How do you factor by grouping `5n^2 + 19n + 12`?

Factor by grouping? This is just normal factoring, no grouping is needed:

`5n^2+19n+12 = (5n+4)(n+3)`

If you want to use factoring by grouping anyway, then you have to break up the middle term with `n` into pieces that sum to `19n` in total.

`=(5n^2+15n) + (4n + 12)`

`=5n(n+3) + 4(n+3)`

`=(5n+4)(n+3)`

When we do factoring by grouping, we want to split the `b` term (coefficient on `n`) into two terms so we can factor two separate expression.

`5n^2+color(blue)(19n)+12` can be rewritten as

`5n^2+color(blue)(15n+4n)+12`

Notice, the blue terms are equivalent, so I didn't change the meaning of this expression.

Now, we can factor!

`color(red)(5n^2+15n)+color(lime)(4n+12)`

We can factor a `5n` out of the red expression, and a `4` out of the green expression. We get

`color(red)(5n(n+3))+color(lime)(4(n+3))`

which can be rewritten as

`(5n+4)(n+3)`

Hope this helps!

Given:

`5n^2+19n+12`

Use an AC Method to find the split we need.

Look for a pair of factors of `AC = 5 * 12 = 60` with sum `B=19`.

The pair `15, 4` works in that `15 * 4 = 60` and `15+4=19`.

Using this pair to split the middle term, we can then factor by grouping:

`5n^2+19n+12 = (5n^2+15n)+(4n+12)`

`color(white)(5n^2+19n+12) = 5n(n+3)+4(n+3)`

`color(white)(5n^2+19n+12) = (5n+4)(n+3)`