Factoring by the box method Example 3

Factoring 6x^2+17x-14 by the box method

[grvdbkv]

This was the most organized and helpful video I have seen on this concept, thank you so much!

[halfjaw]

THANK YOU. I'm in Community College and my professor never even mentioned this method, what a lifesaver.

[GettoJ]

This seems so much easier the way you taught it in this video than it did in my actual class. Thanks a bunch.

[kamilamexicano]

You have no idea how much you have helped me sir .Thank you so much!!!

[JustCrazy-fujk]

omg thank you sir thank you soo much holy cow! why don't teachers teach like you?

[jasonmathisjefferson]

this is so great . i didnt understand much about the factoring concept, but u really helped alot plz keep doing this ur the best !!!

[phoebemckain]

What sorcery is this? How did I not discover this method earlier?!

[MajesticRecaps]

This is a smartest way in solving factors

[ashengamage]

This video gives me what I wanted . Thanks !

[elithemiar]

My teacher didn't teach us this until the 'reverse foil with grouping' I had issues until I found this. It helps with the number combinations

[fubaralakbar]

Nice! Here's a one-size-fits-all (as long as it can be factored and has a solution) method that I use:

Start by computing all the factor pairs of your c-term.

6x^2+17x-14
14 1
2 7

Now compute the factors of your a-term

6x^2+17x-14
6 1 14 1
3 2 2 7

Now we have to figure out our signs. The 14 is negative, meaning that one and only one of its factors will be negative. This means that we must *subtract* to get our middle term, not add.

So the next step is to multiply out our factor pairs. Remember: there are two ways that each pair can multiplied: straight multiplication, as 6(14)-1(1), OR cross-multiplication, as 6(1)-1(14). We can see that the first will give us 83, and the second -8, so that would not work. Now we try 6(2)-1(7) and 6(7)-1(2), which give us 5 and 40 respectively. Neither of these work either, and now we have exhausted our c-term factor pairs for the first a-term factor pair. 6 and 1 will not work, period. So we move on to the 3 and 2 a-term factor pair. 3(2)-2(7)=6-14=-8, 3(7)-2(2)=17. And we have found our factor pairs: 3x and 2x, 2 and 7.

Now we look at our signage again. We know that one of our c-term factors has to be negative, but which one? Well, we have to subtract here, so we need to start with a number higher than our positive b-term. That would be 21: 3*7. Since the the 3 is positive, the 7 must also be positive, so that we can get positive 21. That means the 2 is negative. So now we have: 3x, 2x, 7, -2.

Here it gets a bit squirrelly, if you don't watch what you are doing. I call this the law of opposites: if you cross multiplied to get your b-term, you must straight-multiply to get the factorization. If you straight-multiplied to get your b-term, you have to cross-multiply to get your factorization. We had to cross multiply to get our 17 (3*7-2*2), so we do it the other way to get our factors:

(3x-2)(2x+7)=6x^2+21x-4x-14

And there you have it. Your middle terms add up, and you can set your binomials to 0 and solve! You'll get x=-7/2 or 2/3.

[animlateef]

omg thankksss i looked at so many videos on the box method, and this is the ojly one i understooodd THIANKS SO

[cxvila]

Come to my school and replace my algebra 2 teacher

[elisavetdemetriou]

very good way as it's quicker from just quessing values.. thanks

[starponys]

Here is a simple "irreducible" polynomial: x^2 + 5. Here is how to factor it: (x+i√5)(x–i√5). (The "i" stands for √-1, which is a rotation of 90° — as opposed to 180° for -1.)

[starponys]

If the polynomial can be factored in the normal sense, the above method works. It's amazing.

But if the polynomial is "prime, " so to speak, you will know. That is when the method won't give a solution. The actual solution, if you insist on "factoring, " would then have irrational coefficients. It would be akin to finding the square root of a prime number.

[nepalnepali]

1:24 because 14 is negative we have to take opposite sign ? what if 6 is also negative?

[mjk]

what if you have a prime number at the end of your equation?

[stevematson]

Very good but what if the middle term was 16X

[ProJoker]

omg thnk u much. didnt underatand this crap at all till this. sooo easy now. would this count as doing algebra, if told to show algebric working. plz reply thnx