Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the zeroes of the polynomial : p(x)=(x-2)^{2}-(x+2)^{2}

Find the zeroes of the polynomial :  p(x)=(x-2)^{2}-(x+2)^{2}

Answers (3)

[0]

Solution:        p(x) = (x - 2)^{2} - (x + 2)^{2}

We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero

i.e., p(x) = 0   

That is,

\\(x - 2)^{2}- (x + 2)^{2} = 0\\ (x^{2} + 2^{2} - 2 \times x \times 2) - (x^{2} + 2^{2} + 2 \times x \times 2) = 0\\ {\because (a - b)^{2} = a^{2} + b^{2} - 2ab}\\ {\because (a + b)^{2} = a^{2} + b^{2} + 2ab}\\ So,\\ x^{2} + 4 - 4x - x^{2} - 4 - 4x = 0\\ - 4x - 4x = 0\\ -8x = 0\\ x = 0

Hence 0 is the zero of the given polynomial

Posted by

infoexpert24

View full answer

Given,
p(x)=(x-2)²-(x+2)²
  0   =(x-2)²-(x+2)²
(x+2)²=(x-2)²
x²+4+4x=x²+4-4x
because (a-b)²=a²+b²+2ab;
(a+b)²=a²+b²-2ab
     4x+4x=0
           8x=0
             x=0
Therefore,zeroes of the given polynomial is 0.

Posted by

NIHARIKA PERIMILLA

View full answer

Given,

p(x)=(x-2)²-(x+2)²

0 =(x-2)²-(x+2)²

(x+2)²=(x-2)²

x²+4+4x=x²+4-4x

because (a-b)²=a²+b²+2ab;

(a+b)²=a²+b²-2ab

4x+4x=0

8x=0

x=0

Therefore,zeroes of the given polynomial is 0.

Posted by

NIHARIKA PERIMILLA

View full answer

Similar Questions

Latest Question