Question

find zeroes of polynomial p(x) = (x-2)^2-(x+2)^2

Answer 1

p(x)=(x-2)²-(x+2)²
This is in the form of a²-b²

p(x)=(x-2+x+2)(x-2-(x+2))
p(x)=2x(x-2-x-2)
p(x)=2x(-4)
p(x)=-8x


p(x)=0
-8x=0
x=0

x=0 is the zero of the polynomial p(x)

Answer 2

Answer:

The zero of the polynomial is 0.

Step-by-step explanation:

Given polynomial p(x) = $(x-2)^{2}-(x+2)^{2}$

$(x-2)^{2}-(x+2)^{2}=0$………………equation 1

Apply the formula $\bold{\left[a^{2}-b^{2}=(a+b)(a-b)\right]}$

Here in the above equation 1

a= (x-2)     $a^{2}=(x-2)^{2}$

b=(x+2)     $b^{2}=(x+2)^{2}$

By substituting the values of a and b in the equation 1

We get= [(x-2) +(x+2)] [(x-2) -(x+2)]=0

[In the above step while simplifying we have to be careful with the + and – signs]

Simplify the above equation

(x-2+x+2)(x-2-x-2)=0

(2x)(-4)=0

-8x=0

x=0

Therefore p(x)= $(x-2)^{2}-(x+2)^{2}$ is 0