Question

Q17. Find the zeroes of (x- 2)2 – (x+2)2

Answer 1

Answer:

solution:-

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)=0−8x=0x=0

Hence,

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

Answer 2

Answer:

p(x)= (x-2)^2 - ,(x+2)^2

This in the form of the identity a^2 - b^2= (a+b)(a-b)

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

p(x)=2x(x-2-x-2)

p(x)=2x(-4)

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

The zero of this polynomial is 0 itself

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