Question

Find the zeroes of the polynomial: p(x) = (x – 2)2 – (x + 2)2

Answer 1

Here, we first simplify the polynomial.

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

Here, we see that the polynomial is independent of x.

That means , we can put any value of x. p(x) will always equal -8.

Thus, whatever be the value of x, p(x) will never equal zero. This means that zeros of p(x) do not exist.

So, the set of zeros of p(x) is null set.

Answer 2

Hey there !!!!

Sorry couldn't understand your question .....

If your question is p(x)=(x-2)2-(x+2)2
                          
                             p(x)= 2x-4-2x-4= -8-----Equation1

Then there exists no roots for  p(x)=(x-2)2-(x+2)2 because from equation 1 we can observe that at p(x) is independent of "x".

If your question is :

p(x) =(x-2)²-(x+2)²

p(x) = x²-4x+4-(x²+4x+4)
      
        =x²-4x+4-x²-4x-4

         0=-8x 

x=0 is the zero of polynomial for  p(x)=(x-2)²-(x+2)²

Verification :

substituting x=0 in   p(x)=(x-2)²-(x+2)²
                               p(0) =2²-2² =0
∴ x=0 is a root of  p(x)=(x-2)²-(x+2)²

Hope this helped you.....