Question

plz answer.....
it's urgent...... ​

Answer 1

Answer:

0

Step-by-step explanation:

$(x-2)^{2} - (x+2)^{2}=0\\ x^{2} -4x+4-x^{2} -4x-4=0\\-8x=0\\x=0$

Answer 2

Answer:

To find the zeros,

equate the p(x) to 0.

${(x -2)}^{2} - {(x + 2)}^{2} = 0 = > (x - 2 + x + 2)(x - 2 - x - 2) = 0 = > 2x( - 4) = 0 = > 2x = 0 = > x = 0$

So, zero of p(x) is 0.

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