Question

Factor completely: 3x^2 - 27

Answer 1

@MDoodler

Answer 2

@mathmale

Answer 3

@ParthKohli

Answer 4

@shrutipande9

Answer 5

@AkashdeepDeb

Answer 6

@dan815

Answer 7

hey @hartnn

Answer 8

so i know that i have to factor out the GCF

The terms \(\large\cal\color{lime}{3x^2~and~27~have~a~GCF~of~3}\).

Answer 9

i know that i have to determine also x^2 - 9 is a Difference of Two Squares
so 3(x^2-9) but i dont think that three doesnt have a perfect square

Answer 10

Hey @Law&Order

You can right this as:

\[3x^2 - 27\]

Now what can you take in common?

*This is what I say in my head*
{
Well, I see a 3 an \(x^2\) and a 27, what can I do here?
Idea! 3 factorizes 27, because 3 * 9 = 27 itself. So let me take 3 in common.
Then let me see what I can do.
}

My spidey senses tell me to take 3 in common.

So, we get:
\[3x^2 - 27 = 3(x^2 - 9)\]

*Talking to myself intensifies*
{
This does not look so right.
Maybe I can factorize more, can I, really?
OH YES! I CAN!
}

If you remember, there is an identity for perfect squares: \(a^2 - b^2 = (a+b)(a-b)\)
And we can use it over here.

So we get,
\[3(x^2 - 9) = 3(x-3)(x+3)\]

Getting this? :)
Do you think we can factorize more?

Answer 11

oh i think we can factoize more

Answer 12

but first i have to check my factors

Answer 13

\(\color{red}{\large\ (3(x + 3)(x - 3) = 3[x^2 - 3x + 3x - 9]\ }\)
= 3[x^2 - 3x + 3x - 9]
= 3[x^2 - 9]
= 3x^2 - 27 \(\color{red}{\huge\ ✓}\)

Answer 14

so the answer is
\(\huge\ 3(x + 3)(x - 3)\)

Answer 15

Yes. :)