Let’s find the zeros of the polynomial p(x)=(x-2)^2-(x+2)^2.

**p(x) = (x-2) ^{2} – (x+2)^{2}**

**Note:** For solving this we need 2 formuleas as follow:

(a+b)^{2} = a^{2}+b^{2}+2ab

(a-b)^{2} = a^{2}+b^{2}-2ab

So we will consider a = x and b = 2 and will solve this polynomial

Let’s see:

p(x) = **(x-2) ^{2} – (x+2)^{2}**

= (x^{2}+4-4x) – (x^{2}+4+4x)

= x^{2}+4-4x – x^{2}-4-4x

= x^{2}-x^{2}+4-4-4x-4x

= 0+0-8x

= 0-8x

= -8x

## Find The Zeros of The Polynomial p(x)=(x-2)^2-(x+2)^2

So, p(x) = **(x-2) ^{2} – (x+2)^{2}**

**(x-2)**= 0

^{2}– (x+2)^{2}-8x = 0

x = 0

So, the zeros of the Polynomial p(x)=(x-2)^2-(x+2)^2 will be zero when x = 0.

8×2-22x-21=0 By Factorization Method

Find The Zeros of The Quadratic Polynomial x2-2x-8 = 0

Factorization of 3×2-8x+5

More Answers on Brainly

## Leave a Reply