find zeroes of polynomial p(x) = (x-2)^2-(x+2)^2
Answers
Answered by
762
p(x)=(x-2)²-(x+2)²
This is in the form of a²-b²
p(x)=(x-2+x+2)(x-2-(x+2))
p(x)=2x(x-2-x-2)
p(x)=2x(-4)
p(x)=-8x
p(x)=0
-8x=0
x=0
x=0 is the zero of the polynomial p(x)
This is in the form of a²-b²
p(x)=(x-2+x+2)(x-2-(x+2))
p(x)=2x(x-2-x-2)
p(x)=2x(-4)
p(x)=-8x
p(x)=0
-8x=0
x=0
x=0 is the zero of the polynomial p(x)
Answered by
182
Answer:
The zero of the polynomial is 0.
Step-by-step explanation:
Given polynomial p(x) =
………………equation 1
Apply the formula
Here in the above equation 1
a= (x-2)
b=(x+2)
By substituting the values of a and b in the equation 1
We get= [(x-2) +(x+2)] [(x-2) -(x+2)]=0
[In the above step while simplifying we have to be careful with the + and – signs]
Simplify the above equation
(x-2+x+2)(x-2-x-2)=0
(2x)(-4)=0
-8x=0
x=0
Therefore p(x)= is 0
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