Factorize it completely using identities:3a^7-48a^3?

Answer 1

3a^7 – 48a^3
GCD of 3 and 48 is 16,
GCD of a^7 and a^3 is a^3
therefore GCD of above expression is 3a^3, take it common outside the bracket.
3a^7 – 48a^3 = 3a^3(a^4 – 16) Noe this of the type (a^2 – b^2) = (a– b)(a+b)
3a^3(a^4 – 16) = 3a^4( a^2 – 4)(a^2 +4)
( a^2 – 4) can again factorise as (a – 2)(a + 2)
3a^7 – 48a^3
= 3a^4( a^2 – 4)(a^2 +4)
= 3a^4(a – 2)(a + 2)(a^2 +4)

Answer 2

3a^7 - 48a^3
The GCD of this expression is 3a^3. Factor it out.

3a^7 - 48a^3 = 3a^3(a^4 - 16)

a^4 - 16 = (a^2)^2 - 4^2
This is in the form x^2 - y^2 = (x + y)(x - y)
a^2 - 16 = (a^2 + 4)(a^2 - 4)

a^2 - 4 can 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 3

3a^7 - 48a^3

First, factor the greatest common monomial.

3a^3 (a^4 - 16)

Now, factor as a difference of squares.

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

Factor as a difference of squares again. The sum of squares has no factorization.

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

Answer 4

3a^3 is a common factor
3a^3( a^4-16)=
in the bracket you have a difference of squares:
3a^3(a^2+4)(a^2-4)=
continue to apply the difference of squares formula for the last bracket, and you get the final answer:
3a^3(a^2+4)(a+2)(a-2)

Answer 5

3a^7 -48a^3

= 3a^3(a^4 -16)

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

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

Answer 6

3a^7-48a^3= 3a^3(a^4-16)
=3a^3{(a^2+4)(a^2-4)} since a^2-b^2=(a+b)(a-b)

Answer 7

3a^7-48a³ =
3a³(a^4 - 16) =
3a³(a² -4)(a² + 4)
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Answer 8

3a^7-48a^3=3a^3(a^4-16)=3a^3(a^2+4)(a^2-4)
=3a^3(a^2+4)(a+2)(a-2)

Answer 9

3a^3(a+2)(a-2)(a+2)(a+2)
make sure you factor everything
3a^3(a+2)^3(a-2)
3a^3(a+2)(a^2-2a+4)(a-2)<--- answer(as far as it goes)

Answer 10

I can give u the right answer but u have already got the answers above, so there is no need to repeat the steps