Question

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

Answer 1

Answer:

(x-2)^-(x+2)^

(a-b)^-(a+b)^

(a^-2ab+b^)-(a^+2ab+b^)

(x^-4x+4)-(x^+4x+4)

x^-4x+4-x^-4x-4

-4x-4x

-8x=0

x=8

I hope you understand my answer

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