multiply

Step by step solution :

Step  1  :

 Multiply 4 by c2-ab 

Trying to factor as a Difference of Squares :

 1.1      Factoring:  c²-ab 

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 :  a¹   is not a square !! 
Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step  1  :

 (((2•(a2))-bc)•2)-(4•(c2-ab)•(b2-ac)) = 0 

 

Step  2  :

 Multiply 4•(c2-ab) by b2-ac 

Trying to factor as a Difference of Squares :

 2.1      Factoring:  b²-ac 

Check :  b²  is the square of  b¹ 

Check :  a¹   is not a square !! 
Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step  2  :

 (((2•(a2))-bc)•2)-4•(c2-ab)•(b2-ac) = 0 

 

Step  3  :

 Multiply 2a2-bc by 2 

Trying to factor as a Difference of Squares :

 3.1      Factoring:  2a²-bc 

Check :  2  is not a square !! 

Ruling : Binomial can not be factored as the
difference of two perfect squares

Equation at the end of step  3  :

 2 • (2a2 - bc) - 4 • (c2 - ab) • (b2 - ac) = 0 

Step  4  :

Simplify 2•(2a2-bc) - 4•(c2-ab)•(b2-ac) 

Pulling out like terms :

 4.1     Pull out like factors :

   -4a²bc + 4a² + 4ab³ + 4ac³ - 4b²c² - 2bc  = 

  -2 • (2a²bc - 2a² - 2ab³ - 2ac³ + 2b²c² + bc) 

Trying to factor by pulling out :

 4.2      Factoring:  2a²bc - 2a² - 2ab³ - 2ac³ + 2b²c² + bc 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  2b²c² - 2ab³ 
Group 2:  2a²bc - 2a² 
Group 3:  -2ac³ + bc 

Pull out from each group separately :

Group 1:   (ab - c²) • (-2b²)
Group 2:   (bc - 1) • (2a²)
Group 3:   (2ac² - b) • (-c)
Looking for common sub-expressions :Group 1:   (ab - c²) • (-2b²)
Group 3:   (2ac² - b) • (-c)
Group 2:   (bc - 1) • (2a²)

Bad news !! Factoring by pulling out fails : 

The groups have no common factor and can not be added up to form a multiplication.

Equation at the end of step  4  :

 -2 • (2a2bc - 2a2 - 2ab3 - 2ac3 + 2b2c2 + bc) = 0 

Step  5  :

Solve -2•(2a2bc-2a2-2ab3-2ac3+2b2c2+bc) = 0 

Equations which are never true :

 5.1      Solve :    -2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 5.2     Solve   2a²bc-2a²-2ab³-2ac³+2b²c²+bc  = 0 

In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.

We shall not handle this type of equations at this time.