I don't know why anyone would need to factor a quadratic equation, but if the exam said factor it, then why memorize the X method when everyone should know the quadratic solution formula for solving ax^2+bx+c=0; x=[-b±√(b^2-4ac)]/2a? The X method is just a modified version of this formula. To illustrate let (2x/3-4/5)(5x/7+2/3)=(10/21)x^2-(8/63)x-8/15=0. The quadratic formula gives x=6/5 and x=-14/15. Thus, (x-6/5)(x+14/15)=0, and this expression may be multiplied by anything without changing it. The coefficient of x^2 is 10/21 and how may it be factored? (2*5)/(3*7). How about (2/3)(5/7)? [(2/3)(x-6/5)][(5/7)(x+14/15)]=(2x/3-4/5)(5x/7+2/3)=0 gives back the original factored form. This was a very complex example, so let's try a simpler one. (2x-3)(5x+4)=10x^2-7x-12=0. The quadratic formula gives roots x=3/2 and x=-4/5. Thus (x-3/2)(x+4/5)=0. Factor the leading coefficient 10 into 2*5 and distribute this as (2*5)(x-3/2)(x+4/5)=[2(x-3/2)][5(x+4/5)]=(2x-3)(5x+4)=0, which matches the original factored form. Compare this method with the X method to see the similarities.
@roger7341 Thanks for the great response. You are absolutely correct that there are many many ways to factor quadratics (and many reasons to: to find the zeros of your parabola to help with manual graphing, to simplify the numerator and denominator of rational functions to help identify zeros, holes, and vertical asymptotes, or simply because you are an Algebra I student and your teacher wants to build your brain by asking you to play with factoring equations in many manners without the use of a graphing calculator. But in terms of the essential similarities of all methods, check out this video that derives the Quadratic Formula by Completing the Square on the Standard form of a Parabola. ruclips.net/video/b45jSYIooVE/видео.html
If you enjoyed that one, I think you'll get a kick out of this way of factoring quadratics when the numbers don't work out in the X method: ruclips.net/video/IwzTkeD4x78/видео.html
There is another method on factoring with leading coefficient. Multiply the leading coefficient of x² to the constant. Do the usual factoring, then divide each constants with the leading coefficient of x². If one constant equals the whole number, retain the quotient as a factor. If the other constant is not divisible with the leading coefficient, copy that leading coefficient to x. Ex: 2x²+11x-6 Multiply 2 to -6, x²+11x-12 Factor out: (x-1)(x+12) Divide all constant by leading coefficient 2 from the original equation: (x-1/2)(x+12/2) Simplify: (x-1/2)(x+6)... Carry over the /2 from x-1/2 and transfer the 2 to x Answer: (2x-1)(x+6)
@barneyDcaller. This method is really intriguing, and seems to work as long as the leading coefficient is a prime number; I'm wonder how it works when the leading coefficient has multiple factors, such as 6x^2 - 7x - 5. You multiply the 6 times the -5 to get -30, factor to get (x-10)(x+3). Divide each constant by 6 and neither is an integer, nor is the answer (6x-10)(6x+3). How do you adjust this method to still work when the leading co-efficient itself has multiple factors?
@@helpwithmathing Yep! It's another method to add to the toolbox. I'm always on the lookout for calculation methods on the same concept, and also different expressions of the same relation. So this 'X' method of validating quadratic roots (to complement the columnar approach) is a good advantage in encapsulating the products, addends, subtrahends, and minuends in a nice, neat little graphic. It's really neat, I like it a lot. ^-^
@angelsmileyprettyprincesss. Thanks so much for asking!! Not only to we need them to add to -10, but we also need them to multiply to positive 24. (-12) x (2) will be -24. So that pairing won't work, leaving us with (-4) x(-6) which both adds to (--10) and multiplies to positive 24. Does that clear that up? If not, ask more questions.
Hi everyone!! Here's the link to the Factoring By Grouping video I mention in the middle of this video: ruclips.net/video/J_-P5OqobV0/видео.html
I am working my way through your videos(very slowly!).
Do you have one on how to check if the quadratic is factorable at all?
Thanks for watching, and excellent question!! I don't have one yet, but check back at the end of the day, and I'll have one posted!!
@vespa2860 Here you go!! "Is My Quadratic Factorable?"
ruclips.net/video/lrR9NZnwFZE/видео.html
Good explanation.
👍👍👍👍
Thank you so much!
Excellent
@Superpkd so glad you found it helpful!
Thanks prof❤
@drisslahlou So glad it was helpful!
Thank you. I need more examples.
So glad to be helpful. Take a look at my factoring play list: there are several different videos to give you more practice with this. :)
Al fin lo entendí, ¡Gracias!
Yay!
I don't know why anyone would need to factor a quadratic equation, but if the exam said factor it, then why memorize the X method when everyone should know the quadratic solution formula for solving ax^2+bx+c=0; x=[-b±√(b^2-4ac)]/2a? The X method is just a modified version of this formula. To illustrate let (2x/3-4/5)(5x/7+2/3)=(10/21)x^2-(8/63)x-8/15=0. The quadratic formula gives x=6/5 and x=-14/15. Thus, (x-6/5)(x+14/15)=0, and this expression may be multiplied by anything without changing it. The coefficient of x^2 is 10/21 and how may it be factored? (2*5)/(3*7). How about (2/3)(5/7)?
[(2/3)(x-6/5)][(5/7)(x+14/15)]=(2x/3-4/5)(5x/7+2/3)=0 gives back the original factored form. This was a very complex example, so let's try a simpler one.
(2x-3)(5x+4)=10x^2-7x-12=0. The quadratic formula gives roots x=3/2 and x=-4/5. Thus (x-3/2)(x+4/5)=0. Factor the leading coefficient 10 into 2*5 and distribute this as (2*5)(x-3/2)(x+4/5)=[2(x-3/2)][5(x+4/5)]=(2x-3)(5x+4)=0, which matches the original factored form. Compare this method with the X method to see the similarities.
@roger7341
Thanks for the great response. You are absolutely correct that there are many many ways to factor quadratics (and many reasons to: to find the zeros of your parabola to help with manual graphing, to simplify the numerator and denominator of rational functions to help identify zeros, holes, and vertical asymptotes, or simply because you are an Algebra I student and your teacher wants to build your brain by asking you to play with factoring equations in many manners without the use of a graphing calculator. But in terms of the essential similarities of all methods, check out this video that derives the Quadratic Formula by Completing the Square on the Standard form of a Parabola. ruclips.net/video/b45jSYIooVE/видео.html
this was great, thank you!!!
I'm thrilled you found it helpful! Thanks for letting me know.
@@helpwithmathing Yes!! This way of solving is great, and your teaching skills are on point!!! 😊😊 Thank you so much!!
If you enjoyed that one, I think you'll get a kick out of this way of factoring quadratics when the numbers don't work out in the X method: ruclips.net/video/IwzTkeD4x78/видео.html
There is another method on factoring with leading coefficient. Multiply the leading coefficient of x² to the constant. Do the usual factoring, then divide each constants with the leading coefficient of x². If one constant equals the whole number, retain the quotient as a factor. If the other constant is not divisible with the leading coefficient, copy that leading coefficient to x.
Ex: 2x²+11x-6
Multiply 2 to -6, x²+11x-12
Factor out: (x-1)(x+12)
Divide all constant by leading coefficient 2 from the original equation: (x-1/2)(x+12/2)
Simplify: (x-1/2)(x+6)... Carry over the /2 from x-1/2 and transfer the 2 to x
Answer: (2x-1)(x+6)
@barneyDcaller. This method is really intriguing, and seems to work as long as the leading coefficient is a prime number; I'm wonder how it works when the leading coefficient has multiple factors, such as 6x^2 - 7x - 5. You multiply the 6 times the -5 to get -30, factor to get (x-10)(x+3). Divide each constant by 6 and neither is an integer, nor is the answer (6x-10)(6x+3). How do you adjust this method to still work when the leading co-efficient itself has multiple factors?
I dig this method.
@quandarkumtanglehairs4743 Terrific! Thanks for watching and glad it was helpful!
@@helpwithmathing Yep! It's another method to add to the toolbox. I'm always on the lookout for calculation methods on the same concept, and also different expressions of the same relation. So this 'X' method of validating quadratic roots (to complement the columnar approach) is a good advantage in encapsulating the products, addends, subtrahends, and minuends in a nice, neat little graphic.
It's really neat, I like it a lot. ^-^
Fantastic
Ready for more? Check out factoring using the x method, even when the squared coefficient is greater than 1! ruclips.net/video/r8JJ50wdCJA/видео.html
Mathematics 10C!
@brandonlillo9849 Thanks for watching and boosting!
Why not( -12x ) +2x? i.e-12x+2x
@angelsmileyprettyprincesss. Thanks so much for asking!! Not only to we need them to add to -10, but we also need them to multiply to positive 24. (-12) x (2) will be -24. So that pairing won't work, leaving us with (-4) x(-6) which both adds to (--10) and multiplies to positive 24. Does that clear that up? If not, ask more questions.