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Is x(4x^(2)+8x+12) completely factored? If not, what other factoring

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The expression
x(4x^2 + 8x + 12)is not completely factored. To factor it further, we can use the distributive property. Let's start by factoring out the common factor x from the expression. When we do this, we get x times
(4x^2 + 8x + 12).

Now, let's look at the expression inside the parentheses, which is
4x^2 + 8x + 12.To factor this trinomial, we can look for two numbers whose product is equal to the product of the first and last term
(4x^2 * 12 = 48x^2) and whose sum is equal to the coefficient of the middle term (8x).

In this case, the numbers we're looking for are 6 and 8 because 6 * 8 = 48 and 6 + 8 = 14. So, we can rewrite the expression as
x(4x^2 + 6x + 8x + 12).Next, we can factor by grouping. We group the first two terms (4x^2 and 6x) together and the last two terms (8x and 12) together.

Taking out the common factor from the first group gives us
2x(2x + 3).Taking out the common factor from the second group gives us
4(2x + 3). Now, we can see that we have a common binomial factor of (2x + 3) in both terms. Finally, we can rewrite the expression as
x(2x + 3)(2x + 4).In conclusion, the completely factored form of
x(4x^2 + 8x + 12) is x(2x + 3)(2x + 4).

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