How do you factor #2m + mn + 14 + 7n#?

Answer 1

The expression can be factored by creating groups of two terms:

#(2m + mn) + (14 + 7n)#

#m# is a common factor to both the terms in the first group, and #7# is a common factor to both the terms in the second group

# = m(2+n) + 7(2+n)#

The binomial #2+n# is common to both the terms above:

# = (2+n)(m+7)#

# = color(green)((n+2)(m+7)#

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Answer 2

Isaiah Anthony

The answer is #(m+7)(2+n)# .

Problem: Factor #2m+mn+14+7n#

Since there isn't a GCF, factor using grouping.

#(2m+mn)+(14+7n)#

Factor out #m# from the first expression.

#m(2+n)+(14+7n)#

Factor out #7# from the second expression.

#m(2+n)+7(2+n)#

#(2+n)# is the GCF.

The factorization in its entirety is:

#(m+7)(2+n)#

Apply the FOIL method to verify.

#(m+7)(2+n)=2m+mn+14+7n#

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Answer 3

Abigail Allen

To factor the expression (2m + mn + 14 + 7n), you can group the terms:

Group the terms with a common factor: [ (2m + mn) + (14 + 7n) ]

Factor out the common factors from each group: [ m(2 + n) + 7(2 + n) ]

Notice that both terms now have a common factor of (2 + n). Factor out (2 + n): [ (2 + n)(m + 7) ]

So, the factored form of (2m + mn + 14 + 7n) is ((2 + n)(m + 7)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.