Find the zeroes of the polynomial:p(x)=(x-2)^ | vedclass

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(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.Given $p(x) = (x-2)^{2} - (x+2)^{2}$.Using the algebraic identity $a^2 - b^2 = (a-b)

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$0$

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(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.Given $p(x) = (x-2)^{2} - (x+2)^{2}$.Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$,where $a = (x-2)$ and $b = (x+2)$:$p(x) = [(x-2) - (x+2)] \cdot [(x-2) + (x+2)] = 0$$p(x) = (x - 2 - x - 2) \cdot (x - 2 + x + 2) = 0$$p(x) = (-4) \cdot (2x) = 0$$-8x = 0$$x = 0$Thus,the zero of the polynomial is $0$.

Find the zeroes of the polynomial:
$p(x)=(x-2)^{2}-(x+2)^{2}$

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Find the zeroes of the polynomial:$p(x)=(x-2)^{2}-(x+2)^{2}$