Question

24.Find the zeroes of the polynomial p(x)=(x-2)2 - (x+2)2
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Answer 1

Answer:

The zero of p(x) is 0.

Step-by-step explanation:

$p(x) = (x-2)^2 - (x+2)`^2\\\\\\= (x^2 + 2^2 - 2(x)(2)) - (x^2 + 2^2 + 2(x)(2)\\\\\\- \text{(Using:)} \: (a-b)^2 = a^2 + b^2 - 2ab \: \text{And,} (a+b)^2 = a^2+b^2 + 2ab\\\\\\= x^2 + 4 - 4x - x^2 - 4 -4x\\\\\\= x^2 - x^2 + 4 - 4 - 4x - 4x\\\\\\= -8x$

To find the zeroes, let p(x) = 0.

Then, -8x = 0

⇒ x = 0/-8 = 0

Therefore, the zero of p(x) is 0.

Answer 2

Answer:

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

x^2-4x+4 -(x^2+4x+4)=0

= -8x=0

= x=0