Question

please solve this algebraic expression by factorisation by grouping. ab(c^2+d^2)-a^2cd-b^2cd

Answer 1

ab(c2-d2)-a2cd-b2-2b  

Final result :

 -a2cd + abc2 - abd2 - b2 - 2b

Step by step solution :

Step  1  :

Trying to factor as a Difference of Squares :

1.1      Factoring:  c2-d2  

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =  

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.  

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  c2  is the square of  c1  

Check :  d2  is the square of  d1  

Factorization is :       (c + d)  •  (c - d)  

Equation at the end of step  1  :

 ((ab•(c+d)•(c-d)-a2cd)-b2)-2b

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  -a2cd + abc2 - abd2 - b2 - 2b  =  

 -1 • (a2cd - abc2 + abd2 + b2 + 2b)  

Final result :

 -a2cd + abc2 - abd2 - b2 - 2b