find the zero of the polynomial f(x)=x^2-9
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Answer:
p(x) = x²-9
p(x) = x² - 3²
p(x) = (x-3)(x+3)
(x-3)(x+3) = 0
x-3 = 0
x = 3
and
x+3 = 0
x = -3
Therefore,the zeroes of the given polynomial are 3 and -3.
Hope it helps.....
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