Factor completely 3x^2+9x-3 3(x^2+3) 3(x^2+3x-1) D 3x(x^2+3x-1) Prime

To factor the quadratic expression \(3x^2 + 9x - 3\) completely, we first look for any common factors among the terms. In this case, we can factor out a 3 from each term:<br /><br />\[ 3x^2 + 9x - 3 = 3(x^2 + 3x - 1) \]<br /><br />Next, we need to factor the quadratic expression \(x^2 + 3x - 1\). We look for two numbers that multiply to \(-1\) (the constant term) and add up to \(3\) (the coefficient of the linear term). However, there are no such integer pairs that satisfy these conditions.<br /><br />Since \(x^2 + 3x - 1\) cannot be factored further using integers, the completely factored form of the original expression is:<br /><br />\[ 3(x^2 + 3x - 1) \]<br /><br />So, the correct answer is:<br /><br />\[ 3(x^2 + 3x - 1) \]