Me again, dumb question. If the "a" coefficient has more than 2 factors (such as, 4x^2), how would you know if the factorization should be (4x + ___) (x + ___), or (2x + ___) (2x + ___) ? Thanks.
@chrishelbling3879 There are no dumb questions!! The brilliance of this method is it takes care of that for you in the execution. No more trial and error. Just factor the biggest thing out of both terms when you get to the factor by grouping portion and it will always work out. Take a peek at this earlier video I made that shows an example with 4x^2 as its leading term (forgive the camera going out of focus a bit: I learned as I went along). If you don't want to watch the whole video you can jump to 6min in for this example. Happy watching! ua-cam.com/video/r8JJ50wdCJA/v-deo.html
That's the beauty (and power) of this method. In the beginning you multiply a and c. That means you have already dealt with the x^2 coefficient and all its factors. Believe it or not! No more hassling with multiple possibilities where the x^2 coefficient is concerned. By splitting the x term into two parts, you can convert the trinomial into two polynomials. Here's an example: 6x^2 - 5x - 4 First multiply a times c =-24 Then figure what combination of factors of 24 will add or subtract to -5. Those will be 3 and -8. Now convert the -5x into -8x and +3x to make: 6x^2 +3x -8x -4 Group into two polynomials: (6x^2 + 3x) + (-8x -4) Factor these. The GCF for the first polynomial is 3x. The GCF for the second polynomial is -4. So factored, you get 3x(2x + 1) - 4(2x+1) So far, everything is the same as in the video. Now you factor out the GCF of these two polynomials. That's 2x+1. (Or just use the distributive property.) That gives (2x+1)(3x-4) Now what? HEY! You're done!!
Thanks for alerting me to the two mis-speaks!! So glad what I wrote on the whiteboard is correct. I’ll see if I can go into the UA-cam editor and fix those or use subtitles to note the things I said incorrectly but wrote correctly!
Pardon my comment, but it strikes me you call this the "X method," yet the variable is also an "x," do your students ever get confused by this nomenclature? Perhaps you might consider a goofier name for the process, like the Criss-cross method, or the Burried Treasure method, or whatever? But hey, if your students follow with you OK, then good job.
you know, some people call it the AC method, but that sounds so clinical to me. The X Method sounds exciting to me, but I take your point about x also being the name of the variable. I like the Criss-cross Method, maybe I'll adopt that. Thanks for the musings!
Thank you v much Mum .... you' great professor.
@naderhumood1199. Oh I'm so glad it helped!!
Me again, dumb question. If the "a" coefficient has more than 2 factors (such as, 4x^2), how would you know if the factorization should be (4x + ___) (x + ___), or (2x + ___) (2x + ___) ? Thanks.
@chrishelbling3879 There are no dumb questions!! The brilliance of this method is it takes care of that for you in the execution. No more trial and error. Just factor the biggest thing out of both terms when you get to the factor by grouping portion and it will always work out. Take a peek at this earlier video I made that shows an example with 4x^2 as its leading term (forgive the camera going out of focus a bit: I learned as I went along). If you don't want to watch the whole video you can jump to 6min in for this example. Happy watching!
ua-cam.com/video/r8JJ50wdCJA/v-deo.html
@@helpwithmathing trust me, there are dumb questions. :D
That's the beauty (and power) of this method. In the beginning you multiply a and c. That means you have already dealt with the x^2 coefficient and all its factors. Believe it or not! No more hassling with multiple possibilities where the x^2 coefficient is concerned.
By splitting the x term into two parts, you can convert the trinomial into two polynomials. Here's an example:
6x^2 - 5x - 4
First multiply a times c =-24
Then figure what combination of factors of 24 will add or subtract to -5. Those will be 3 and -8.
Now convert the -5x into -8x and +3x to make:
6x^2 +3x -8x -4
Group into two polynomials: (6x^2 + 3x) + (-8x -4)
Factor these. The GCF for the first polynomial is 3x. The GCF for the second polynomial is -4. So factored, you get 3x(2x + 1) - 4(2x+1)
So far, everything is the same as in the video.
Now you factor out the GCF of these two polynomials. That's 2x+1. (Or just use the distributive property.)
That gives (2x+1)(3x-4)
Now what? HEY! You're done!!
@@Astrobrant2 Well said!
@@Astrobrant2
f(x) = 6x² - 5x - 4 ⇔
f(x) = ([6x]² - 5[6x] - 24)⅙ ⇔
f(x) = ([6x + 3][6x - 8])(⅓•½) ⇔
f(x) = (2x + 1)(3x - 4)
■
Excellent
So glad you found this helpful!!
✌️❣️
@MathsSirg Thanks for the support!
2 errors in the commentating, you said 12 x squared, (5 30 in)and at the end, you said X - 4 (6 16)
Thanks for alerting me to the two mis-speaks!! So glad what I wrote on the whiteboard is correct. I’ll see if I can go into the UA-cam editor and fix those or use subtitles to note the things I said incorrectly but wrote correctly!
@SuperJemser OK! Used Subtitles to put in the corrections. Thanks for the keen ear and sorry for the mis-speak.
* XX - 6X - 16
16 = 2 x 8 = 4 x 4
8 - 2 = 6 = ( b)
* 8 X 8 - 6 x 8 - 16
64 - 48 - 16 = 0.
* 2 x 2 -6 x 2 - 16 # 0
(-2)(-2) - 6(-2) - 16
4 + 12 - 16 = 0
X' = 8 , X" = - 2 .
Thank you so much for these demonstrations!
Pardon my comment, but it strikes me you call this the "X method," yet the variable is also an "x," do your students ever get confused by this nomenclature? Perhaps you might consider a goofier name for the process, like the Criss-cross method, or the Burried Treasure method, or whatever? But hey, if your students follow with you OK, then good job.
you know, some people call it the AC method, but that sounds so clinical to me. The X Method sounds exciting to me, but I take your point about x also being the name of the variable. I like the Criss-cross Method, maybe I'll adopt that. Thanks for the musings!
3x² + 11x- 4 = 0 -> x² + 11x -4.3 = 0 or x² + 11x -12 = 0 12 = 12x1 and 12-1 =11 so (x+12) . (x-1) = 0
dividing the answer with 3 x = - 12/3 = -4 or x = 1/3
Absolutely. Check this out, and see if I am teaching the same method you are here: ua-cam.com/video/XFxyiJjeCvg/v-deo.html
f(x) = 6x² - 5x - 4 ⇔
f(x) = ([6x]² - 5[6x] - 24)⅙ ⇔
f(x) = ([6x + 3][6x - 8])(⅓•½) ⇔
f(x) = (2x + 1)(3x - 4)
■
3x² + 11x - 4 = 0 ⇔
3•(3x² + 11x - 4) = 3•0 ⇔
(3x)² + 11(3x) - 12 = 0 ⇔
Let U = 3x
U² + 11U - 12 = 0 ⇔
(U - 1)(U + 12) = 0 ⇒
U - 1 = 0 i.e. 3x - 1 = 0
x = ⅓
or
U + 12 = 0 i.e. 3x + 12 = 0
x = -4