How to factorize x^4 + 5x^2 + 9 ?

x⁴ + 5x² + 9

= x⁴ + 5x² + 9 + x² - x², adding and subtracting x²

= x⁴ + 5x² + x² + 9 - x², using commutativity

= (x⁴ + 6x² +9) - x² , adding 5x² and x²

= [ (x²)² + 2.x².3 + 3² ] - x²,  using (A^m)ⁿ = A^mn and factorizing 6

= (x² + 3) ² - x², using (A + B)² = A² + 2AB + B²

= (x² + 3 - x)(x² + 3 + x), using (A - B)² = (A - B)(A + B)

= (x² - x + 3)(x² + x +3), Ans. , using commutativity

How to factorize 9x^2 - 4a^2 + 4ay - y^2

9x² - 4a² + 4ay - y²
= 9x² - (4a² - 4ay + y²), taking ' - ' common
= (3x)² - (2a - y) ² , since A² - 2AB + B² = (A - B)²
= [3x - (2a -y)][3x + (2a - y)], using A² - B² = (A - B)(A + B)
= (3x - 2a + y)(3x + 2a - y) Ans. , using commutativity

Factorization of (x - 2)(x + 2) + 3

(x - 2)(x + 2) + 3
= (x^2 - 2^2) + 3, since (A -B)(A + B) = A^2 - B^2
=  x^2 - 4 + 3
= x^2 - 1
= x^2 - 1^2, since 1^2 = 1
= (x - 1)(x + 1) Ans. using A^2 - B^2 = (A -B)(A + B)

How to factorize x^2 + 1/x^2 - 11

x^2 + 1/x^2 - 11

= x^2 + 1/x^2 - 9 - 2

= (x^2 - 2 + 1/x^2) - 9 , using commutativity and associativity

= (x^2 - 2x * 1/x + 1/x^2) - 3^2, since x * 1/x = 1

= (x - 1/x)^2 - 3^2, using A^2 -2AB + B^2 = (A - B)^2

= (x - 1/x + 3)(x - 1/x -3), Ans. using A^2 - B^2 = (A + B)(A - B)

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.

Factorization of (a^2-b^2)(c^2-d^2)-4abcd

Dear Learners, I present an easy method to solve the given problem.
(a^2-b^2)(c^2-d^2)-4abcd
=a^2(c^2-d^2)-b^2(c^2-d^2)-4abcd, using distributive property in this and the next step
=a^2c^2-a^2d^2-b^2c^2+b^2d^2-4abcd
=a^2c^2+b^2d^2-2abcd-a^2d^2-b^2c^2-2abcd, using commutativity
=(a^2c^2+b^2d^2-2abcd)-(a^2d^2+b^2c^2+2abcd), using (A+B)^2=A^2+2AB+B^2,etc.
=(ac-bd)^2-(ad+bc)^2
=(ac-bd+ad+bc)(ac-bd-ad-bc) Ans.
To arrive at the last step, I used the property A^2-B^2=(A+B)(A-B)

Factorization of x^6+63x^3-64

Solution: Let x^3=a
So, x^6+63x^3-64
=(x^3)^2+63x^3-64,using A^MN=(A^M)^N
=a^2+63a-64,by substitution
=a^2+64a-a-64
=a(a+64)-1(a+64)
=(a+64)(a-1),taking (a+64) as common
=(x^3+64)(x^3-1),substituting back
=(x^3+4^3)(x^3-1^3)
=(x+4)(x^2-4x+4^2)(x-1)(x^2+1x+1^2),using A^3+B^3=(A+B)(A^2-AB+B^2),etc
=(x+4)(x^2-4x+16)(x-1)(x^2+x+1) Ans.