Question

Factor by grouping.......................

Answer 1

For this case we have the following xpresion:

$2y ^ 2 + 3y + 4y + 6$

We can rewrite it as:

$2y ^ 2 + 7y + 6$

Where:

$a = 2\\b = 7\\c = 6$

We must factor, for this we follow the steps below:

Step 1:

The term of the medium must be rewritten as the sum of two terms, whose sum is 7 and the product is $a.c = 2 * (6) = 12$:

Then, the term of the medium, fulfilling the two previous conditions, can be written as:

$4y + 3y$

We check:

$4 * 3 = 12\\4 + 3 = 7$

So, we have:

$2y ^ 2 + 4y + 3y + 6$

Step 2:

The maximum common denominator (the largest integer that divides them without leaving residue) of each group is factored

$2y ^ 2 + 4y + 3y + 6\\2y (y + 2) + 3 * (y + 2)$

Step 3:

We take common factor $(y + 2)$:

$(y + 2) (2y + 3)$

Thus, the expression $2y ^ 2 + 3y + 4y + 6$ can be factored as:

$(y + 2) (2y + 3)$

Answer:

$(y + 2) (2y + 3)$