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Current time:0:00Total duration:7:12

Video transcript

we have other videos on individual techniques for factoring quadratics but I would like to do in this video is get some practice figuring out which technique to use so I'm going to write a bunch of quadratic so I encourage you to pause the video try to see if you can factor that quadratic yourself before I work through it with you so the first quadratic is 6x squared plus 3x so pause and see if you can factor this so this one might jump out at you that both of these terms here have a common factor both of them are divisible by 3 6 is divisible by 3 and so is 3 and both of them are divisible by X so you can factor out a 3x so if you factor out a 3x 6x squared divided by 3x you're going to have a 2x left over there and then 3x divided by 3x you're going to have a 1 and that's about as much as we can actually factor it and you can verify that these two expressions are the same if you distribute the 3x 3x times 2x is 6x squared 3x times 1 is 3x and that's all we would do we would be done that's all you can do to really factor that and as we'll see in this example trying to factor out a common factor was all we had to do but as we'll see in future examples that's usually a good first step do all of check whether the terms have a common factor and if they do it never hurts to factor that out so let's do another example let's say I have the quadratic 4x squared minus 4x minus 48 pause the video and try to factor that as much as you can alright so the first thing you might have noticed is that there is a common factor amongst the terms all of them are divisible by 4 4 is clearly divisible by 4 and 48 is also divisible by 4 so let's factor a 4 out this would be the same thing is 4 times x squared minus X minus 12 I just divided each of these by 4 and I factored it out you can distribute the 4 and verify that these two expressions are the same now are we done well no we can factor what we have inside the parentheses we can factor this further now how would we do that so over here the key realization is all right I have a 1 as the coefficient on my second degree term I've written it in standard form where I have the second degree and then if there's a first degree term and then I have my constant term or my zero degree term and if I have a 1 coefficient right over here then I say okay are there two numbers whose sum equals the coefficient on the first degree term on the X term so are there two numbers that add up to negative one you didn't see a 1 here before but we it's implicit there negative x is the same thing as negative 1x so there are two numbers a plus B that is equal to negative 1 and whose product is equal to negative 12 this is a technique that we do in other videos and here the key is to realize that hey maybe we can use it here so a times B is equal to negative 12 and there's a couple of key realizations here it's like okay if I have two numbers and their product is going to be a negative that means one of them that means are going to have different signs one's going to be positive and one's going to be negative if they have the same sign then this would be positive so let's think about the factors of 12 and especially think about them in terms of different sign combinations so you could think about 1 and 12 and whether you're thinking about negative 1 and 12 negative 1 plus 12 would be positive 11 if you want the other way around if you went negative 12 and one would be negative 11 but either way that doesn't work 2 & 6 negative 2 & 6 would be 4 negative 6 and 2 would be negative 4 so that doesn't work 3 & 4 let's see negative 3 & 4 would be positive 1 but 3 and negative 4 works out you add these two together you take their product you clearly get negative 12 and then you add them together you get negative 1 so we can write inside the parentheses so let me write so there's going to be 4 times so we can factor that as 2 binomials the first is going to be X plus the first is going to be X plus 3 and then the next is going to be we could say X plus negative 4 we could say X - four and we're done and if any of this seems intimidating to you I encourage you to watch the videos on introduction to factoring polynomials the key here is to recognize the method so once again at first try to factor out any common factors we did that in both examples then we saw here hey if we have a leading 1 coefficient here on the second degree term and we have it written in standard form well let's think of two numbers that add up to this coefficient and whose product is equal to the constant term and in this case it was 3 and negative 4 we were able to factor it this way we proved that in other videos let's do another example we can't get enough practice and like always pause the video and see if you can work through it yourself 3x squared + 30 + 75 all right I'm assuming you had a go at it so you might immediately see that all the terms are divisible by 3 so let's factor 3 out so it's gonna be 3 times x squared + hope so this should be an X here my apologies pause the video again and see if you can do it with lookit now that I wrote the actual the right thing there alright so as you imagine it's nice to factor out a 3 first so it's 3 times x squared plus 10x plus 25 and so you might immediately say all right well let's use the technique we head here we have a leading one coefficient it's written in standard form can I think of two numbers that add up to 10 so a plus B is equal to 10 and whose product a times B is equal to 25 and this would work and if you look at the factors of 25 you'd say all right well you know this thing here is positive this is positive so I'm dealing with two positive numbers and we get 25 it's either 1 + 25 or 5 & 5 & 5 & 5 match this 5 plus 5 is equal to 10 5 times 5 is equal to 25 and so just using the exact technique we just did you'd say okay this is 3 times and the stuff in parentheses would be X plus 5 times X plus 5 or you could say 3 times X plus 5 squared so some of you might have immediately said well I don't have to do that exact technique I could have immediately recognized this as a perfect square because I have a square constant right over here and that's a good sign that hey maybe I should explore whether this is going to be a perfect square polynomial so this is a perfect square and if I were to take the square root of it and this coefficient is twice that square root well that's a good sign that I'm probably dealing or that I am dealing with a perfect square but either way whether you recognize it as a perfect square or whether you use the technique that we use in the second problem either one of those would get you to the appropriate answer