Factorization of x^4 + 4

x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2, adding and subtracting 4x^2 (the value remains unchanged)
= (x^4 + 4x^2 + 4) - 4x^2
= (x^2 + 2)^2 - 4x^2, since (A + B)^2 = A^2 + B^2
= (x^2 + 2)^2 - (2x)^2
= (x^2 + 2 -2x)(x^2 + 2 +2x), using (A^2 - B^2) = (A - B)(A + B)
= (x^2 - 2x + 2)(x^2 + 2x +2) Ans. , using commutativity

Factorize 3x^5 - 48x

3x^5 - 48x
= 3x(x^4 - 16), taking 3x common
= 3x[(x^2)^2 - 4^2]
= 3x[(x^2 - 4)(x^2 + 4)], since A^2 - B^2 = (A - B)(A + B)
= 3x[(x^2 - 2^2)(x^2 + 4)]
= 3x[(x - 2)(x + 2)(x^2 + 4)], using, again, A^2 - B^2 = (A - B)(A + B)
= 3x(x - 2)(x + 2)(x^2 + 4) Ans.

How to factorize x^4 + 3x^2 + 4

Since 3x^2 can be written as 4x^2 - x^2, we can write

x^4 + 3x^2 + 4 = x^4 + 4x^2 + 4 - x^2

= (x^2 + 2) ^ 2 - (x)^2 , using (A + B)^2 = A^2 + 2AB + B^2 where A = x^2 and B = 2

= (x^2 + 2 + x) (x^2 + 2 - x) , using A^2 - B^2 = (A + B)(A - B)

= (x^2 + x + 2) (x^2 - x + 2)  Ans., using commutativity

How to factorize x^3 + 3x^2 + 3x - 7

Sometimes by adding or subtracting a term we may get a perfect cube or a perfect square, so in the present case we can take
 x^3 + 3x^2 + 3x - 7
= x^3 + 3x^2 + 3x + 1 - 1 - 7 ,by adding and subtracting 1 ,
so that the value of the expression does not change
= x^3 + 3x^2 + 3x + 1  - 8
= (x^3 + 3x^2 + 3x +1) - 8, by associativity
= (x+1)^3 - 2^3, since (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3
= (x + 1 - 2) [ (x+1)^2  + (x+1)2 + 2^2 ], since A^3 - B^3 = (A-B) (A^2 + AB + B^2)
= (x - 1) ( x^2 + 2x + 1 + 2x +2 +4) , using (A + B) ^2 = A^2 + 2AB + B^2
= (x - 1) (x^2 + 4x + 7) Ans., adding like terms

Factorization of a^7 + ab^6

a^7 + ab^6 = a (a^6 + b^6) , using distributive law

= a [ (a^2) ^ 3 + (b^2) ^ 3) ], using A^(M*N) = (A^M) ^ N

=a [ (a^2 + b^2) (a^4 - a^2 b^2 + b^4) ], using A^3 + B^3 = (A+B) ( A^2 -AB + B^2)

= a (a^2 + b^2) (a^4 - a^2 b^2 + b^4) Ans.

Factorization of a^6 - b^6

Using the property A^(M*N)=(A^M)^N,we get

a^6 - b^6 = (a^3)^2 - (b^3) ^ 2

= (a^3 - b^3)(a^3 + b^3) , applying A^2 - B^2 = (A - B)(A + B)

= [(a - b)(a^2 + ab + b^2)][(a + b)(a^2 - ab + b^2)] , using A^3 - B^3 = (A - B)(A^2 + AB + B^2)

= (a - b) (a + b) (a^2 + ab + b^2) (a^2 - ab + b^2) Ans.

The last step was arrived at by using associativity and commutativity of multiplication.