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Sal factors 5rs+25r-3s-15 as (s+5)(5r-3). Created by Sal Khan and Monterey Institute for Technology and Education.

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• Can you use FOIL to do this problem?
• So...FOIL "undoes" factoring and Factoring "undoes" foiling.
• I dont get this extra credit question my teacher gave me, and i am not sure if sal have videos about this type of question:
a^2 x^2 + 4ab +b^2
thank you very much!
• you just need to apply the power and the add the exponents
(1 vote)
• Doesn't FOIL stand for: First, Outside, Inner, and Last?
• Yes it does. Basically what that means is that when multiply to polynomials. In the case on (x+1)(y+2), you multiply the first (x and y), the outer (x and 2), the inner (1 and y), and the last (1 and 2). In the end you get x * y + x * 2 + 1 * y + 1 * 2. This can be simplified to xy+2x+y+2.
• how can you do the with variables which have at least variable in common. for ex; 25x-5x-15xy-3y
how can i factor 15 and 3? I know they're both divisible by 3 but, i don't know where to put the x.
• This is slightly tricky since you have a term with an xy. So here's what I'd do. I'd factor out the 5x's and leave the y's.
5x(5 - 1 - 3y) - 3y
5x(4 - 3y) - 3y
If it was just -15xy-3y alone, I would just leave the x since the second term doesn't have an x in it.
-3y(5x - 1)
Like that.
• Isn't the title of this video potentially misleading? The title is "Factoring two-variable quadratics: grouping", but the polynomial in the video 5rs+25r-3s-15 has no second degree exponent and is therefore a linear polynomial rather than a quadratic.
• It must have been factoring two variable polynomials: quadratics or grouping method.
• what does foil mean?
(1 vote)
• FOIL is a method for mulitplying polynomials within parentheses. For example:

(2x+4)(3xy+9)

You mulitply the terms inside the parentheses using FOIL, which is First Outside Inner Last.

(2x+4)(3xy+9)
2x(3xy)+2x(9)+4(3xy)+4(9)
6x^2y+18x+12xy+37

Hope this helps! :D
• I'm curious about how useful this type of technique is. It seems that the problems it can solve are very artificial - they conveniently happen to have a common factor (in this case, s + 5) that can easily be factored out. How often, when solving real-world problems that weren't designed specifically to illustrate this technique, are you going to be fortunate enough to have something where this technique works out so nicely? I feel like you'd almost always just use the quadratic formula. Can't think of any situation outside of homework where this video's technique would be useful, but if anyone can, please enlighten me.
• As far as I know, this would work on any quadratic equation. Yes, the quadratic formula would be quicker than this method, but in my opinion, this is much easier, and much more fun.
(1 vote)
• how to solve this? x^2-y^2/xy - xy-y^2/xy-x^2
(1 vote)
• You can't solve it, as it doesn't equal anything. You can simplify it to -2y/x -xy.
• How does this apply to the real world?
(1 vote)
• Any kind of engineering, computer software, and a few other random jobs require you to know this. Not this specifically, but factoring polynomials in general.
• This equation can't be factored that easily, because you would have to add some components first. By adding components I mean taking something like `+ a - a` to the equation. As `a - a = 0`, this doesn't change the value of the equation. In this case I add `- b*y + b*y - y + y`, so I can factorize the equation:
``b^2 - 2*b + 1 - y^2= b^2 - b - b + 1 - y^2 - b*y + b*y - y + y= b(b - y - 1) + y(b - y - 1) - 1 (b - y - 1)= (b + y - 1)(b - y - 1)``