Introduction to grouping

Sal factors t^2+8t+15 as (t+3)(t+5), first using the "sum-product" method, then using the grouping method.

Introduction to grouping

Discussion and questions for this video
He factors (t+3) out. You can treat t*(t + 3) + 5*(t+3) as though it has parenthesis around it (t*(t + 3) + 5*(t+3)). If you do that you can see that you have two things (t and 5) multiplied by (t+3) so you can factor out a (t+3). This leaves you with a (t+3) multiplied by what is left inside the parenthesis which is (t+5). So the expression ends up being (t+3)(t+5)
You didn't specify at which time but I guess that you meant somewhere at 3:30 where we have:
(t²+3t) + (5t+15)
t(t+3) + 5(t+3) → We factored out t from (t²+3t) and 5 from (5t+15)
(t+3) (t+5) → now this is part that you were asking for, you're wondering where did the other (t+3) go, actually if you look at t(t+3) + 5(t+3) carefully you might realize that they both have (t+3) so we can just factor that out and add their coefficients in another parenthesis as another variable.
And that is why (t+3) (t+5) is our final answer!
3/2x + 4 = -9. To solve for x, we can do a few things. I would suggest one method though, let me show you...
3/2x + 4 = -9
subtract 4 from both sides.
3/2x = -13
Now, multiply both sides by 2/3.
(2/3)*3/2x = (2/3)*(-13/1)
6/6x = -26/3
6/6 = 1, so
x = -26/3
and we are done. Tell me if that helped or not.
Why do the (t+3)'s you factored out of the trinomial cancel each other, and if they did wouldn't they leave a 1 instead of a a single (t+3) remaining? sorry if this was a badly worded question btw
When doing the KhanAcademy lessons for Factoring Polynomials 1 ( I was given this problem to factor.

x2−10x+24 (x to the second power minus by ten x plus twenty four)

I wrote down this answer and it solves the right way,

(x + 2)(x - 12)

But KhanAcademy said that I got it wrong and that the answer was actually,


Can anybody tell me where I went wrong? I just don't get it. They both solve the same way.
So the equation they want is x^2-10x+24 Correct?

If you multiplied yours out, it would be x^2-10x-24.
(x+2)(x-12) = x*x-12x+2x-24 = x^2-10x-24 because 2 times -12 is -24.

While the answer provided
(x-6)(x-4) = x*x-6x-4x+24 = x^2-10x+24. Negative 6 times negative 4 is positive 24.Which is what they wanted.

You got really close, just gotta watch that minus sign on the last term.
what if the problem is x^2-2x-8. the factors of 8 dont add up to 2? like in the video 5 and3 add up to 8
-4*2 = -8
and -4+2 = -2
So x²-2x-8 can be factored.

There are other methods such as "completing the square" and the quadratic formula that can be used when you can't easily factor the expression.
Here is a video on completing the square

And on the quadratic formula

I hope that helps make it click for you.
To factor the expression ax² + bx + c, first solve the equation ax² + bx + c = 0 to find the roots of the polynomial (this can always be accomplished using the quadratic formula). Suppose ß and µ are roots of said equation. Then both (x - ß) and (x - µ) must be factors of the expression. We may then write ax² + bx + c = a(x - ß)(x - µ).

Note: ß and µ need not be distinct, and they may even be complex conjugates of each other.
I want to see an example of factoring a trinomial in the type of ax(squared) + bx + c where a is greater than 1 and a, b, c do not have a common factor greater than 1.
Julie - Chuck's answer leads you to the quadratic formula, which is a great one-size-fits-all solution to polynomials like those in this video. But it doesn't answer your question: if you're looking for a way to factor that's analogous to the way above, where "a" is not 1, check out this video:

The quadratic formula becomes more important when you encounter quadratic polynomials that you can't factor easily - often roots and irrational values result.
This question is a bit off topic but can anyone explain how I would factor an expression such as this one
x^2+(3y+4)x+(2y+3)(y+1) I have been stuck on this question for a while and any help would be useful, Thank you in advance.
The sum of two numbers is 3y + 4 and the product of two numbers is (2y + 3)(y + 1)
So this equation can be factored as (x + 2y + 3)(x + y + 1)
This problem is of the form (x + a)(x + b) where a = 2y + 3 and b = y + 1
What if at 3:40 you had t(t+3) + 5(t-3). How do you deal with one of the values being negative?
ok, now what if the question is x^2 -x -8 = 0 ?
I'm just having trouble coming up with the two numbers for the factoring...
Sal is an awesome teacher explaining it in short times and no questions needed for his methods
You find factors for the last term and then add them together and the ones that add together to equal the middle coefficient are the ones that you put after x.

Eg. t^2+8t+15
The factors of 15 are 5 and 3, and 1 and 15. 5 and 3 add together to make 8, which is the middle coefficient, so the factored equation is (x+3)(x+5)
I know this is not a quadratic equation, but how would I factor a polynomial such as this one: 2p^3 + 5p^2 + 6p + 15 ?
There are a variety of methods, the one that you try first is factoring by grouping. But note it won't always work. Here's how to do it with your example:
2p³ + 5p² + 6p + 15
Step 1: Group
(2p³ + 5p²) + (6p + 15)
Step 2: Factor each group's GCF
p²(2p + 5) + 3(2p + 5)
Step 3: If there is a common factor, factor that out of the sum:
(2p + 5)(p² + 3)
Step 4: If the quadratic factor can be factored, do so:
(p²+3) has no real factors
So the fully factored form is:
(2p + 5)(p² + 3)
You would have to expand it out using the distributive property. Then you could simplify it into a trinomial. That particular problem cannot be factored using real numbers, so a trinomial would be its simplest form.
if a polynomial cannot be factored completely and you use the quadratic formula to solve for the variable should you simplify and factor the beginning polynomial as far as you can before you use the quadratic formula?
so the final answer, the thing you would write down and box is "(t+3)(t+5)" right?
It does not make a difference. Just do the same thing that you normally do. Ex: -x^2+5x-6
=> (-x+2)(x-3).
First lets notice that all of the coefficients are divisible by 3. We can factor that out to get 3(2x^2-15x-8).
So now we have ax^2 + bx + c and we're looking for something with the form of (mx + p)(nx + q) where a = mn, b = pn + mq, and c = pq.
a = 2 so m and n have to be 1 and 2. We now have 3(2x+p)(x+q).
b = p + 2q and c = pq. c also equals -8 so p and q must be { -1, 1, -2, 2, -4, 4, -8, 8}. If you test out all of the option you'll see that 8 * -1 = -8 (c) and -1 + 2 * 8 = -15 (b).
Our answer is 3(2x - 1)(x + 8)
How does he get from t(t + 3) + 5(t + 3) to (t + 3)(t + 5)

I'm interested why and how the factoring out of t + 3 works, exactly which places are factored out.
Answering my own question with the help of my GF, important is the following math rule:
a * b + a * c = a (b + c)

*Why is this?*
(a * b) + (a * c)
____________ * a = (b + c) * a

So to get from t(t + 3) + 5( + 3) to (t + 3)(t + 5) we can do the following:

t(t+3) + 5(t+3)
____________ * (t+3) = (t+5) * (t+3)
Am I performing a correct arithmetic operation if I factor like following:
I haven't seen anything like it, so I got a bit confused.
A trinomial follows the form a^2+2ab+b^2, where a and b do not equal 0. When factored, the trinomial equals (x+a)(x+b). A and b are the variables commonly used to write out the formula for factoring a trinomial, but really any two symbols can be used.
I need some homework help.
The problem is 20t^2 - 12t - 32
I know the answer is 4(5t-8)(t+1) but I don't understand how my teacher got to that answer.
It is kind of a homework question, practice question. I have worked on it for two days!! x(4x-5) =114. I cannot find anything that factors out. 4x-5x-114=0 What adds to 5 and when multiplies gives you 114??
Hi Patty,
Nikola is correct in the factoring but I wanted to point out where I think you were having trouble. It sounds like you were forgetting about the 4 in front of the x^2 when you were looking for factors. Remember that for Ax^2 + Bx + C = 0, when we are looking to break down the expression, we multiply A*C to get a value. In this case, A*C = -456. We then break down the -456 into two factors that, when added together, give us B, in this case, -24 and 19 are the two factors we want. We then rewrite the original equation, using our two factors to replace B:
4x^2 - 24x + 19x - 114 = 0
We can then group and factor:
(4x^2 - 24x) + (19x - 114) = 0
From the first group we can factor out a 4x and from the second group we can factor out a 19:
4x(x - 6) + 19(x - 6) = 0
At this point, since we have like terms (x-6), we can combine the 4x and the 19:
(4x + 19)(x - 6) = 0
From there, if you are solving for x, you would set each factor equal to zero:
x - 6 = 0
x = 6
4x + 19 = 0
4x = -19
x = -19/4
So your roots would be x = -19/4 or x = 6
Hope that helps.
need a more clearer question, what the topic? You can multiply an entire equation by some factor to match up other factors, say in elimination.

x^2+4x+5 can be multiplied by say 2 to get 2[x^2+4x+5] = 2x^2+8x+10 or by whatever you want to change the coeffients to.
The FOIL Method is a process used in algebra to multiply two binomials. The lesson on the Distributive Property, explained how to multiply a monomial or a single term such as 7 by a binomial such as (4 + 9x).
The leading term, first term, `x^2` has no explicit coefficient since, by convention, we do not write the one, but it is there.
A coefficient is the constant number in front of a variable that is multiplying by it. So if you have "5x", 5 is the coefficient.
i was looking for an example where i could find how to solve quadratics when a is more than one. Anybody know where I could find that? has a worksheet creating tool where it gives you questions for free and you can choose the topic and how many questions you need
I'm only confused about problems where "a" (in ax^2 + bx + c) does not have a value of 1. How do you take this into account when factoring?

Without using the quadratic formula.
If you have ax^2 + bx + c = 0 and you want to factor, the first thing I would do is divide both sides by a. Here is an example: 4x^2 + 12x + 9 = 0. Dividing both sides by 4 yields: x^2 + 3x + (9/4) = 0. You might be able to recognize this as a perfect square: The square root of 9/4 is 3/2 and (3/2)x + (3/2)x = (6/2)x = 3x. So we have x^2+3x+(9/4) = (x + 3/2)^2
ok, what if there is a variable in the middle like X to the second,minus x, minus 20?
You can still factor x2-x-20. the -x would be the same as -1x. I prefer plugging these things into the quadratic equation for then you get exact answers without the guess and check method.
equals what? you have to complete the square or use the quadratic formula if you wanted to find t
Well, to be fair to teachers, they do have questions asked all the time. . . . but as for the 'taught it better' part, that depends on the teacher. A bit of a redundant statement, even by my standards. And the awesome thing is that we can ask our questions in the comments and know that we won't be shunned by the world for asking a redundant question, or a simple question that many people know but a few just can't grasp.
Hi everyone, for factoring quadratic expressions how do you know what numbers to put in the parenthesis. I understand that you find numbers that multiply to whatever x and add up to the other number but does anyone have tips for finding out these numbers??
You can factor the last number (15 in the example) to see which of its factors add up to the middle number (8 in this case). The factors of 15 are {15,1} and {5,3}. Since 5 +3 = 8, you can try them to see if they work.
(t + 5)(t+3).= (t^2 + 8t +15)
Also, since the middle number was positive, we didn't need to consider the negative factors of 15, namely {-15,-1} and {-5,-3}. If the middle number was negative (-8) then {-5,-3} would have been the answer since
(t - 5)(t - 3).= (t^2 - 8t +15)
At 4:00, is that the answer for this problem? I still don't get it how does that when mutiplied do we get t squared? Is it the the one in purple with green or the one you were talking about t +3 times t+5. I still don't understand.
The 1 doesn't have to be put since there is only one "t", but if there were 3 t's, then you would need to put 3t, because 1t=t, you do not need to worry about the leading coefficient.
Sal is just showing us the same problem in "color Coded" form. he is showing us that if you replaced a and b with the real numbers in this problem you will get the same "form" of answer.
How would you factor an expression with numbers that are more than squared? Something along the lines of X to the 3rd + bX squared + cX. How would that work? Btw just as a side do you get a computer to type in exponents?
Ok, I run into an issue when trying to factor my own equation: X^2+4x-3.

I cannot seem to get both the product of -3 and the sum of 4. I am running into many problems like this. does the method in this video not apply here??
You are correct in that this method does not work for all quadratics. Your expression's roots are not rational numbers so you would need to solve for the roots with the quadratic formula.
Thank you! That just never even clicked in my mind! Glad I found out before EOC's though!
what if you have a problem such as 9x^2+72x+142=0 where none of the numbers you mutiply together to get 142 and have a sum of 72
Ok, so I have a question. If i had a problem like this: x^2+x-90, what would i do? because the middle term only has a variable x. Can someone help please?
ok, so here goes nothing! so we know that we have to find a value for a and b. a times b should be equal to 15 and their sum should be equal to 8. so a and b are 5 and 3. this makes the equation: t(t+3)+5(t+3). so since there are two (t+3)'s then we can factor it out to this: (t+3)(t+5) and that would be the answer! does that help? hope ie does!
Can you factor this out with monomials or does it just work with binomials?
I'm working on Factoring Quadratics 2 and I got a question that looks like this: -2x^3-18x^2+20x
I'm confused about the -18 and the +20. I tried watching a video to know how to factor that part, but all of the videos either show two negatives or two positives. Can anyone help? Thanks!
```-2•x^3 - 18•x^2 + 20•x

-2•x•(x^2 + 9•x - 10)

[-2•x•(x + 10)•(x - 1)]```
There's no specific time in the video but I was wondering why, for integers a and b and the form Ax^2 + Bx + C, ab = AC and not ab = C. Isn't it just C. Sorry if this is confusing
This may seem like a dumb question, but what is the mathematical definition of factoring?
I'm starting to get the concept of the factorizing idea. But what if I had to imply on harder polynomials, for example this one in my book:
The answer is (9p^2-q^2)^2

Note that you can treat the p^4 as (p^2)^2 and q^4 as (q^2)^2.

It may be simpler to look at it this this way.
Make the definition of
p^2 = a
and the definition of
q^2 = b.

Then the original expression can be written as 81a^2 - 18ab + b^2.

This factors into (9a - b)(9a - b).

Then make the substitution a = p^2 and b = q^2. This will give you
(9p^2 - q^2)(9p^2 - q^2)
(9p^2 - q^2)^2
I've been swearing over this for a couple of days now. :) Is it really not possible to solve a second degree polynomial (equal to 0) without grouping? I either seem to go in loops, or end up with x = +/- 1, which is false in most cases. Can't seem to find any videos about it either, it just goes unsaid. I'm guessing it's due to the raised var being added to the var, or that there's two solutions, but it really really bugs me. Any clarifying thoughts or ideas about what I'm missing?

Here's the last go I had without factoring much or grouping:

Bear in mind, I did learn all about factoring, I just want to understand why I can't do this.
There are usually 2 solutions to the Quadratic Equation where it is equal to zero. It is much easier to find solutions by grouping. While your method results in rearranging the expression several different ways, it is very difficult to come up with solutions. It is very helpful to factor it out into groups. (This is further helpful when you need to multiply, divide, or otherwise further process the polynomial, perhaps using another polynomial (i.e., you will more easily be able to cancel certain groups out).
So, to solve where x^2 -6x +5 = 0, the factors (x-5)(x-1) will tell you your answers: what values of x result in the expression to equal zero? If x = 5, the first factor turns to zero, making the whole thing zero; and where x =1 it is also zero. So the solutions are x = 1 and x = 5.

I see an error in one of your steps which makes all that follows incorrect:
x-6=-5/x does not equal x-5/x=6. It should be x+5/x =6
Either way, you still have either a fraction of an x or a factor of an x to deal with at some point. You could do trial and error with plugging in numbers to find where x + 5/x =6 and come up with the 2 solutions, but there are cases where it is more difficult. I think that (x-5)(x-1) is nicer to deal with in the long run.
Mathematical convention. a and b are coefficients. The accepted general form of a quadratic equation is `ax^2 + bx + c = 0`
Using x and y would be confusing because the expression already has avariable named x.
Factoring or factorial?

Factoring is when you take the components of a number, such as the factors of 4 are 1,4 and 2,2.

Factorial is multiplying a number by all previous numbers and is denoted by an exclaimation point. For example, 4! is 4*3*2*1.
I am working on Factoring Trinomials using FOIL. However, I am having a problem finding a video to explain that. Can you point me in the right direction? Thanks!
I am sorry, but Sal doesn't have a video on this topic. I found someone else who does a really good job on this.
If you meant "a^2 + 11a + 18", AKA "a² + 11a + 18", then here are the steps:

first, find two numbers that multiply to 18, but add to 11. In this case, the two numbers are "9" and "2", because "9 * 2 = 18", and "9 + 2 = 11". So now, you can factor:

(a + 9) * (a + 2)

If you want to double-check that, you can just FOIL it out to see that it works.

Next, if we want to find when the whole thing equals zero (aka a solution), set that equal to zero:

(a + 9) * (a + 2) = 0

This will only equal zero when "a + 9" equals zero, or when "a + 2" equals zero. So the solutions are "a = -9" and "a = -2".

I hope this helps.
The answer to that question is 4x+2x^2+1.
Here are the steps I did:
4x (2x-1) + (2x-1)^2 (Original Problem)
4x (2x-1) + 2x^2 + 1 ( I applied the two squared to the second pair of parentheses.)
8x-4x + 2x^2 + 1 ( I distributed the 4x into the parentheses.)
4x + 2x^2 + 1 ( I combined like terms.)
Hope that helps!
Find factors of 15 (2 numbers that multiply to create 15) and those same 2 numbers need to add to -8
-3 and -5 work
This let's you factor the polynomial into: (T - 3) (T - 5)
I would like to know what video would inform me on how to do a problem like y=(x+9)². It is a quadratic equation.
I don't get from 3:35 to the end. how did he factor it out to where from t(t+3) + 5(t+3) to the answer being (t+3)(t+5) ? Please Help!
okay this follows the distributive property, he distributed the t+3 out because
(t+3)(t+5) can be rewritten as t(t+3) + 5(t+3) if you distribute the t+3...did that make sense? its like when you have 15+30 and you can make it (5)(3)+(5)(6) = (5)(3+6)
I understand how to factor these equations but I don't understand when the leading coefficient is greater than 1. For example, a problem I received in class was 4x^2+12x-7. How would this be solved? Someone please help me. Thanks.
Okay, so i understand this, but what if the second number in the trinomial was posotive and the third one was negative? How would you solve the problem x squared +14x-49???
This is a Great video but the entire thing was confusing to me can you help me Khan
Lots of people here would love to help you, but please ask a more specific question. What was the first thing in the video that confused you?
is the a and b step really necessary i feel that there is a easier way to factoring if so please tell me
Once you become better at factoring quadratics, you'll realize that the "a" and "b" step is practically useless, as you asked. Basically, for simple quadratics without a coefficient in front of the squared variable, you just have to find two numbers that multiply to equal the constant, or number without the variable. Also, the two numbers must be added to equal a sum that is the coefficient in front of the second term. Hopefully, that will make sense.
Perhaps that quadratic equation cannot be factored. Try using the Quadratic Formula instead and see what kind of solutions you get for x. If they involve irrational numbers, then the equation is "prime" and cannot be factored.
i can't use what sal is doing in the problem in my homework..