Question

find the zero of the polynomial p(x)=(x-2)²-(x+2)².​

Answer 1

Step-by-step explanation:

$p(x) = {( x - 2)}^{2} - {(x + 2)}^{2} \\ put \:p( x ) = 0 \\ {(x - 2)}^{2} - {(x + 2)}^{2} = 0 \\ { (x - 2)}^{2} = {(x +2)}^{2} \\ x² - 4x + 4 = x² + 4x + 4 \\ -4x = 4x \\ - 4x - 4x = 0 \\ -8x =0 \\ x = 0$

Therefore the zero of the polynomial is at x=0.

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Answer 2

Step-by-step explanation:

Given:

p(x) = (x - 2)² - (x + 2)²

To find:

zeroes of the polynomial.

Solution:

We have

p(x) = (x - 2)² - (x + 2)²

Finding a zero of the polynomial, is the same as solving the equation p(x) = 0, we get

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

⇒ (2x) (-4) = 0

⇒-8x = 0

⇒x = 0

Hence, 0 is a zero of the polynomial p(x) = (x - 2)² - (x + 2)²