x2 + 3x + 9 has no real solutions. Applying the quadratic formula, we get two imaginary solutions: (negative three plus or minus three i times the square root of three) divided by two.
x = -1.5 + 2.598076211353316i
x = -1.5 - 2.598076211353316i
where i is the square root of negative one.
(x - 9)(x + 6)
9X2 = 3X(3X) Best I can see to factor here.
(x + 4)(x + 9)
x - 3
9x^2-6xy+y^2 = (y-3x)^2 (y-3x)^2-81 -> ((y-3x)+9)((y-3x)-9)
3(x + 3)(x2 - 3x + 9)
x2-3x+2.25 is the same as 4x2-12x+9 and is (2x-3)(2x-3) when factored
First, it is important to regroup, so I am going to rearrange this equation: (x4 - 7x2 - 18 - 3x3 + 27x I can now factor the first three terms and the last two terms: (x4 - 7x2 - 18) becomes (x2 - 9)(x2 + 2) -3x3 + 27x becomes -3x(x2 - 9); so the new equation looks like: (x2 - 9)(x2 + 2) - 3x(x2 - 9) From here, factor out what is common, in this case- x2- 9. Therefore, you will have (x2 - 9)(x2 + 2 - 3x), which can be rearranged to (x2 -9)(x2 - 3x + 2). Further factoring reveals (x + 3)(x - 3)(x - 1)(x - 2) as the final answer.
9x(x2+3x-10) * * * * * 9x3 + 27x2 - 90x = 9x(x2 + 3x - 10) = 9x(x + 5)(x - 2).
-3x3 - 6x2 + 189x = -3x(x2 + 2x - 63) = -3x(x + 9)(x - 7)
(x - 9)(x + 6)
If that's + 81, the answer is 3(x + 3)(x2 - 3x + 9) If that's - 81, the answer is 3(x - 3)(x2 + 3x + 9)
(x - 3)(x2 + 3x + 9)
(x - 3)(x2 + 3x + 3)
4x3+12x2+3x+9
9x2 + 27x - 36 = 9(x2 + 3x - 4) = 9(x + 4)(x - 1)
-51