Question

Factor completely 3x^2+9x-3
3(x^2+3)
3(x^2+3-1)
3x(x^2+3x-1)
prime​

Answer 1

Answer from the answer data to choose from

3(x² + 3x - 1)

Factor completely

`3\left(x+(3-√(13))/(2)\right)\left(x+(3+√(13))/(2)\right)`

Explanation:

`3x^2+9x-3=(3)(x^2)+(3)(3x)-(3)(1)\\\\=(3)(x^2+3x-1)\\\\\text{For}\ x^2+3x-1\ \text{use the quadratic formula}\\\\x=(-b\pm√(b^2-4ac))/(2a)\\\\x^2+3x-1\to a=1,\ b=3,\ c=-1\\\\x=(-3\pm√(3^2-4(1)(-1)))/(2(1))=(-3\pm√(9+4))/(2)=(-3\pm√(13))/(2)=-(3\pm√(13))/(2)\\\\3x^2+9x-3=3\left(x+(3-√(13))/(2)\right)\left(x+(3+√(13))/(2)\right)`