- Intro to grouping
- Factoring by grouping
- Factoring quadratics by grouping
- Factoring quadratics: leading coefficient ≠ 1
- Factor quadratics by grouping
- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Creativity break: How can we combine ways of thinking in problem solving?
Intro to grouping
Sal introduces the method of grouping, which is very useful in factoring quadratics whose leading coefficient is not 1. Created by Sal Khan and CK-12 Foundation.
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- Does this kind of factoring work for all problems?(81 votes)
- It works for problems with 3 terms. With 4 terms you need to factor by grouping(83 votes)
- What about a problem with four terms and no common factor where you have to factor by grouping, like: 3x^3 - x^2 +18x - 6? My book only gives examples with 3 terms and a common factor and the gives problems at the end of the chapter like that one...(21 votes)
- In this case you factor as he did after he went through his little process to create four terms, but you don't do that little process. You group the terms: (3x^3 - x^2) + (18x - 6) and factor out what you can from each term: x^2(3x - 1) + 6(3x - 1). Now you go on and factor out the common factor: (3x - 1)(x^2 + 6). I hope this answered your question, I was a little iffy on what exactly you meant.(30 votes)
- At9:35, wouldn't ' fx * hx ' be equal to ' (f*h)*x^2 ' (f times h times x-squared) instead of just ' fhx ' ?
Take ' (2*3) * (2*4) ' , for example: that would be ' (3*4)*2^2 ' (3 times 4 times 2-squared), NOT ' (3*4)*2 ' , right?(24 votes)
- Indeed it should be fhx^2. There's an annotation pointing out the error a few seconds later, though.(17 votes)
- 1:23Why does he multiply 4 by -21?(20 votes)
- He is solving using this trick. The mystery numbers (a,b) must be multiplied to get -84 (4 times -21) and when added, will get 25 (the middle number).(13 votes)
- Does Sal mean "a x c" since the equation is ax^2+bx+c and does he mean simply "b" because 4+25= 29. I learned this in class but I wanted to make sure. So can someone clear this up you can find it in1:00to1:42(14 votes)
- This video explains more clearly why a and b are used: https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-quadratic-expressions/v/factoring-quadratic-expressions(7 votes)
- What is the significance of learning this method of grouping? Why not learn how to factor these quadratics with the quadratic formula?(4 votes)
- When you practice this and other methods of grouping/factoring, your intuition of how to factor and what factors are likely will grow so that you won't need to use the quadratic formula, you will be able to "see" or "intuit" what the possible factors could be. Surprisingly, a high number of students make careless errors using the quadratic - it is best to use it only when all else fails.
Developing your mathematical intuition is the best thing you can do for yourself when learning math.(18 votes)
- Did Sal mistook at9:35? I think fx times hx is fhx^2 not fhx(5 votes)
- You are correct, but this mistake has been corrected with a text box in lower corner within 2 seconds of the error. It does not show up if you are on full screen.(8 votes)
- The only thing I don't fully understand is how it works out that we multiply "c" or the constant in the polynomial by the coefficient of x^2. I didn't understand when Sal was explaining using all letters at the 9 minute mark. I tried reverse engineering by starting with a quadratic with an "ax" other than 1, but I still don't see how it means that the last term needs to be multiplied by the leading coefficient. The last term is not made up by multiplying it with any x term when expanding. Example: In 3x^2+17x+10. Why would the 10 need to be multiplied by 3? I expanded this from (3x+2)(x+5) and I still don't see it. Please help?(4 votes)
- (3𝑥 +2)(1𝑥 + 5) = (3 ∙ 1)𝑥² + (3 ∙ 5 + 2 ∙ 1)𝑥 + (2 ∙ 5)
As we multiply the coefficient of the 𝑥²-term with the constant term we get (3 ∙ 1) ∙ (2 ∙ 5),
which we can also write as (3 ∙ 5) ∙ (2 ∙ 1)
Now compare this to the coefficient of the 𝑥-term, 3 ∙ 5 + 2 ∙ 1
We have found that the product of the coefficient of the 𝑥²-term and the constant term is also the product of two other numbers whose sum is equal to the coefficient of the 𝑥-term.(7 votes)
- When factoring a trinomial in the form of ax^2 + bx + c, why do we multiply the "a" by the "c" ? I have been struggling to understand the logic behind this process for a very long time. Please help! :((5 votes)
- Do you know FOIL (first, outside, inside, last)? To get the O and I, we have to multiply factors of a times factors of c to get these two terms. If O and I are factors, then the number that get these two factors is a*c. In other words, I*O =a*c and I + O = b. We always have to multiply a*c, we just ignore that fact if a = 1 because multiplying by 1 does not change a number.
Hope this helps(1 vote)
- I've rewatched this video so many times. I still don't understand why this works or what I'm even supposed to do in order to factor out an expression like this. Why does ab need to be equal to -84? Why do we even multiply a and b together, or 4 and -21? I'm totally lost, could someone please point me in the right direction with a better resource/less confusing explanation?(3 votes)
- The point is that we need to split the middle term into two middle terms, with coefficients a and b, in such a way that 1) the four-term expression is equivalent to the trinomial, and 2) the four-term expression can be factored by grouping.
To satisfy 1), we clearly need to make a+b equal to the coefficient on the middle term of the trinomial. In this particular problem, we need a+b=25.
To satisfy 2), we need to have the ratio of the first term to the second term equal to the ratio of the third term to the fourth term (otherwise, there won't be a common factor to factor out in the final step of factoring by grouping!). This means that we need the product of the two middle coefficients (i.e. ab) equal to the leading coefficient times the constant term. So in this particular problem, we need ab=4(-21).
Have a blessed, wonderful day!(4 votes)
In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as the leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or negative 1 where this 4 is sitting. All of a sudden now, we have this 4 here. So what I'm going to teach you is a technique called, factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. To some degree, it'll become obsolete once you learn the quadratic formula, because, frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique. And then at the end of this video, I'll actually show you why it works. So what we need to do here, is we need to think of two numbers, a and b, where a times b is equal 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a plus b, need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is. So we go, 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or, if you had negative 4 and 21, you'd get positive 17. Doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference. Because that's essentially what you're going to do, if one is negative and one is positive. Too far apart. Let's see you could do 3-- I'm jumping the gun. 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 42 is negative 40-- too far apart. 3 and-- Let's see, 3 goes into 84-- 3 goes into 8 2 times. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. Goes exactly 8 times. So 3 and 28. This seems interesting. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25. Now, we've found our two numbers. But it's not going to be quite as simple of an operation as what we did when this was a 1 or negative 1. What we're going to do now is split up this term right here. We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there, and you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here, and the negative 3 to go there? And it actually does matter. The way I thought about it is 3 or negative 3, and 21 or negative 21 , they have some common factors. In particular, they have the factor 3 in common. And 28 and 4 have some common factors. So I grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we, literally, group these so that term becomes 4x squared plus 28x. And then, this side, over here in pink, it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21, or the negative 21, because they're both divisible by 3. And I grouped the 28 with the 4, because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x-- 4x squared divided by 4x is just x-- plus 28x divided by 4x is just 7. Now, this second term. Remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus, x plus 7 times negative 3. So we can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial. But you could view this could be like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I can just leave this as a minus sign. Let me delete this plus right here. Because it's just minus 3, right? Plus negative 3, same thing as minus 3. So what can we do here? We have an x plus 7, times 4x. We have an x plus 7, times negative 3. Let's factor out the x plus 7. We get x plus 7, times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping. And we factored it into two binomials. Let's do another example of that, because it's a little bit involved. But once you get the hang of it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, which is equal to 6. And we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. What are the-- well, the obvious one is 1 and 6, right? 1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Or let me not even assign them. The numbers here are 1 and 6. Now, we want to split this into a 1x and a 6x. But we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squar ed here, plus-- and so I'm going to put the 6x first because 6 and 6 share a factor. And then, we're going to have plus 1x, right? 6x plus 1x equals 7x . That was the whole point. They had to add up to 7 . And then we have the final plus 1 there. Now, in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times-- 6x squar ed divided by 6x is just an x. 6x divided by 6x is just a 1. And then, the second group-- we're going to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm afraid to use the a's and b's. I think that'll confuse you, because I use a's and b's here. They won't be the same thing. So let me use completely different letters. Let's say I have fx plus g, times hx plus, I'll use j instead of i. You'll learn in the future why don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx which is fhx. And then, fx times j. So plus fjx. And then, we're going to have g times hx. So plus ghx. And then g times j. Plus gj. Or, if we add these two middle terms, you have fh times x, plus-- add these two terms-- fj plus gh x. Plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative 1 coefficient here, we look for two numbers that add up to this, whose product is equal to the product of that times that. Well, here we have two numbers that add up-- let's say that a is equal to fj. That is a. And b is equal to gh. So a plus b is going to be equal to that middle coefficient. And then what is a times b? a times b is going to be equal to fj times gh. We could just reorder these terms. We're just multiplying a bunch of terms. So that could be rewritten as f times h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So a plus b will be equal to the middle coefficient. And a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now, I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is to teach you to factor things a little bit more completely. And this is a little bit of a add-on. I was going to make a whole video on this. But I think, on some level, it might be a little obvious for you. So let's say we had-- let me get a good one here. Let's say we had negative x to the third, plus 17x squared, minus 70x. Immediately, you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to third power. And the first thing you should realize is that every term here is divisible by x. So let's factor out an x. Or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times-- negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. The x's cancel out. And now, you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1. We just have to find two numbers whose product is 70, and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70. You add them up, you get negative 17. So this part right here is going to be x minus 10, times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out. And that'll get it into a form that you might recognize. Hopefully, you found this helpful. I want to reiterate what I showed you at the beginning of this video. I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient. But to some degree, you're going to find out easier ways to do this, especially with the quadratic formula, in not too long.