=0$

$x^2-4x+8=0$

By using the formula: x = −b ± √(b^2 − 4ac)/ 2a.

$x^2-4x+8=0$

$x= \frac{4+\sqrt{16-4(8)} }{2}$  (or)  $x=\frac{4-\sqrt{16-32} }{2}$

Therefore;

$x= \frac{4 + \sqrt{16}\sqrt{-1} }{2}$    (or) $x=\frac{4-\sqrt{16}\sqrt{-1} }{2}$

$x=\frac{4+4i}{2}$  (or)  $x=\frac{4-4i}{2}$

Answer:       2+2i (or) 2-2i

Therefore the given function has two zeros of the polynomial

II Method:

we obviously know that the given function is quadratic therefore it has two solutions or we must say it has two zeros of the polynomial

Answer 2

Answer:

0

Step-by-step explanation:

use discriminant method

don"t end on quadratic eq.

also use your own brain