To factor the quadratic expression \(3x^2 - 6x + 9\), we start by factoring out the greatest common factor, which in this case is \(3\):
\[ 3(x^2 - 2x + 3) \]
Next, we need to factor the quadratic expression \(x^2 - 2x + 3\). To factor this, we look for two numbers that multiply to the constant term (which is \(3\)) and add to the linear coefficient (which is \(-2\)). However, there are no two real numbers that satisfy these conditions, indicating that \(x^2 - 2x + 3\) is not factorable over the real numbers.
To further analyze \(x^2 - 2x + 3\), we can find its roots using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = 3\):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(3)}}{2(1)} = \frac{2 \pm \sqrt{4 - 12}}{2} = \frac{2 \pm \sqrt{-8}}{2} = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2} \]
Since the roots are complex, the quadratic does not factor into real linear factors.
Therefore, the factored form of the expression over the reals is:
\[ 3(x^2 - 2x + 3) \]
And since \(x^2 - 2x + 3\) cannot be factored further over the reals, the final answer is:
\[ 3(x^2 - 2x + 3) \]
If you wanted to factor it over the complex numbers, you would express it as:
\[ 3\left(x - (1 + i\sqrt{2})\right)\left(x - (1 - i\sqrt{2})\right) \]
However, note that for many purposes, the expression is usually left in the real factors as shown earlier.
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