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Factor completely 3x3 + 12x2 + 18x.

x2 + 4x + 6
3(x3 + 4x2 + 6x)
3x(x2 + 4x + 6)
x(3x2 + 12x + 18)

1 answer

To factor the expression \(3x^3 + 12x^2 + 18x\) completely, we can start by factoring out the common factor.

  1. Identify the common factor: The common factor in the expression is \(3x\).

  2. Factor out \(3x\): \[ 3x^3 + 12x^2 + 18x = 3x(x^2 + 4x + 6) \]

Now we have factored the original expression into \(3x(x^2 + 4x + 6)\).

Now, we need to check if \(x^2 + 4x + 6\) can be factored further. To do this, we can calculate its discriminant:

\[ \text{Discriminant} = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 6 = 16 - 24 = -8 \]

Since the discriminant is negative, \(x^2 + 4x + 6\) has no real roots and cannot be factored over the real numbers.

So the complete factorization of \(3x^3 + 12x^2 + 18x\) is: \[ \boxed{3x(x^2 + 4x + 6)} \]