Factor completely 3x ^2 - 27

First, factor out a 3.

3 (x² - 9)

In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.

In this case, we have x² + 0x - 9. (the 0x is a placeholder)

We want two numbers that add to 0 and multiply to get - 9.

Obviously, these numbers are 3 and - 3.

Now we have 3 (x² + 3x - 3x - 9).

Let's factor.

3 (x (x+3) - 3 (x+3))

3 (x-3) (x+3)

There are multiple shortcuts which you could make here, FYI:

Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number) (x+other number).

Whenever you have a difference of squares, like a²-b², that factors to (a+b) (a-b).

Both terms are multiply of 3, so you can factor 3 out;

3 (x² - 9)

Now, we can see that x² - 9 is difference of two perfect squares

Recall that a² - b² = (a+b) (a-b)

So we have x²-9 = (x+3) (x-3)

So we can factor more here; 3 (x²-9) = 3 (x+3) (x-3)

Final answer: 3 (x+3) (x-3)