SOLUTION: How do you find the roots of the polynomial," x^4-81=0"? I don't understand this because supposedly with a degree of four there are exactly four roots to this polynomial, but there

Question 574632: How do you find the roots of the polynomial," x^4-81=0"? I don't understand this because supposedly with a degree of four there are exactly four roots to this polynomial, but there are only two terms.
Answer by mathsmiles(68) (Show Source): You can put this solution on YOUR website!
Tricky!

Let's not solve for X right away, but X^2. Does that help?
X^4-81=0
(x^2 - )(X^2 + ) = 0 (We need to find two factors of 81.)
9x9 = 81 Aha!
(X^2 - 9)(X^2 + 9) = 0 Now factor as you would normally, but using X^2 as terms

Left paren first:
X^2 - 9 = 0
X^2 = 9
Sqrt(X^2) = Sqrt(9)
X = 3 or -3

Right paren:
X^2 + 9 = 0
X^2 = -9
Depending on your grade, either this is impossible - or an irrational possibility. ;-) I'm going to guess you're not into irrational numbers yet. But if you are, you can still take it from here. By the way, this is how we get more than the two solutions we did above.

Checking:
X^4 - 81 = 0
(3)^4 - 81 = 0
81 - 81 = 0 Correct!

AND
X^4 - 81 = 0
(-3)^4 - 81 = 0
81 - 81 = 0 Correct!
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