Question

find the zeros of the polynomial x minus 2 whole square + 4

Answer 1

(x-2)²2 +4=x²+4-4x+4=x²+8-4x

Answer 2

Answer:

Zeros of polynomial $2(1+i),2(1-i)$

Step-by-step explanation:

Given : Polynomial $(x-2)^2+4$

To find : The zeros of the polynomial.

Solution :

Zeros of the polynomial is the roots of the polynomial.

So first solve the polynomial,

$(x-2)^2+4$

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

$=x^2-4x+8$

Solving the quadratic equation by discriminant method

Form = $ax^2+bx+c=0$

$D=b^2-4ac$

Solution $x=\frac{-b\pm\sqrt{D}}{2a}$

Comparing the quadratic equation with general form,

a=1 , b=-4, c=8

$D=b^2-4ac$

$D=(-4)^2-4(1)(8)$

$D=16-32$

$D=-16$

Solution $x=\frac{-b\pm\sqrt{D}}{2a}$

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

$x=\frac{4\pm4i}{2}$

$x=2\pm2i$

$x=2(1\pm i)$

Therefore, zeros of the polynomial is $x=2(1+i),2(1-i)$