Math, asked by vijayanagha23, 18 days ago

m^3 - n^2 + m - m^2 n^2 factorization

Answers

Answered by shrashti25
2

Answer:

Group 1:  -mn2-n3 

Group 2:  m3+m2n 

Group 3:  m2-n2 

Pull out from each group separately :

Group 1:   (m+n) • (-n2)

Group 2:   (m+n) • (m2)

Group 3:   (m2-n2) • (1)

Looking for common sub-expressions :

Group 1:   (m+n) • (-n2)

Group 2:   (m+n) • (m2)

Group 3:   (m2-n2) • (1)

(1) + (2): (m+n)  •  (-n2+m2)  ,then adding (3)                -------------------

We end up with :

               (m+n+1)  •  (m2-n2) 

Which is the desired factorization

Trying to factor as a Difference of Squares:

 1.2      Factoring:  m2-n2 

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 :  m2  is the square of  m1 

Check :  n2  is the square of  n1 

Factorization is :       (m + n)  •  (m - n) 

Final result :

(m + n) • (m - n) • (m + n + 1)

Answered by GunvantSingh
1
  1. m³ + m - n²- m²n²
  2. m(m²+1) - n²(m²+1)
  3. (m²+1)(m-n²)

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