2006-12-11 10:15:42 · 8 answers · asked by swimming_dramastar19 4 in Science & Mathematics ➔ Mathematics
It doesn't factor in REAL terms. If it did, you could solve x^2 + 81 = 0, or x^2 = - 81 in real terms. Clearly, that can't be done.
(When you are asked to factor a polynomial, it usually means the same as "break it down into as many basic REAL factors as possible." That is the traditional mathematical convention.)
IF you're allowed COMPLEX factors (and I stress once again that most people would cry FOUL!"), then you'd have:
x^2 + 81 = (x + 9i)(x - 9i),
corresponding to the two COMPLEX roots x = -/+ 9i respectively.
Of course, if the "+" in the equation SHOULD have been a "-" sign, then:
x^2 - 81 = (x + 9)(x - 9).
Both of these "factorings" are examples of the algebraic identity:
x^2 - a^2 = (x + a)(x - a),
whether 'a' is real or complex.
Live long and prosper.
P.S. I am not a regular "guy" --- and I've got an alien card to prove it!
2006-12-11 10:17:26 · answer #1 · answered by Dr Spock 6 · 1⤊ 0⤋
To answer both people above me, the first person meant it doesn't factor evenly, which it doesn't. And I agree, use the quadratic formula
x=[-b+or-sqrt(b^2-4ac)]/2a where the equation is ax^2+bx+c=0 (btw in this case a=1, b=0, and c=81, so I don't think there are any real answers :(
2006-12-11 18:22:16 · answer #2 · answered by americanmimeboy 4 · 0⤊ 0⤋
since we have a sum, this is not the difference of two squares.
x^2 and 81 have no common factors other than 1 , so, this polynomial is simplified (in lowest terms) and is prime.
note, prime numbers and prime polynomials are similar. they only have one and themselves as factors.
also, this is not an equation (it is an expression), so, the quadratic formula is not necessary here. simplify expressions and solve equations.
if an x is factored, we do not have a polynomial. it is possible to factor if we use complex numbers ... (x+9i)(x-9i)...only if complex numbers are allowed in our definition of the polynomial.
2006-12-11 18:19:05 · answer #3 · answered by mr green 4 · 0⤊ 0⤋
use the quadratic formula
x = [ -b ± â(b^2 - 4ac) ] / 2a
(x^2+81)
a=1
b=0
c=81
x = [0 ± â(0^2 - 4(1)(81)]/2(1)
x = [ ± â(-324)]/2
x = [ ± â(-1)â
â(324)]/2
x = [ ± iâ
(18)]/2
x = ± 9i
checking
(x+9i)(x-9i)= (x^2 +9ix -9ix -81i^2) =
(x^2 -81i^2) = x^2 -(81)(-1) =
(x^2 + 81) check
2006-12-11 18:28:09 · answer #4 · answered by rm 3 · 1⤊ 0⤋
x(x+81/x)
2006-12-11 18:20:49 · answer #5 · answered by deep_purple84 2 · 1⤊ 0⤋
quadratic formula that sucka!
and in response to guy above, everything factors
2006-12-11 18:17:36 · answer #6 · answered by Anonymous · 0⤊ 0⤋
(x + 9i)(x - 9i)
2006-12-11 19:28:06 · answer #7 · answered by Sherman81 6 · 1⤊ 0⤋
it will be (x+9i)(x-9i)
2006-12-11 18:21:09 · answer #8 · answered by raj 7 · 1⤊ 0⤋