# How do you solve #3w^2 + 28w + 9 = 0# by factoring?

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To solve the quadratic equation (3w^2 + 28w + 9 = 0) by factoring, you first need to find two numbers that multiply to (3 \times 9 = 27) (the product of the leading coefficient and the constant term) and add up to (28) (the coefficient of the linear term). These numbers are (3) and (9).

Then, you rewrite the middle term using these numbers, breaking it into two terms:

[3w^2 + 27w + w + 9 = 0]

Next, factor by grouping:

[3w(w + 9) + 1(w + 9) = 0]

Now, you can see that both terms have a common factor of ((w + 9)), so you can factor that out:

[(3w + 1)(w + 9) = 0]

Now, you have a product of two factors equal to zero, so you can set each factor equal to zero and solve for (w):

[3w + 1 = 0 \implies w = -\frac{1}{3}]

[w + 9 = 0 \implies w = -9]

Therefore, the solutions to the equation (3w^2 + 28w + 9 = 0) are (w = -\frac{1}{3}) and (w = -9).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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