Find the zeroes of the polynomial p(x)=(x-2)2 – (x+2)2 .
Answers
Answered by
53
Given polynomial,
p(x)=(x-2)²-(x+2)²
(x-2)²-(x+2)²=0
[a²-b²=(a-b)(a+b)]
(x-2-(x+2))(x-2+x+2)=0
(x-2-x-2)(2x)=0
(-4)(2x)=0
2x=0/-4
2x=0
x=0/2
x=0
Therefore,zero of the polynomial is 0.
Hope it helps
p(x)=(x-2)²-(x+2)²
(x-2)²-(x+2)²=0
[a²-b²=(a-b)(a+b)]
(x-2-(x+2))(x-2+x+2)=0
(x-2-x-2)(2x)=0
(-4)(2x)=0
2x=0/-4
2x=0
x=0/2
x=0
Therefore,zero of the polynomial is 0.
Hope it helps
Answered by
4
Given polynomial,
p(x)=(x-2)²-(x+2)²
(x-2)²-(x+2)²=0
[a²-b²=(a-b)(a+b)]
(x-2-(x+2))(x-2+x+2)=0
(x-2-x-2)(2x)=0
(-4)(2x)=0
2x=0/-4
2x=0
x=0/2
x=0
Therefore,zero of the polynomial is 0.
Hope it helps
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