Help please: Find the real zeros and their multiplicities for the function using factoring g(x)=x^6-81x^2
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In general, a number k is a zero if (x-k) is a factor. If (x-k)^a is a factor, then its multiplicity is a. i.e. a number's multiplicity is like the number of times it is a factor.
So we need to factor g(x) as far as possible, and see what we get:
g(x) = x^6 - 81 x^2
= x^2(x^4 - 81)
Now here is a nice trick to remember: (x^2 - a^2) = (x-a)(x+a). So we can factor further:
= x^2(x^2 - 9)(x^2 + 9)
= x^2(x - 3)(x + 3)(x^2 + 9)
= (x-0)^2(x - 3)(x + 3)(x^2 + 9)
x^2 + 9 has no real zeroes, you can easily check this from the quadratic formula, or simply by thinking about it: x^2 + 9 = 0 means x^2 = -9, but squares of real numbers are never negative.
(x-0)^2 is a factor, so 0 is a zero with multiplicity 2
(x-3) is a factor, so 3 is a zero with multiplicity 1
(x+3) is a factor, so -3 is a zero with multiplicity 1
So we need to factor g(x) as far as possible, and see what we get:
g(x) = x^6 - 81 x^2
= x^2(x^4 - 81)
Now here is a nice trick to remember: (x^2 - a^2) = (x-a)(x+a). So we can factor further:
= x^2(x^2 - 9)(x^2 + 9)
= x^2(x - 3)(x + 3)(x^2 + 9)
= (x-0)^2(x - 3)(x + 3)(x^2 + 9)
x^2 + 9 has no real zeroes, you can easily check this from the quadratic formula, or simply by thinking about it: x^2 + 9 = 0 means x^2 = -9, but squares of real numbers are never negative.
(x-0)^2 is a factor, so 0 is a zero with multiplicity 2
(x-3) is a factor, so 3 is a zero with multiplicity 1
(x+3) is a factor, so -3 is a zero with multiplicity 1