Ask Question
8 November, 02:56

Factor completely 3x ^2 - 27

+1
Answers (2)
  1. 8 November, 03:10
    0
    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)
  2. 8 November, 06:39
    0
    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).
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Factor completely 3x ^2 - 27 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers