Strategy in factoring quadratics (part 1 of 2)
There are a lot of methods to factor quadratics, which apply on different occasions and conditions. After learning all of them in separate, let's think strategically about which method is useful for a given quadratic expression we want to factor.
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- Have we found all possible strategies for factoring? Or is that even possible?(18 votes)
- Considering modern math, we might have, but there could also be more efficient ways of doing this other than "guess and check". Hopefully we will find more efficient ways!(17 votes)
- At5:50Sal shows the "perfect square" techinque. Is it possible to factor a perfect square when the constant is negative. For example, 3x^2+30x-75?(11 votes)
- No. The reason why we can't is because that would require the second term of the square (the b in
(a+b)^2) to be a negative number when squared. Although this is possible if b is a complex (or imaginary) number, factoring usually only uses real numbers. The square of any real number is always positive (try it: A x A is always positive), so the last term cannot be negative.
There is one major exception: If the leading coefficient (the 3 in your example) is also negative, then when you factor it out, the constant term will become positive (because a negative number divided by a negative number results in a positive number) and you can factor it out normally.
If this was confusing or you still have questions, feel free to comment them below and I'll try my best to answer them!
Hope this helps!(7 votes)
- Please excuse my naivety , but what is the purpose of factoring?? How does it help??(6 votes)
- It will give you the x intercepts of the quadratic if there are any.(6 votes)
- So when you add -12 and 1 you get -11, and when you add -1 and 12 you get 11, does this mean it will always be that way? Or are there times when that doesn't happen?(6 votes)
- it will always be that way, check out the lessons on positive negative numbers to get the intuition behind it.(4 votes)
- I factored "3x^2 + 30x + 75" and I got (3x + 15)(x + 5).
Do we have multiple answers? Am I wrong?(4 votes)
- Well, you could have factored a 3 in the beginning to get 3(x^2 + 10x + 25) which is a perfect square 3(x+5)^2. So you have two answers, x = -5 and x = -5, but they are the same, so there is only one unique answer.(3 votes)
- Hi, when you are doing the a+b and a.b shoudn't you find -b/a and c/a respectively?
- can we also factor the 6x^2+3x
like this 6(x^2+3x)
- No, that does not work because when you distribute, you will get 6x^2+18x. The right way to factor it is by taking the greatest common factor out, which is 3x. It would be 3x(2x+1).(5 votes)
- Has anyone taken the algebra 1 EOC?
I'm terrified!!(3 votes)
- Is there any way to systematically solve for a+b and a*b when factoring? The "guess and check" methodology of thinking about what could possibly fit in those slots is rather slow going.(3 votes)
- I have a question...when we don't have any common factors and we have to multiply the a term and c term, how can we find the factors of big numbers like 168 for example that will have two factors that add up to the b term?(3 votes)
- When you have to factor a quadratic equation with really big numbers where nothing jumps out at you immediately, one strategy is to take the product of 'a' and 'c', and then find factors of that using something like a factor tree. Group the prime factors together by multiplying until you have just two numbers, and then add them together and see if they will sum to 'b'. If not, you'll have to find a different combination. It is long and repetitive, but starting from the prime factors at the bottom of a factor tree could help you stay organized and make sure you check every combination of factors.
For 168, we can break it down into 2*2*2*3*7. Possible combinations you can make could be 2 + (2*2*3*7) = 86, (2*2*3) + (2*7) = 26, (2*3*7) + (2*2) = 46, or anything else.(1 vote)
- [Instructor] We have other videos on individual techniques for factoring quadratics, but what I would like to do in this video is get some practice figuring out which technique to use, so I'm gonna write a bunch of quadratics, and 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 three, six is divisible by three and so is three, 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 gonna have a 2x left over there and then 3x divided by 3x, you're going to have a one. And that's about as much as we can actually factor, and you can verify that these two expressions are the same if you distribute the 3x, 3x times 2x is 6x squared, 3x times one is 3x, and that's all we would do. We would be done. That's all you can really do to 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 four. Four is clearly divisible by four and 48 is also divisible by four. So let's factor a four out. This would be the same thing as four times x squared minus x minus 12. I just divided each of these by four and I factored it out. You can distribute the four 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 alright, I have a one as a 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 one 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 one here before but it's implicit there. Negative x is the same thing as negative 1x. So are there two numbers, a plus b, that is equal to negative one 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 be 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 they're gonna have different signs. One's going to be positive and one's going to be negative. If they had 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 one and 12, and whether you're thinking about negative one and 12, negative one plus 12 would be positive 11. If you went the other way around, if you went negative 12 and one it would be negative 11, but either way that doesn't work. Two and six, negative two and six would be four, negative six and two would be negative four, so that doesn't work. Three and four. Let's see, negative three and four would be positive one, but three and negative four 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 one. So we can write inside the parentheses, so let me write, so this is gonna be four times, so we can factor that as two binomials. The first is going to be x plus, the first is going to be x plus three and then the next is going to be, we could say x plus negative four or we could say x minus 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. And then we saw here, hey if we have a leading one 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 three and negative four, we were able to factor it this way. We prove 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. Three x squared, plus 30, plus 75. Alright, I'm assuming you had a go at it. So you might immediately see that all of the terms are divisible by three, so let's factor three out, so it's gonna be three times x squared plus, oh whoops, this should be an x here, my apologies. Pause the video again and see if you can do it (laughs) now that I wrote the actual right thing there. So as you imagine it's nice to factor out a three first, so it's three times x squared, plus 10x, plus 25, and so you might immediately say alright, let's use the technique we had here. We have a leading one coefficient, it's written in standard form. Can I think of two numbers that add up to ten, 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, alright, well this thing here is positive, this is positive, so I'm dealing with two positive numbers and to get 25 it's either a one and 25 or five and five, and five and five match this. Five plus five is equal to 10. Five times five is equal to 25. And so just using the exact technique we just did, you'd say okay this is three times and the stuff in parentheses would be x plus five times x plus five, or you could say three times x plus five 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 used in the second problem, either one of those would get you to the appropriate answer.