+0

# Factoring

0
7
3
+1406

Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions:

2mn - 21m + 6n + 15m - 5n + ___

Dec 19, 2023

#1
+259
-1

To factor the given expression, we need to find a constant to add such that we can group the terms into two pairs with a common factor in each pair. Let's analyze the existing terms:

2mn - 21m: Both terms have a common factor of m.

6n + 15m: Both terms have a common factor of 3.

-5n: This term stands alone with no obvious way to pair it with another.

Therefore, we need to add a constant that can be paired with -5n and also fit into the existing grouping.

Two possibilities come to mind:

-9n: This creates two pairs: (2mn - 21m) + (-9n + 6n) and (15m + 3) + (-5n). Both pairs have common factors of m and 3, respectively.

-15n: This creates two pairs: (2mn - 21m) + (-15n) and (15m + 6n + 3). The first pair has a common factor of m, and the second pair combines the remaining terms conveniently.

Both options are valid, and the choice depends on what feels more elegant or convenient in the context of your specific problem. Here, adding -15n seems more concise:

2mn - 21m + 6n + 15m - 5n - 15n = (2mn - 21m + 15m) + (6n - 5n - 15n) = 6m + (-24n) = 6(m - 4n).

Therefore, the blank should be filled with -15n, resulting in a factored expression of 6(m - 4n).

Dec 19, 2023
#2
+33600
+1

As follows:

Dec 19, 2023
#3
+126978
+1

Simplify  as

2mn - 6m +n +  ____

2m ( n - 3)  + (n - 3)  =

(2m + 1) ( n - 3)  =

2mn - 6m + n  - 3  =

2mn - 6m + n +  (-3)

Dec 19, 2023