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

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Answer 1 · 2018-05-09T21:12:41

#=> (5n+4)(n+3)#

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)#

Answer 2 · 2018-05-09T21:17:02

#(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!

Answer 3 · 2018-05-09T21:47:23

Use an AC method to split the middle term, then factor by grouping to find:

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

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)#