m^3 - n^2 + m - m^2 n^2 factorization
Answers
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)
- m³ + m - n²- m²n²
- m(m²+1) - n²(m²+1)
- (m²+1)(m-n²)