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Why are you not able to factor the polynomial: 7m + 6mn + 3m^2 + 14n by grouping?

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3 Answers

This is factorable by grouping. One thing to consider is that terms are commutative, meaning they can be rearranged. 

7m + 6mn + 3m2 + 14n 

Rearrange terms and group 1st two and 2nd two==> (7m + 14n) + (6mn + 3m2)

From each group, factor out the GCF ==> 7(m + 2n) + 3m(2n + m)

Notice that the quantities in bold are identical (this let's you know you did the technique correctly). So now, the GCF of this new expression is the quantity (2n + m), so let's factor that out.

(2n + m)(7 + 3m) ==> note that if you distribute the bold quantity, you'd get the above expression again.

 

It looks like you can:

3m2 + 6mn + 7m + 14n

3m(m + 2n) + 7(m + 2n)

(3m + 7)(m + 2n)

3m^2 +3mn + 7m +14m:
   Just look at the common factor between terms
    
    3 m^2 + 3mn  + 7m +14n
 
      greatest common factor between 1st and 2nd/ also between 3rd & 4th is 3m, therefore
     
    3m(m +n) +7 ( m+n)
 
      now (m+n) is GCF between 2 terms:
 
           (m+n)(3m+7)