Man, I am in college retaking a course and was never taught this properly in high school. No one has ever taught me this method. You just taught me in 11 minutes what my professor hasn't two years in a row. Thank you. You deserve my tuition, not her.
I've gotten frustrated doing this so many times because it was too confusing for me but now it's like quick magic! This is perfect for visual learners thank you so much!!!
Thank you for the help. My algebra teacher teaches me the X thing where on the top you multiply to get that number and the bottom add to get it. It’s useful, but not when it comes to this! Thanks! What I had to solve was 10x^2+13x+4 divides by 2x+1 -> (5x+4)(2x+1) Answer: 5x+4
I finally get it thanks to the box method! Thank you so much, my head was scrambled egg before this after watching lots of other videos, now it seems so easy with this method!!!! Many many thanks!
Wow this was very helpful. I was wondering could you post a few more examples. I factoring quadratic equations is literally keeping me from passing college algebra. This is my 3rd attempt and I want to pass.
Thank you so much for making this video! It was so incredibly helpful for me! I was really confused, but now I am confident that I know how to do this! Please keep making awesome videos like this!
I'm an A Level student studying in the UK and this method is amazing. I've always struggled with Quadratics. If you could, could you post more examples that would be awesome!
Question: If the 15x was negative you would drag the negative sign correct? So you would have a negative on both sides. I did this and my answer was wrong. 5y2-3y-14 = (5y-7)(y-2) but the correct answer was (5y+7)(y-2) why is this? Thank you in advance. This video REALLY HELPED ME OUT.
From my link in the comments, please watch my video & from that method you'd get the following: 21b^2-48b-45... all three terms are divisible by 3 so GCF that out, giving you 3(7b^2-16b-15)... multiply the new a & c coefficients so 7 * -15 = -105... factors of -105 that add to -16 would be 5 & -21... divide them (as a fraction by the leading a-coefficient of 7, thus 5/7 & -21/7... reduce the fractions, thus 5/7 & -3/1... in the binomial factors, the denominators go in front of the variables & the numerators go after, thus (7b+5) & (1b-3)... put everything together physically, thus 3(7b+5)(b-3) is your factored expression. Same process for your 12z^2+7z-12 problem (always write in descending degree) but there's no GCF to worry about first... 12 * -12 = -144... factors of -144 that add to 7 are 16 & -9... put them over the leading a-coefficient of 12, thus 16/12 & -9/12... reduce both fractions, thus 4/3 & -3/4... denominators in front of the letters, numerators after, thus (3z+4)(4z-3) is your final factored answer. Distribute/FOIL both answers & double-check. IF both of your problems were equations set = to zero, then your solutions would simply be the opposites of those reduced fractions that made the factorings, thus your first problem's solutions in this case would've been b=-5/7 & b=3... and the second problem's solutions would've been z=-4/3 & z=3/4. :-)
If you go to my video link in the comments & try this problem that way, then it would be: 12 * -2 = -24... factors of -24 that add to -5... -8 & 3... divide those (as a fraction) by the leading a-coefficient of 12, thus -8/12 & 3/12... reduce those fractions, thus -2/3 & 1/4. Factored answers have the form of (ax+b)(cx+d). From those reduced fractions (-2/3 & 1/4), the denominators go in front of the variable, the numerators after, thus (3x-2)(4x+1), or with your variables (3p-2)(4p+1). Redistribute/FOIL that & see that it'll get you your original problem. If your problem was 12p^2-5p-2=0, then your solutions are simply the opposites of the two reduced fractions, thus p=2/3 & p=-1/4. :-)
MrMathTutorials, thank you for the swift solution. I honestly have doubted my capabilities with certain math areas due to frustration. Your video was on point, I'm definitely going to check out more of your videos also a word of advice, your videos would be more noticed if you looked into Tyler Dewitt's science channel, it could really help Sincerely, J
Hello. Can I also ask, what if I will extract this quadratic equation: 2x²=50 where b=0 How?? I followed your progress but in the checking it's wrong My answer: (2x+5) (x-5) Checking answer: 2x² -5x -50
When setting up the equation to solve it this way, you would have to set it equal to zero, and then take out any greatest common factor in the binomial. Since it would be 2x^2 -50 =0 at that point, you would factor out a 2 since that's their common term. This would leave the equation as 2(x^2-25)=0, and you would only factor what remains in the parentheses. After factoring that, you would get 2(x-5)(x+5), which if multiplied, gets you back to your original polynomial.
Thank you Mr. Math! I have been searching for help on how to do this for what seems like hours! I have been stuck in my Aleks program for two days due to this type of problem! Thank you Thank you.
Then the quadratic trinomial can't be factored... it would be prime & already simplified as an expression. You could solve it as a quadratic equation set = to zero if you use completing the square or quadratic formula.
@@bruhfridc1957 In the video, he's talking about ax^2 + bx + c, which is an expression. Actually, ax^2 + bx + c is not the same as ax^2 + bx + c = 0. So, the video is on factoring quadratic trinomials; it's common student misconception, though.
Yeah, it does, but i just figured out, whenever the sign before c is negative, switch the sign located on the top of the box (not the sign on the side. And this is once you've dragged out the signs from the numbers, that's when you switch the sign. Also, if the sign before b is negative, then also switch the sign, but the sign on the side of the box.)
Go to my video link in the comments to view a better way. Following the method in my video, your problem would work out as such: Factor out a GCF of -1 from all 3 terms, thus -1(3x^2-8x+4)... multiply the new a & c-coefficients, thus 3 * 4 = 12... factors of 12 that add to -8 are -6 & -2... divide those numbers (as a fraction) by the new a-coefficient of 3, thus -6/3 & -2/3... reduce the fractions, thus -2/1 & -2/3... for the binomial factors, the denominators go before the variable & the numerators go after them, thus (1x-2) & (3x-2)... now put everything physically together, thus -1(x-2)(3x-2) is your final, factored form. Applying the -1 to either factor can also be acceptable, thus either (-x+2)(3x-2) or (x-2)(-3x+2) can be "correct" answers as well. IF your problem was set = to zero & you had to solve, then your solutions would simply have been the opposites of those reduced fractions that made the binomial factors, thus x=2 & x=2/3. :-)
Man, I am in college retaking a course and was never taught this properly in high school. No one has ever taught me this method. You just taught me in 11 minutes what my professor hasn't two years in a row. Thank you. You deserve my tuition, not her.
I've gotten frustrated doing this so many times because it was too confusing for me but now it's like quick magic! This is perfect for visual learners thank you so much!!!
i have to watch this video because my math teacher doesnt know how to teach
also the box method is so helpful thank very much
Or you don't know how to learn.
@@myimorata7678 No such thing as a bad student. Only a bad teacher.
I was a god at math before they started including the alphabet
Same man it used to be my favorite subject now sucks
Lol..i can relate as well
HAHAHA
It used to be my favorite, with science as a close second. Now science easily wins, and math is like my second to last(I really don't like ELA)
🤣🤣🤣
When you put the 5 down I was like... “I don’t get it; this is probably a new way/formula” So glad you changed it to 1. 🙂
lmao
I’m an AP Calculus student this year and I just realized that I never quite got down how to factor when a isn’t one. Thanks!
an 11 minute video just summed up a whole semester of algebra 2 lmao
I am learning this in Algebra 1
I am learning this in Algebra 1
@@longing_hiraeth me too
DerKelee 😂😂😂😂
Wow this saved me I am a good math student but factoring always seemed so random to me, this way is concrete and works every time.
Came in clutch for college level pre calculus (havent taken a math in 2 years) so thank you so much for this.
This explanation really needs patience and attention, yet it serves as a true solution to what students have had difficulties with
As soon as I saw "5" there I moved to the comments section.🤣🤣🤣
Thank you for this video, it's really helped me brush up on my (non-existent rip) factoring knowledge! :)
SUPER HELPFUL
extremely helpful. I have not taken algebra in years and I've watched every tutorial and yours was the easiest to understand!!!!
Try to solve it in a simple way..that's why are so many people are afraid of math because of that kind of solution..
I agree
thank you so much dude. saved my life frfr.
Thank you for the help. My algebra teacher teaches me the X thing where on the top you multiply to get that number and the bottom add to get it. It’s useful, but not when it comes to this! Thanks! What I had to solve was 10x^2+13x+4 divides by 2x+1 -> (5x+4)(2x+1) Answer: 5x+4
Thank God for you
Thanks man you helped me really alot with that 11 min video.thank you with love from egypt❤️❤️
I finally get it thanks to the box method! Thank you so much, my head was scrambled egg before this after watching lots of other videos, now it seems so easy with this method!!!! Many many thanks!
Wow this was very helpful. I was wondering could you post a few more examples. I factoring quadratic equations is literally keeping me from passing college algebra. This is my 3rd attempt and I want to pass.
Omg!! Thank you this method has saved me so much time as opposed to doing so much guessing.
Thank you so much for making this video! It was so incredibly helpful for me! I was really confused, but now I am confident that I know how to do this! Please keep making awesome videos like this!
i kept getting the answers wrong, all because i kept forgetting to simplify. thank you man!
I really like this method. Thank you
Bless your soul
I'm an A Level student studying in the UK and this method is amazing. I've always struggled with Quadratics. If you could, could you post more examples that would be awesome!
Thank you so much for sharing this method!!!!
Be careful when using the box method if the GCF for all three terms is not 1. You must factor out the GCF first or you answer will not be correct.
Awesome Video Thanks
Question: If the 15x was negative you would drag the negative sign correct? So you would have a negative on both sides. I did this and my answer was wrong. 5y2-3y-14 = (5y-7)(y-2) but the correct answer was (5y+7)(y-2) why is this? Thank you in advance. This video REALLY HELPED ME OUT.
I know there is probably a simple answer to this but, what is the only variable is y? do I just do the same thing?
Thank you!
Well, Don't you put =0 since it is a Quadratic Equation?
This amazing
This is madness! 28x^2-46x+16. The correct answer is (4x-2)(7x-8). I keep getting (4x-2)(7x-16).
dorianjepsen it's divisible by coefficient of x² or a
factor out the 2 first
2(7x−8)(2x−1) -- That is the correct answer.
Absolute madness.....
ur a god
Where did you get the 15?
thank u so much sir, hope u were my homeschool teacher
if you could cut to the point, your videos would improve (in my point of view)
keep up the good work
many thanks
Thanks for this it really means alot
can you possibly do these 2 problems so I can understand better. I have never seen this method before and I love it.
21b^2-48b-45
and then
7z+12z^2-12
From my link in the comments, please watch my video & from that method you'd get the following: 21b^2-48b-45... all three terms are divisible by 3 so GCF that out, giving you 3(7b^2-16b-15)... multiply the new a & c coefficients so 7 * -15 = -105... factors of -105 that add to -16 would be 5 & -21... divide them (as a fraction by the leading a-coefficient of 7, thus 5/7 & -21/7... reduce the fractions, thus 5/7 & -3/1... in the binomial factors, the denominators go in front of the variables & the numerators go after, thus (7b+5) & (1b-3)... put everything together physically, thus 3(7b+5)(b-3) is your factored expression. Same process for your 12z^2+7z-12 problem (always write in descending degree) but there's no GCF to worry about first... 12 * -12 = -144... factors of -144 that add to 7 are 16 & -9... put them over the leading a-coefficient of 12, thus 16/12 & -9/12... reduce both fractions, thus 4/3 & -3/4... denominators in front of the letters, numerators after, thus (3z+4)(4z-3) is your final factored answer. Distribute/FOIL both answers & double-check. IF both of your problems were equations set = to zero, then your solutions would simply be the opposites of those reduced fractions that made the factorings, thus your first problem's solutions in this case would've been b=-5/7 & b=3... and the second problem's solutions would've been z=-4/3 & z=3/4. :-)
this guy is god
Wow! This helps a lot. Could you please make another video with another example. Such as 12p^2-5p-2?
If you go to my video link in the comments & try this problem that way, then it would be: 12 * -2 = -24... factors of -24 that add to -5... -8 & 3... divide those (as a fraction) by the leading a-coefficient of 12, thus -8/12 & 3/12... reduce those fractions, thus -2/3 & 1/4. Factored answers have the form of (ax+b)(cx+d). From those reduced fractions (-2/3 & 1/4), the denominators go in front of the variable, the numerators after, thus (3x-2)(4x+1), or with your variables (3p-2)(4p+1). Redistribute/FOIL that & see that it'll get you your original problem. If your problem was 12p^2-5p-2=0, then your solutions are simply the opposites of the two reduced fractions, thus p=2/3 & p=-1/4. :-)
MrMathTutorials, thank you for the swift solution. I honestly have doubted my capabilities with certain math areas due to frustration. Your video was on point, I'm definitely going to check out more of your videos
also a word of advice, your videos would be more noticed if you looked into Tyler Dewitt's science channel, it could really help
Sincerely, J
THANK YOU SO MUCH!!!
I love you bro
Hello. Can I also ask, what if I will extract this quadratic equation:
2x²=50 where b=0
How?? I followed your progress but in the checking it's wrong
My answer: (2x+5) (x-5)
Checking answer: 2x² -5x -50
By the way, thanks
When setting up the equation to solve it this way, you would have to set it equal to zero, and then take out any greatest common factor in the binomial. Since it would be 2x^2 -50 =0 at that point, you would factor out a 2 since that's their common term. This would leave the equation as 2(x^2-25)=0, and you would only factor what remains in the parentheses. After factoring that, you would get 2(x-5)(x+5), which if multiplied, gets you back to your original polynomial.
Thank you Mr. Math! I have been searching for help on how to do this for what seems like hours! I have been stuck in my Aleks program for two days due to this type of problem! Thank you Thank you.
When youtube teaches you better than an actual school it's honestly so embarrassing.
thank you for this
You're a life saver. Truly!
Can anyone help with 6X(square 2) - 7X + 12. thanks in advance.
whos here cuz of corona :(
INSANE
I got a different answer. I got (x+4)(2x+2). When FOILed it comes back to the original quadratic of 2x^2+10x+8. How is his answer right?
Thanks
U can factor your 2x+2 into 2(x+1) there you go :)
Is it expression or equation clear this🎉
Now I can explain a method of factoring :D
What if the factors of A•C dont add to B and they are not all divisible by the leading coefficient
Peacok then you have to use the quadratic formula I think
Then the quadratic trinomial can't be factored... it would be prime & already simplified as an expression. You could solve it as a quadratic equation set = to zero if you use completing the square or quadratic formula.
THANK YOU!!!!!
i have a simple way to do it it will be so easy than that square
i will make it and send it it will be easy for students
I like factoring quadratics with grouping
Yeah same it's way easier
did not work for me
I thought each parenthesis after factoring has to have the same thing
Don’t quadratic equations all equal 0?
I have a problem what will i do when c is equals to zero?
solve exactly the same just make sure it adds up to zero
ooooooooooooooooh i get it now
In our country we were taught this in 9th class
This is the most complicated way ever, anyone else struggling with this go watch the hegartymaths video as that method is so much more simpler.
it's not an "equation"; it's just an algebraic expression.
Anna Amosova it’s an equation because it is ax^2 + bx + c = 0
@@bruhfridc1957 In the video, he's talking about ax^2 + bx + c, which is an expression. Actually, ax^2 + bx + c is not the same as ax^2 + bx + c = 0. So, the video is on factoring quadratic trinomials; it's common student misconception, though.
My problems don't work for these methods.
You'll have to use other methods like the using the quadratic equation or completing the square and a few others I forgot.
We learned a way called bottoms up, and it's the quickest way I have learned to do this process.
what? i got (x + 4).(2x + 2) for the first one
8x^2+26x+20. The correct answer is (2x+4)(4x+5). I've followed your method and keep getting (2x+4)(8x+10). Why?
factor out the 2 first
The correct answer can't be (2x+4)(4x+5) cause you gotta factor it. It'd be 2(13x+42)
No, the correct, complete factoring will be 2(x+2)(4x+5)... take the GCF of 2 out of your (2x+4).
i love u
What happened to 14x???
Wonder why it isn't this easy to learn in class.
this is not an equation
now is right i didn't know you put there 1 sorry
I agree
-3x^2+8x-4 doesnt work with this.
Yeah, it does, but i just figured out, whenever the sign before c is negative, switch the sign located on the top of the box (not the sign on the side. And this is once you've dragged out the signs from the numbers, that's when you switch the sign. Also, if the sign before b is negative, then also switch the sign, but the sign on the side of the box.)
Wait what? What sign of what number should I change?
Go to my video link in the comments to view a better way. Following the method in my video, your problem would work out as such: Factor out a GCF of -1 from all 3 terms, thus -1(3x^2-8x+4)... multiply the new a & c-coefficients, thus 3 * 4 = 12... factors of 12 that add to -8 are -6 & -2... divide those numbers (as a fraction) by the new a-coefficient of 3, thus -6/3 & -2/3... reduce the fractions, thus -2/1 & -2/3... for the binomial factors, the denominators go before the variable & the numerators go after them, thus (1x-2) & (3x-2)... now put everything physically together, thus -1(x-2)(3x-2) is your final, factored form. Applying the -1 to either factor can also be acceptable, thus either (-x+2)(3x-2) or (x-2)(-3x+2) can be "correct" answers as well. IF your problem was set = to zero & you had to solve, then your solutions would simply have been the opposites of those reduced fractions that made the binomial factors, thus x=2 & x=2/3. :-)
Alarm
This method doesn't work on all quadratic equations, unfortunately.
I want my zeros...
what
the first one is wrong
How to factorize Quadratic Equations with Complex Solutions?
For example x**2 + 4*x + 5 = 0
This method is useless and only can confuse students!
I agree
Such a stupid method!
These are quadratic EXPRESSIONS not EQUATIONS. Ugh. You couldn't even get the terminology right.
u took too long.