Question

factor the expression of 9x2-4y2

Answer 1

\[ x^2-49\]

Answer 2

This is the difference of "2 perfect squares". -> x^2 is a square (obviously) and 49 is some particular integer squared. That part will be up to you to find it.

And when you have 2 perfect squares:

a^2 - b^2 that factors to (a - b)(a + b)

Answer 3

@romanortiz65 do you know how to factor?

Answer 4

isnt it (x-7)(x+7)

Answer 5

yup

Answer 6

Good job! You got it!

Answer 7

\[x^2 - 49 = (x+7)(x-7)\]

Answer 8

All good now, @some_someone ?

Answer 9

what about 9x2-4y2

Answer 10

lol yeah i said he was right before :P

Answer 11

I was just writing it again haha :P

Answer 12

?

Answer 13

Anyway, 9x2-4y2 can be viewed as:

(3x)^2 - (2y)^2

So, if you use the formula I wrote out for
a = 3x and b = 2y

you will get the answer.

Answer 14

(x-3)(x-2)?

Answer 15

no

Answer 16

(a - b)(a + b) = (3x - 2y)(3x + 2y)

Answer 17

ok so for 4x2-4

Answer 18

it would be

Answer 19

(4x-4)(4x+4)

Answer 20

nope.

Answer 21

wtf damn i suck at this

Answer 22

\[(4x-4)(4x+4) = 16x^2 + 16x - 16x - 16 \]

Answer 23

(4x-2)(x+2)

Answer 24

\[(4x-2)(x+2) = 4x^2 + 8x -2x - 4 = 4x^2 +6x - 4\]

Answer 25

???????????????????

Answer 26

Remember, you have to get these in the form of a^2 - b^2 first.

4x^2 - 4 = (2x)^2 - 2^2 = (a - b)(a + b)

(2x - 2)(2x + 2)

But your not done yet because you can further factor.

2(x - 1)2(x + 1) = (2)(2)(x - 1)(x + 1) = 4(x - 1)(x + 1)

With the integer factors, you "bring them back together" to get the "4".

Answer 27

so final answer would be 4(x-1)(x+1)

Answer 28

Yes, that's correct. If you have a new one, just close this question and write it in a new post.