Factoring trinomials with a non-1 leading coefficient by grouping

u12_l2_t1_we3 Factoring trinomials with a non-1 leading coefficient by grouping

[ascrub]

I love how this youtube channel makes consistently timeless content. Not a million views on the upload date or week, but as time goes on it racks up.

[nousername]

My teacher was literal trash at teaching me this. Thank you!!

[jazzb]

Thank you so much. I think I can raise my math grade by quite a bit now. Honestly, your channel is a life savor.

[smooths]

I had forgotten (or never learned!) this basic skill from my school days, and then watched this video 2 hrs before my university end of semester exam yesterday to help with limits (calculus) questions...Thanks from Australia!!

[football.highlight.heaven]

I am proud of all you hard working students grinding! Keep up the good work!

[meggoegoo]

Your videos are always so clear and helpful, thank you

[nevin]

THANK YOU MAN!!! I missed a day of school and this helped me out

[sabotage_yt]

this is the most useful yt channel ever thank you so much for always helping me through years of math

[LumpiaBoi]

Thank you so much! I suffered through a whole quarter with this topic and now everything Thank you!

[xBrookeyNx]

Thank you for this! for some reason I don't understand it in class but the way you explain it is so much easier to understand.

[MrSupernova]

Hello Sal,

I remember watching this video last November for my pre calc course and trying to understand what the heck was going on and I could not grasp it. Today, I'm taking my second college level calc course and I can factor no problem since I have practiced much.

I believe a big part of my confusion in regards to this video is the use of a*b on their own. Factoring at the time for me was alien and I could not understand that a*b was part of the equation: a^2+2ab+c    or    a^2-2ab+c.

I get it now but I think it would be beneficial if you illustrate the rules to solve the equations first before jumping into it. Anyway, thank you for all your videos as I have benefited from them greatly while taking college level math.

[firgos]

my teacher did really good but i needed someone else to teach me it helps me remember and understand thank you.

[ericsunderland]

Thank u so much my teacher so bad this help so much thx

[readysetshrimps]

My teacher sent me this video. Straight to the point.

[viritrilbiaslayerofargus.]

that trick about the middle term being the clue that tells difference in how far apart the two factors(that we are looking for) are from eachother was the last missing piece of the puzzle I need to know!!

Now I can factor out quadratic trinomials with a leading coefficient greater than one.

[pgbrandon]

The video that explains why you find the factors of a x c is titled "Factor by Grouping and Factoring Completely."  It is around the 9 minute mark in the video.  Note that it should read "(fh)x^2", not "(fh)x".

[Doskey]

great tutorials! they help alot. the most frustrating part of algebra for me is definitely factoring, and remembering the different techniques in different situations. it drives me insane lol

[ScoobaMusic]

Thank you so much, you just helped me help my brother with his homework, it's only been a year but I couldn't for the life of me remember how to do it

[mrwhatleyohs]

Look at the videos on grouping. If you look at the two pieces 2y(2y-3) + 5(2y-3), what is the GCF? Believe it or not, (2y-3) is the GCF because it is what the two pieces have in common. When you pull the GCF and put it into the front like you are used to doing, you have (2y + 5) left over to put to the right, so it looks like (2y-3)(2y+5). Does this help?

[pgbrandon]

With all due respect to Grifir3, the order of the decomposition of the middle term does not matter. We start with 4y^2 + 10y - 6y -15. We now factor the first two terms and the last two terms. 2y(2y+5) -3(2y+5). Note that we have to factor -3 out of the last two terms so we end up with the same factor as the first set, (2y+5). We now combine the 2y and the -3 to end up with (2y-3) (2y+5), exactly the same result as before, with the order of the terms reversed, but still equal, as axb = bxa.