Factorise completely: 3a² - 12b²?

Question

I've tried doing 3(a+4b)(a-4b), but that doesn't work. Notice it not only says "factorise", but it says "factorise completely".

Answer 1

Try 3a2 - 12b2 = 3(a2 - 4b2) = 3(a - 2b)(a + 2b)

Answer 2

Complete factors of:

3a² - 12b² = 3(a² - 4b² )
= 3(a + 2b)(a - 2b).

Answer 3

You had the first step right...factor out the 3 to get:

(3)*(a^2 - 4b^2)

Now the second factor is the difference of two squares, so take the square root of both terms and get:

(3)*(a+2b)*(a-2b)

(You're earlier mistake was not taking the square root of the 4)

Answer 4

3 is a common to both so
= 3(a² - 4b²)
= 3(a-2b)(a+2b)

when u expand u get

3(a² +2ab-2ab-4b²)
3(a² -4b²)
= 3a² -12b²

Answer 5

3a^7 - 48a^3 The GCD of this expression is 3a^3. factor it out. 3a^7 - 48a^3 = 3a^3(a^4 - sixteen) a^4 - sixteen = (a^2)^2 - 4^2 that's interior the style x^2 - y^2 = (x + y)(x - y) a^2 - sixteen = (a^2 + 4)(a^2 - 4) a^2 - 4 might properly be further factored a^2 - 4 = (a + 2)(a - 2) answer : 3a^7 - 48a^3 = 3a^3(a^2 + 4)(a + 2)(a - 2)

Answer 6

Take the 3 out. You are left with

3(a^2 - 4b^2) = 3(a^2 - (2b)^2) = 3(a - 2b)(a + 3b)

Answer 7

3a² - 12b²
3(a^2-4b^2)
3(a+2b)(a-2b)

Answer 8

the square root of 4 is 2 not 4:
3(a+2b)(a-2b)

Answer 9

3(a^2 - 4b^2) = 3(a-2b)(a+2b)

Answer 10

3a^2-12b^2=
3(a^2-4b^2)=
3[(a-2b)(a+2b)]

Answer 11

3a² - 12b²

3(a² - 4b²)

3(a - 2b)(a + 2b)

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