Get ready for Algebra 2
- Factoring quadratics as (x+a)(x+b)
- Factoring quadratics: leading coefficient = 1
- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Factoring quadratics with a common factor
- Factoring completely with a common factor
- Factoring quadratics with a common factor
- Factoring simple quadratics review
Factoring completely with a common factor
We can factor quadratics by first pulling out a common factor so the result looks like a(x+b)(x+c). Created by Sal Khan.
Want to join the conversation?
- Does a have to be greater than b when a≠b? For example, 8(x+2)(x+4) versus 8(x+4)(x+2).(4 votes)
- I didn't really understand what you were asking, but 8(x+2)(x+4) and 8(x+4)(x+2) are exactly the same thing. Remember that the order in which you multiply two or more things together doesn't change the final product.
a*b*c = b*a*c = c*a*b
Hope this helps.(16 votes)
- To use this method does a have to be greater than 1?(5 votes)
- It doesn't need to be greater than 1, but if it wasn't greater than 1 then it would be pointless. If the only greatest common factor was 1 for the expression, then you can't really factor the expression therefore making the expression prime.(3 votes)
- in1:50shouldn't be 4((x-3)*(x+1))? i'm thinking:
also, i found strange how in the alternative path of the second example "3(-x^2+7x-10)". it seem like there is no answer to: a+b=7; a*b=-10. did i miss something?(3 votes)
- At2:53Sal says that you can factor the thing(what should I call it? A trinomial multiplied by a constant), he says you can factor with the "x squared" being negative. So how can you factor the trinomial with the x squared leading term being negative? Thank you.(3 votes)
- Answer for2:19
-3x^2+21x-30 = (-3x+6)(x-5)
Find 2 number that gives 90 when multiplied and 21 when added.
15 and 6
Is this still correct?
When checked it still gives me the first -3x^2+21x-30. But the final answer way different than the one shown in the video(2 votes)
- You are correct, but yours is not the completely factored form of this polynomial. Sal factors out 3 at the very beginning, and you can see this in yours: in -3x+6, 3 is common.
Hope this helps.(3 votes)
- How do we 'reverse' the process? eg at2:03, how do we apply the distributive property to end up with the original trinomial 4x^2-8x-12? Does the 4 multiply just the first set of brackets or both of them ? i.e. 4(x-3) and 4(x+1) or 4(x-3) and (x+1) remains as it is before expanding?(2 votes)
- To ´reverse´ the process the 4 is multiplied only to the first set of brackets. Once you multiply the 4 to the first set of brackets multiply the remaining 2 brackets together to get back to the original trinomial. Think of 4 as just a factor. For example, if you had 2x3x4, you would multiply from left to right or use the commutative property to multiply whichever numbers seem more compatible. The trinomial in the video is the same except with variables, and variables are just unknown numbers. You could have 4(x-3) first then (4x-12)(x+1) to get 4x^2-8x-12, or 4(x+1) first then (4x+4)(x-3) to get 4x^2-8x-12. You could even multiply (x+1) and (x-3) together first then multiply (x^2-2x-3) by 4 to get the original trinomial 4x^2-8x-12. All of these different combinations of multiplication work due to the commutative property.(3 votes)
- How do I do this with numbers that have no common factors? My maths teacher says it is possible but I do not understand how.(2 votes)
- You're just trying to get rid of the number in front of x^2. You just divide all the terms by that number. This will turn up as a fraction if they don't have a common factor.
4x^2 +3x +25
(x^2)/4 +(3x)/4 +(25)/4
x^2 +3/4x +25/4
This is super hard to factor though so i would recommend choosing a different method, like the quadratic formula: https://www.khanacademy.org/math/math2/xe2ae2386aa2e13d6:quad-2/xe2ae2386aa2e13d6:quad-formula/v/using-the-quadratic-formula
or completing the square: https://www.khanacademy.org/math/math2/xe2ae2386aa2e13d6:quad-2/xe2ae2386aa2e13d6:completing-sq/v/solving-quadratic-equations-by-completing-the-square(1 vote)
- Let me be more clear about my question: when the question asks for thee product of two binomials the question is x^2+8x+15 answer 4 and 6 or(x+3)and(x+5) how is this entered into the answer box ?(2 votes)
- what is the answer to 6x^2 -18x-60.
I got 6(x-60)(x+57)
followed the steps in the video best I could(1 vote)
- If by "answer," you mean finding the roots, no, you are not correct.
First, take out the 6 because everything is divisible by 6. Thus:
Then, you can solve for the roots from there.
You did not take out the 6 from all the numbers, only the X. Additionally, you made a mistake with the roots. I would recommend rewatching the video.
Keep up the good work!(2 votes)
- how do you enter the answer(2 votes)
- [Instructor] So let's see if we can try to factor the following expression completely. So factor this completely, pause the video and have a go at that. All right, now let's work through this together. So the way that I like to think about it, I first try to see is there any common factor to all the terms, and I try to find the greatest of the common factor, possible common factors to all of the terms. So let's see, they're all divisible by two, so two would be a common factor, but let's see, they're also all divisible by four, four is divisible by four, eight is divisible by four, 12 is divisible by four, and that looks like the greatest common factor. They're not all divisible by x, so I can't throw an x in there. So what I wanna do is factor out a four. So I could re-write this as four times, now what would it be, four times what? Well if I factor a four out of four x squared, I'm just going to be left with an x squared. If I factor a four out of negative eight x, negative eight x divided by four is negative two, so I'm going to have negative two x. And if I factor a four out of negative 12, negative 12 divided by four is negative three. Now am I done factoring? Well it looks like I could factor this thing a little bit more. Can I think of two numbers that add up to negative two, and when I multiply it I get negative three, since when I multiply I get a negative value, one of the 'em is going to be positive and one of 'em is going to be negative. I can think about it this way. A plus B is equal to negative two, A times B needs to be equal to negative three. So let's see, A could be equal to negative three and B could be equal to one because negative three plus one is negative two, and negative three times one is negative three. So I could re-write all of this as four times x plus negative three, or I could just write that as x minus three, times x plus one, x plus one. And now I have actually factored this completely. Let's do another example. So let's say that we had the expression negative three x squared plus 21 x minus 30. Pause the video and see if you can factor this completely. All right now let's do this together. So what would be the greatest common factor? So let's see, they're all divisible by three, so you could factor out a three. Let's see what happens if you factor out a three. This is the same thing as three times, well negative three x squared divided by three is negative x squared, 21 x divided by three is seven x, so plus seven x, and then negative 30 divided by three is negative 10. You could do it this way, but having this negative out on the x squared term still makes it a little bit confusing on how you would factor this further. You can do it, but it still takes a little bit more of a mental load. So instead of just factoring out a three, let's factor out a negative three. So we could write it this way. If we factor out a negative three, what does that become? Well then if you factor out a negative three out of this term, you're just left with an x squared. If you factor out a negative three from this term, 21 divided by negative three is negative seven x. And if you factor out a negative three out of negative 30, you're left with a positive 10, positive 10. And now let's see if we can factor this thing a little bit more. Can I think of two numbers where if I were to add them I get to negative seven, and if I were to multiply them, I get to 10? And let's see, they'd have to have the same sign 'cause their product is positive. So let's see A could be equal to negative five, and then B is equal to negative two. So I can re-write this whole thing as equal to negative three times x plus negative five, which is the same thing as x minus five, times x plus negative two, which is the same thing as x minus two. And now we have factored completely.