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# 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.

## Want to join the conversation?

• Does this kind of factoring work for all problems?
• It works for problems with 3 terms. With 4 terms you need to factor by grouping
• 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...
• 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.
• At , 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?
• Indeed it should be fhx^2. There's an annotation pointing out the error a few seconds later, though.
• Why does he multiply 4 by -21?
• 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).
• 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 in to
• What is the significance of learning this method of grouping? Why not learn how to factor these quadratics with the quadratic formula?
• 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.
• Did Sal mistook at ? I think fx times hx is fhx^2 not fhx
• 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.
• 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?
• (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.
• 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! :(
• 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 have tried solving the practice problems but I don't quite get it. I know how to solve the problems but, it won't come out right. How do I solve the problems so I get them right? A problem like -3x^2+17x-20. How do I solve that?
• First, I recommend factoring out (-1). My teacher has found that a negative leading coefficient is inconvenient.

-(3x^2-17x+20)

Next, multiply together the leading coefficient (3) and the constant (20) to get 60. Then find what adds to -17 and multiplies to 60. In this case, it would be +/- 20 and +/- 3 (+/- means plus or minus). In order to find out which number is positive and which is negative, first find the largest number (20), and tag the sign of the second coefficient (-17) to it. So you would get:

-(3x^2-20x+3x+20)

From here, you may rearrange the two middle terms to factor by grouping. Ignore factored out -1 for now.

3x^2+3x -20x+20
3x(x+1) -20(x+1)

= -(3x-20)(x+1)

Sorry for the super long answer, and let me know if I made a mistake! 😲