Algebra II (2018 edition)
- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Intro to grouping
- Factoring quadratics by grouping
- Factor quadratics by grouping
- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Factor polynomials: quadratic methods
- Factoring two-variable quadratics
- Factoring two-variable quadratics: rearranging
- Factoring two-variable quadratics: grouping
- Factor polynomials: quadratic methods (challenge)
Sal factors 4y^2+4y-15 as (2y-3)(2y+5) by grouping. Created by Sal Khan and Monterey Institute for Technology and Education.
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- i really can not bare the curiosity of why the tecnic Sal uses(in problem 4y^+4y-15,
a+b=4*15)makes sense. why do you have to multiply the product of y^ and 15?
i get that if you multiply that way you can solve the problems but i'm pretty sure that the mathmetician who dicovered this didn't figure out is suddenly with no thought.
i know it works but can someone please tell me WHY it works?(46 votes)
- Any equation with a factored form of (ax+b)(cx+d) will multiply, by distribution, to get acx^2 + (ad + bc)x + bd.
You can then multiply the coefficient of x^2 and the constant (ac*bd) like the instructor suggests.
Notice that this is all multiplication a*c*b*d, therefore, using the commutative property, ac*bd=ad*bc.
ad*bc is the product of the two numbers, ad and bc, who's sum is the middle term of the trinomial.(25 votes)
- Can somebody tell me why grouping works? I get that it's factoring, but how do we know to split the middle number into factors of the multiple of the first coefficient and the last constant? Are there any videos or links explaining this?(12 votes)
- When a number is written such that,
It can also be factorize as
as we factorize it we get first factor as ab
and the 2nd and 3rd factor as ax+bx.
So we're kinda just doing the reverse of it for quadratic polynomial like these by finding two number which satisfy both ab and ax+bx.
Hope it helps :D(5 votes)
- Do you have any tutorial videos on factoring by GCF?(11 votes)
- Yes he does, it starts with Factoring and the Distributive Properties, and then he does two more videos to hit the point home, in Factoring and the Distributive Properties 2, and Factoring and the Distributive Properties 3. Enjoy and have fun with it!(8 votes)
- I still don't get where the minus 60 is coming from...(11 votes)
- For equations in the form
ax^2 + bx + c, you need to find two numbers that multiply together to get
a*cand add together to get
For more see: https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-by-grouping/v/factor-by-grouping-and-factoring-completely(2 votes)
- What would you do if it was a positive 15 instead of a negative 15?(7 votes)
- no, that equals 4y^2 + 16y + 15
You can not factor that except for taking a 4 out of it. If you try to solve for the zeros you cant do that either. You end up with a negative square root in the quadratic equation.(8 votes)
- How do you continue factoring this: 2x(x+4)+5(x+2)(5 votes)
- so at the part where he had the 2 factors -6 and 10 how did you know which factor to put down first(2 votes)
- The order doesn't matter. The only thing thatt matters is that the 2 terms need to add back to the original value (in this case 4y). I'll reverse the 2 terms and redo the problem in the video so that you can see.
4y^2 + 10y - 6y - 15
2y (2y + 5) - 3 (2y + 5)
(2y + 5)(2y - 3)
These are the same 2 factors that Sal created in the video.
Hope this helps.(5 votes)
- I don’t really get this. I really need help on the skill “Compound Inequalities”, but I feel like Sal is talking way too fast and I don’t get what he is saying. Help me please!(3 votes)
- Is it possible for an equation to not be possible to factor?(2 votes)
- Yes if you are talking about the set of real numbers. Factors are related to the x intercept, and it is possible to draw a parabola which does not intersect or touch the x intercept. In that case, the discriminant b^2 - 4ac would be negative.(2 votes)
We're asked to factor 4y squared plus 4y, minus 15. And whenever you have an expression like this, where you have a non-one coefficient on the y squared, or on the second degree term-- it could have been an x squared-- the best way to do this is by grouping. And to factor by grouping we need to look for two numbers whose product is equal to 4 times negative 15. So we're looking for two numbers whose product-- let's call those a and b-- is going to be equal to 4 times negative 15, or negative 60. And the sum of those two numbers, a plus b, needs to be equal to this 4 right there. So let's think about all the factors of negative 60, or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative, so when you take two numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. All I did is expand this into these two numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y. Negative 15 divided by five is 3. And so we have 2y times 2y minus 3, plus 5 times 2y minus 3. So now you have two terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3, so this is equal to 2y minus 3, times 2y, times that 2y, plus that 5. There's no magic happening here, all I did is undistribute the 2y minus 3. I factored it out of both of these guys. I took it out of the parentheses. If I distribute it in, you'd get back to this expression. But we're done, we factored it. We factored it into two binomial expressions. 4y squared plus 4y, minus 15 is 2y minus 3, times 2y plus 5.