SOLUTION: Factor this problem completely 6x^2+28x+16

Question 232375: Factor this problem completely
6x^2+28x+16
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
First take out a common factor of 2:

Now, by trial and error, factor this by starting with 3x*x:

The Last times Last has to be 8, which is probably going to be 2*4 or 4*2. The combination that works is 2 *4 like this:

For additional help with this VERY important topic of FACTORING, please visit my own website. Do a "Bing" or "Google" search for my last name "Rapalje". Look for "Rapalje Homepage" near the top of the search list. From my Homepage, look for "Basic, Intermediate, and College Algebra: One Step at a Time." Select "Basic Algebra" and look in "Chapter 2" for several topics on Factoring, especially "Factoring the Common Factor", "Factoring Trinomials", and "Advanced Trinomial Factoring." These sections are supported by my "MATH IN LIVING COLOR" pages in which I solved problems IN COLOR!

In addition, I have FREE videos posted of me teaching this topic (a few years ago before I retired!). To see the videos, from my Homepage, look for "Rapalje Videos in Living Color". The videos and the non-traditional explanations are all FREE! I'm not selling anything!!

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

RELATED QUESTIONS
factor the algebraic expressions completely 6x^2 - 28x +... (answered by lyra)

6x^2 - 28x + 16 (answered by jim_thompson5910)

Factor:... (answered by jim_thompson5910)

Factor completely 28x^2 -29x + 6/-28x^2 + 50x... (answered by jsmallt9)

Factor Completely: 4x^2-28x+49 Thanks... (answered by smartness)

Please factor completely 4x^2 - 28x +... (answered by nerdybill)

Please help me factor this problem completely: {{{ 16-9y^2... (answered by jim_thompson5910)

Factor completely: x^3 � 2x^2 + 6x � 12 . Factor completely: x^2 + 17x + 16... (answered by stanbon,edjones)

Hello, I need some help with factoring. I have a couple of problems that are giving me a... (answered by rfer)