2007-04-13 05:53:37 · 7 answers · asked by a100 1 in Science & Mathematics ➔ Mathematics
Sure, (x-√3)(x+√3)
2007-04-13 05:57:42 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋
No. The only form of factoring that might technically be done is with the distributive law to create : 1(x^2 - 3) but since you've failed to actually accomplish anything but add a step, it's wasted effort.
2007-04-13 13:04:58 · answer #2 · answered by Bo 2 · 0⤊ 0⤋
sq rt of 3 is not an integer, so no, we can not factor this polynomial
however, if you had x^2 - 4 we could factor into (x - 2)(x + 2)
2007-04-13 12:58:22 · answer #3 · answered by Ana 4 · 0⤊ 0⤋
If you have an expression of the form x^2 - r=0 for a non-negative real number r, Use the rule
a^2 - b^2 = (a + b)(a - b)
Take x=a, b=sqrt(3) and ...
x^2 - 3 = x^2 - sqrt(3)^2 = [x + sqrt(3)][x - sqrt(3)]
2007-04-13 12:59:46 · answer #4 · answered by Amit Y 5 · 0⤊ 0⤋
No, you would need to use the quadratic equation to solve. You can see this be realizing that the answers will be the positive and negative for the square root of three. So clearly no factoring possible.
2007-04-13 12:58:29 · answer #5 · answered by Jeffrey O 3 · 0⤊ 0⤋
You can factor it, but you wont get integers.
Using a^2 - b^2 = (a+b)(a-b)
you get x^2 - 3 = (x+sq root 3)(x- sq root 3)
2007-04-13 12:58:21 · answer #6 · answered by Happy to help 2 · 0⤊ 0⤋
Not in the ring of integer numbers, no.
2007-04-13 12:58:29 · answer #7 · answered by Alexander 6 · 0⤊ 0⤋