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Factoring quadratic expressions

Factoring Quadratic Expressions
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Factoring quadratic expressions

Discussion and questions for this video
Can both a and b equal the same for example x^2+20x+100 can both a and b equal 10?
Yes the number does not have to be different.
A coefficient is a term in mathematics which is basically the number that we used in this expression to multiply a variable. so X has a coefficient of 1, and 2X has a coefficient of 2, and so on.
watch closely in the video at 01:35, Sal (the teacher) is referring to the middle coefficient (which is of the variable " B ") as 10, meaning its 10 times B or just 10B.

further information > http://en.wikipedia.org/wiki/Coefficient
how come the constant term is the product of a and b?
Let me try (and please know that Sal already tried at 1:49 )...
Because when I you have a quadratic in intercept form `(x+a)(x+b)` like so, and you factor it (basically meaning multiply it and undo it into slandered form) you get: `x^2 + bx + ax + ab`. This of course can be combined to: `x^2 + (a+b)x + ab`. So when you write out a problem like the one he had at 5:39 `x^2 + 15x + 50`, 50, which is your "C" term (the third term) _and_ is your constant, 50 is the product of a and b (ab). This can be shown here: ```
x^2 + 15x + 50 is equal to:
(x+5)(x+10) =
x^2 + 10x + 5x + 50 =
x^2 + 15x + 50 {which is what we started with.}
```
Thank you very much for asking this question, i was wondering it for a long time and now that I know it I am glad that I am not the only one who was confused and I am glad to be of possible service. Please let me know if this helped anyone.
Why would you need to factor a quadratic in the first place?
So that we can........
1.) Find the value of a missing variable, such as x.
2.) Cancel out something that is on the numerator and the denominator to simplify rational expressions.
For instance if we have something like....
x²-2x-80/x²+16x+64
(x-10)(x+8)/(x+8)(x+8)→As we can see we can now cancel out an (x+8) here to simplify it to.....
x-10/x+8
There are a lot of videos on factoring by grouping. We have to learn to do it by trial and error. Are there videos for that too? How would you find 8x^2 -14x+3?
Even simpler (so as to avoid the fractions), if the quadratic is in the form:
`ax^2 + bx + c`
where a is not 1, you can do the following:

Multiply `a` and `c` to get a new number (say `d`). Now, you must find two numbers that add up to `b`, and multiply to `d.`
`x + y = b
x*y = d = a*c`

So in your example above, you would see that `d = a*c = 8*3 = 24.` Two numbers that add up to `b = -14` and multiply to `d = 24` are `x = -12`, and `y = -2.`

Then you can simply group and factor to get the correct answer.
how will i know what x is in the equation x^2-5x+24?
You cannot work out what x is in the expression x^2-5x+24. This is because expressions cannot be solved as they do are not equal to anything. E.g they are like statements.

If x^2-5x+24 was equal to a number or another expression then we would be able to work out the value of x.

For example:
x+1 (x could be equal to any number)
BUT
x+1 = 3 (x can only be equal to 2)

I hope that helps
Are there suppose to be alot of steps to this ?
i think there are around like seven steps maybe
What if the numbers are prime and have no factors?
All quadratics can be factored, but not all of them can be factored with rational numbers or even real numbers. If a quadratic cannot be factored into rational factors, it is said to be irreducible. However, it is always possible to factor a quadratic, if you allow irrational or complex factors.

Here's how to factor ANY quadratic expression in the form: ax² + bx+c.
Let d = b² - 4ac
(If d is not a positive perfect square, then the quadratic is "irreducible".)
The factors are:
a [x + ¹⁄₂ₐ (b + √d)] [x + ¹⁄₂ₐ (b - √d)]
If it makes for convenient numbers, you may use the distributive property to multiply a into one (but not both) of the factors.
How do you factor a polynomial when the highest power is greater than ^2?
Long Division of Polynomials or Synthetic Division.
What would be considered a constant?
A fixed value.

In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number.

Example: in "x + 5 = 9", 5 and 9 are constants

If it is not a constant it is called a variable.

-- taken from "math is fun"
what if we have:
7x^2-31x-20
How do we do that?
Neha,
To factor 7x^2-31x-20
First multiply the first and last numbers. 7*20 = 140.
Factor 140 to its primes 7*5*2*2
Now find two numbers that multiply to 140 and subtract to 31
You need a number that is more than 31 so start with
7*5=35 and 2*2 = 4 And we got it first guess 35-4 = 31
We know the x values are
(7x ± ? )(1x ± ?)
You know the 7 need multiplies by the 5 so the second ? is 5
(7x ± ? )(1x ± 5) The other ? must be 4
(7x ± 4 )(1x ± 5)
The 20 is a -20 so one sign must be + and the other must be -
The middle factor is -31 not +31 so the 7*5 must be negative so use -5
(7x ± 4 )(1x - 5) And as we already said the other sign must be +
(7x + 4 )(x - 5)

7x^2-31x-20 factors to (7x + 4)( x - 5)

I hope that helps make it click for you.
At 14:49, can you do (-x-3)(x+8) instead of -1(x-3)(x+8)?
No. The distributive property is not being properly used.

You are trying to multiply -1 by (x-3). Using the distributive property, you'll get -x+3 as a result. So it should be (-x+3)(x+8).
Do the numbers have to be whole, or could you end up with an answer like (x+2.5)(x+5.7)?
Yes, that could certainly happen. The quadratic expression x^2 + 4x + 3.75 factorises into (x + 1.5)(x + 2.5), for example. But (at least at first) you will probably find that the examples you are given factorise into integers.
Do we use FOIl for this problem?HELP
Once you have a quadratic formula in (x-a)(x-b) form, you don't need to use FOIL. If you do, you just get the original equation. The answers are a and b.
I have a question on solving but I can't find a video that deals with the topic (probably because I don't know what this type of problem is called) Could someone point me in the right direction or solve this step by step for me?

Solve for x
x^4-13x^2+36=0
answers given on the sheet are {3, -3, 2, -2}
my first thought was to reduce to x^2-13x+6=0 and solve from there using the quadratic formula but my answers don't match the answers on the handout and I only got two when we are supposed to find four, how do you get four?

Thanks!
simplify your life and from the very beginning, substitute a different variable in to stand for x^2 the whole time until you've got it solved and are ready for the final answer.
Say, g=x^2. So, just deal with this equation: g^2-13g+36=0.
Factor: Two numbers that multiply to be 36 and add up to be -13 are -9 and -4.
(g-9)(g-4) = 0. Now, set each part equal to zero.
g-9=0 ; g-4=0
so, g=9 ; g=4. NOW, bring back the x^2 instead of g again.
x^2 = 9 ; x^2 = 4
so, x can be -3, 3, -2, or 2.
What if you get a polynomial with 4 terms? Even 5?
What I mean is that polynomials with 4 or more distinct terms. Such as:

x^2 + 4x + 2xy + 10y

In which there are no like terms.
So, what would a real life problem for using this be? How would you use it with other parts of math? Please ask me a question if you are confused about what I mean here! Thanks!
age problems, research and other things
Why is it called Quadratic??
Quadatus is latin for square. Quadratic pertains to a square. If you have a term in an expression or equation that is to the second degree or squared, it is called quadratic.
I have no idea how to transform rational functions into graphing form
For exam le -4/4x↑2 +40x+ 10
Do I factor the bottom im so confused I'm not any good at precalculus it frustrates me because people in my class seem to get it I feel horrible and my teacher refuses to help :(
-4/4x² +40x+ 10
First we simplify by cancelling out the 4's to get: 1/x²
The common denominator is x², so we multiply everything by the factor that would given them that denominator -- it is just the same idea as adding fractions. In this case, one term already has that denominator, the other two need to be multiplied by x² / x² (remember what you did with adding ordinary fractions). This will give you:
-1/x² + (40x)( x²/x²) + 10(x²/x²)
= (-1 + 40x³ + 10x²) / x²
= (40x³ + 10x² - 1) / x²
You cannot simplify this any further. If you could tell me what it is equal to, I could finish solving the problem. Of course, if it is just equal to y or f(x) then that is as simple as it gets.
How do you know when to stop factoring a quadratic expression?
Do *un-factorable* polynomials exist?

For example:

```x^2+8x+36```
That is a great question. The anwer is actually yes. There are un-factorable (or irreducible) polynomials and the reason why can go very deep into mathematics. It is something that people study is college and beyond. Another simple polynomial that can not be factored is x^2+1. To see why suppose it factored as (x+a)(x+b). Then we would need a+b=0 and ab=1. The first equations tells us that either both 'a' and 'b' are 0 or one and not the other is negative. From there we see that both equations can not be true at the same time (did that make sense?) The same kind of thing is true about your polynomial.

It turns out that you can always factor polynomials with degree greater than 2 into smaller polynomials but there are many degree 2 polynomials that can not be factored. If you want to know more about this you will probably have to learn about roots of equations and then maybe the imaginary numbers like the square root of -1.
Is this the easiest way of factoring?
The easiest method is the one which works the best for you
Will this method always work?
only if you have a trinomial
What are quadratic expressions?
Expressions of the form ax^2 + bx + c; where a is not 0.
Is this also referred to as " Factoring x^2+bx+c". My book goes over this and calls it so. Why is it called Factoring quadratic Expressions here on KA? I'm confused.
Can you give some factoring examples that have no solution, one solution, and many solutions. And are there any examples that cannot be factored?
Well, all quadratics can be factored if you allow irrational and complex factors. But, if you mean rational factors, there are plenty that cannot be factored:
2x²+x+15 cannot be factored with real numbers.
No quadratic has "many" factors: Either there are two real, one real, or no real factors. If you have irrational or complex factors, you always have two of them (if there are no irrational or complex numbers in the quadratic expression).
As for having just one factor, it will always be the case that you really have two identical factors. For example, x²+6x+9 factors as (x+3)(x+3) also written as (x+3)².

No quadratic has more than two factors (unless you count factoring out a common factor such as:
2x² + 6x + 4 = 2(x² + 3x + 2) = 2(x+1)(x+2)
What's a coeffcient?
A co-effient is the number before the variable. eg. in -3x, -3 is the co-efficient. The variable is x. Variables are used to represent unknown numbers.
i need help with this!
5x^2=15x
please help!
ive got it to where its equal to zero but i cant get past that
5x²=15x
x²=3x
x²-3x = 0
x(x-3)=0
x=0 or
x-3=0
x = 3
Thus, x= 3 or x=0
What Is A Degree?
A degree is a unit of measurement of angles. It is also equal to 1/360th of the circumference of a circle. º is its symbol.
I hope this helps!
What's a coeffcient?
A coefficient is a constant (like "5") that is multiplied by a variable.
what are polynomials?
A group of several terms.
ughh i just got confused again whats 3a^2+7a+2 my techer just mixed up my brain:[ im way to new at this
Try going to math tutoring on tuesdays, wednesday, & thursdays, it helps alittle better .
What is he talking about?
factoring quadratic expressions….. JK… I really don't know.
why do you start factoring out a -1 in the last two equations of this video?
it seems like extra work and you don't explain why you do it, other than "it makes it easier". how does it make it easier?
How do you factor t^2 = 5t?
That helped a lot! Thank you so much!
What level of math would this be considered? Algebra 1? Algebra 2? College Algebra? Thanks!
yea i'm learning this in Algebra 1
What are first and second degree terms?
Examples of first degree terms: 7x, 4, 89z, 23r
Examples of second degree terms: 3x^2, 64y^2 18p^2
Examples of third degree terms: 23x^3, 34^y^2 87o^3
and so on...
how would you solve questions like x squared + 19x + 17, since 17 is a prime # and the sum of it's only 2 factors add up to 18, what would you do instead?
You could complete the square to
x^2 + 19x + (19/2)^2 = (19/2)^2 - 17
or use the quadratic formula.

Either way, you will get two factors with expressions that include the square root of 293.
x2-4x= -4 how do you solve? i need help.
Essentially these types of questions are asking for you to solve for x.
How you would do this:
Add 4 to both sides then you would get x^2-4x+4=0 which you can factor to get
(x-2)(x-2)=0 so then your answer would be x=2 and -2.
Can anyone tell me why it's called a quadratic expression?
It's from the Latin _quadratus_ which means "made square"
Even though I watched the whole video, I still don't get it .. this is hard ! Can someone explain this to me more easier ?
trust me it gets harder can u help me on mine its like up there
What is th purpose of factoring by grouping? I don't understand how, by what advatnge does it give?
Factoring is a way to find the value for x in a quadratic. (Try just solving for x the way you would do a linear equation. It doesn't work, does it :P)
There are several ways to solve for x in a quadratic (completing the square, quadratic formula, factoring, etc.) However, for simpler quadratics, factoring tends to be the quickest and easiest for most people.
Hope this helps!
A coefficient is the number in front of a variable that is intended to be multiplied by that variable. For example:
4x has a coefficient of 4
but,
4 + x has a coefficient of 1
(if nothing is shown, the coefficient is 1)
(x+a)(x+b) can be used to factor all quadratics, correct?
Okay, my problem is 6r^2+16r-32=0 im at 2(3r^2+8r-16) Now what do I do?
2(3r²+8r-16)
That much is correct. As for finding the rest, many people just try to guess the factors, but that takes up time and doesn't always work. Here's how to do it systematically, for when you can't easily guess the factors:
To make this easier to follow, I will just factor the part you've not factored yet (in real work, you should carry the 2 along the whole way, but we'll just ignore it for the moment.)
3r²+8r-16
we will split the middle term into two terms, mr + nr such that:
m + n = 8 (this is the coefficient of the middle term)
mn = (3)(-16) = - 48 (this is the product of the coefficients of the a and c terms).
Looking at the pairs of factors of -48, we see which of them add to 8 (and of course, multiply to be -48). I won't go through the whole list of possible factors that you would have to eliminate, but you'd get 12 and -4.

So, we split 8x into 12x - 4x
This gives us:
3r²+8r-16
= 3r² + 12r - 4r -16
We factor the first two and then the second two:
= 3r(r + 4) - 4r -16
= 3r(r + 4) - 4(r+4)
There now is a common factor, so we factor (r+4) out:
= (r + 4)(3r - 4)
And, don't forget about the 2 we already factored out:
2(r + 4)(3r - 4)
So are there quadratic expressions that can't be factored? (For example if the constant term is a prime and the first degree coefficient is ≠ the constant term ± 1)
When we swap a number from a side to the other side, we make it a negative number, don't we?
I get it now. Thank you.
I have a problem that's X^2 + X + 1 in my text book it just says prime.... what does that even mean?
Prime means that the expression cannot be factored any further because each term in the expression has no GCF.
what is a polynomial?
A polynomial is kAn expression of more than two algebraic terms, esp. the sum of several terms that contain different powers of the same variable(s).
hi! I have a question about this problem x^3-x^2+x-1 what is the answer? I am very confused
how do I factor the following question(x^3-x^2+x-1)
At 14:49, can you simplify the problem -1(x-3)(x+8)? If so, would the answer be (-x+3)(-x-8)? Or would it be (-x+3)(x+8)?
i really do not understand how to factor at all and i have these huge problems i can not do for example -56m^2+117m-55
Well this is a tough one. If you are really struggling I would suggest working on some easier ones first since the explanation I am about to give uses a lot of problem solving skills that are easier to develop with simpler problems.

We are looking for something of the form (nx+a)(mx+b). When we multiply this out we get nmx^2+nbx+max+ab=nmx^2+(nb+ma)x+ab. We get three equalities from this found by matching the constant terms, the x terms, and the x^2 terms. First mn=-56, next nb+ma=117, last ab=-55. The reason this problem is difficult is that you have to choose what to try and making good choices comes with practice. I would say start with ab=-55 because 55 only has two factors. So we just take a guess that a=55 and b=-1. Then nb+ma=117 means that m is probably small and positive so lets let it be 2. Then we have n(-1)+2(55)=117 so n=-7 but unfortunately 3(-7)=-21 is not equal to -56.

We could try other m's but I don't think it will work so lets switch to a=11 and b=-5. Well m is a factor of -56=-2*2*2*7 so lets try m=8. Then -5n+11(8)=117 means n=-29/5 so not right. m=14 gives n=37/5 so that doesn't work either and we went to high with m. If m=7 then n=-8 which works! We are done. The answer is (-8x+11)(7x-5).

As you saw there was a lot of guessing and trying things. It could have taken a lot longer if I was not aware of what would be good guesses and when to stop following a track. This comes with practice.
At 00:54, why does Sal use the expression (x+a)(x+b) to solve the problem? Is that a formula?
No, that's not a formula. That's just your basic form for what your answer should be after factoring. The basic goal is to have a + b equal to the middle x term, and have a x b equal to the last number term.

So, by finding factors of 9: 1, 3, and 9, and then seeing which two add up to 10 ( 1 and 9) you can factor the equation to (x + 9)(x + 1).
In my school, they taught us a technique called "Splitting the middle term" to factor quadratics.
How it goes is, we think of two numbers that add up to the co-efficient of x but also when multiplied those two numbers form the product of the coefficient of x^2 and the constant term.(We ,often, prime factorize the "product" and then go by trial and error to find the "sum").
Then we replace our x term with two other x terms which have the numbers we thought of as respective co-efficients. Then we proceed to reverse engineer the polynomial to make it into the form (x+a)(x+b).

Now, can somebody explain to me why the method Sal is teaching me is so ridiculously similar?

Also, which method is easier? Splitting the Middle Term OR Sal's Factoring Quadratic expressions.
-1(x-12)(x-6). should you write this like -1((x-12)(x-6)) or is it a rule to multiply the parentheses with each other first?
Nicholas,
You get the same answer if you do it either way.
(-1(x-12))(x-6) = -1((x-12)(x-6))
This is because of the associative property of multiplication.
The associative property is (a*b)*c = a*(b*c)
Why is it called quadratic if it is raised to the second power. Shouldn't it be called bidratic or something?
It actually comes from the Latin word "quadratum", which means "square". The reason lies in geometry: x^2 is the surface of a square with sides of length x.
what if the equation was 2x^2 - 13x - 24?
If you are trying to find the roots of this particular quadratic equation to put it onto a graph the work would look like this:

2x^2 - 13x - 24
ac product = -48

Factors of 48:

1 times 48
2 times 24
3 times 16

Because the middle term is negative and so is the ac product, one of the two factors you replace the middle with has to be negative and the other positive. And due to how you solve this problem, the two factors that you replace the original middle term with have to have a difference of the middle term.

2x^2 - 16x + 3x - 24

This is a big problem to factor now so you split up the problem into two separate problems to factor like so:

2x^2 - 16x ------------------->3x - 24
2x(x - 8) ----------------------> 3(x -8)

Put those together and due to the commutative property you can do that and you get:

(2x + 3) (x - 8)

The roots after that would be solved like this:

2x + 3 = 0 -------------------> x - 8 = 0
- 3 -3 ----------------------> +8 +8
2x = -3 ------------------> x = 8
________
2
x = -2/3




x= -2/3 or 8
How can i relate factoring quadratioc equations to real life ?
In this video, Sal doesn't seem to go over problems like this: (a number)x = -x^2-(a number). I am having trouble finding a video on these specific kind of problems, anybody know where I can find another one of these videos on khanacademy, or elsewhere?
Okay, so my problem is x^2-12x+32=0... My problem is why does my answer sheet tell me that the answer is 8 and 4. I got -8, -4. which makes more since to me. I understand how 8 and 4 could be right... but checking it.. It doesn't.. Here--- (x+4)(x+8) -->x^2+8x+4x+32= x^2+12x+32.. Not x^2-12x+32.. BUT, if I do -4,-8 then I get---- (x-4)(x-8)---->x^2-8x-4x+32= x^2-12x+32. Which is the original equation.. Help?
You are confusing the roots with the factors.
x² - 12x+ 32 = 0
(x-4)(x-8) = 0
So the factors, not the roots, are (x-4) and (x-8).
To get to the roots, you find those values of x for which the equation is true. In this case, (x-4)(x-8) = 0 is true whenever either factor equals 0.
Thus the roots (also called solutions) are:
x-4 = 0 OR x-8 = 0
Solving each equation gives us
x = 4 OR x = 8

The distinction between factors and roots might be easier to see in a problem like this: 6x² + x - 35 = 0
This factors to: (2x+5)(3x-7)=0

So, the factors are (2x+5) and (3x-7).

The solutions, or roots, are:
(2x+5)=0
2x = -5
x = -5/2
OR
3x-7 = 0
3x = 7
x = 7/3
So the roots, or solutions, are x=7/3 OR x= -5/2

Hopefully you can see how, though related, the factors and the roots are not the same thing.
hey what if you have like 24 and 10x its 6 and 4 or 8 and 3 how do you tell what one to use?
The examples on this video which starts with negative such as ( -x^2 + 18x -72) , when we factor out with -1 it works just fine , but just for the intuition i try them without factoring out with -1 , i couldnt figure that out . Why i cannot make it ? is it just me or there is really no way without factoring out with -1 . Thanks for help...
You actually make math stuff tolerable
how do we solve it when a is greater than 1 or less than -1
Once you have a quadratic formula in (x+a)(x-b) form, you don't need to use foil. If you do, you just get the original equation. The answers are a and b.
what does coffecient mean?
A coefficient is the number in front of a variable.
5x
/\
At 1:45, one of the numbers is underlined and called a coefficient, what is that? Are they all coefficients or just that one?
find the value of { 64 / 125} power of - 2 / 3 + 1 / {256 / 625} power of 1 / 4 + square root of 25 / 3rd root of 64
factorise x cube -3x square-10x+24
How is there A . B yet in there equation there is no mention of these letters.
Thank you for the answer>
how to factorise f(x)=3x^2+30x+75
What happens when you multiply a positive and a negative for the x squared or A and B quotient? Des it change the equation, or is it just answered "no solution"
There really isn't a difference. You would solve it the same. For example: x^2+2x-4. 2 and -2 when added equal 2, and when multiplied they equal -4. The answer would be (x+2)(x-2).
This question might be a little stupid but what's a polynomial and binomials????
how would you factor 4x^2+13x+3? I understand how to do it when x^2 doesn't have a number in front of it, but not when it does.
You're looking for a solution on the form (ax + b)(cx + d), and you know that
1) ac = 4
2) ad + bc = 13
3) bd = 3

So I start with a guess: a = 4.
Then according to 1) c must be 1.
Solving 2) and 3) gives you b = 1 and d = 3, and that's all there is to it.

4x^2 + 13x + 3 =
(ax + b)(cx + d) =
(4x + 1)(x + 3)
can you show me how to find the x-coordinates of the points of intersection for two functions
What do you do if x is not equal to 1?
Can you have x^3 and the and use the same process?
how do you solve x^2+9x+13=(x+4)(x+5)=m
You can't. (x + 4)(x + 5) = x^2 + 9x + 20.
At 15:10, why can't a * b be taken directly as (-1)(-72) ? Is that a wrong method?
How can I factor 25a^2+9b^2+30ab
14:00
I never factor them out like that... I do a prime factorisation. (I find that much easier and faster)
Is that a bad way to come to my answers?
At 15:26 he writes x squared plus 18x minus 72, but then he writes x squared minus 18x plus 72.
what do you mean by factoring out the negative one at 13;00 ?
The idea is that he doesn't want the x squared to be a negative, since that would make the entire equation more difficult, so the simplest way to solve it and make it like the previous equations he'd been solving for us, was to make the negative x squared into a positive. He did that by factoring out the -1, making the first term positive and making it exactly like the other ones that you had seen. Hope this helped!
I just want to make sure I have this correct, as I was never taught to pull the -1 out. For example, the last equation, would (-x+6)(x-12) be an equally correct solution?
What if the coeffiencent x^2 ( the first term) gets a number?

5x^2+x+10

What happens?
Divide the expression by 5.Sorry,I'm not sure whether its a equation or an expression. You'll get : x^2+x/5+2. Btw you do understand that the roots of the expression are irrational? So you'll have to use the discriminant method. Factorization is just hit and trial :P.
if there is a number is front of the equation like 3x (squared) -3x -16 what do i doabout the number in front of the X squared
How do I factorize 3x^2+5x-2
when the quadratic equation has a coeeficient in from of the number that is being squaerde what are the steps for that do you just what ever times to get the middle number and adds up to c and ax squared?
I still don't get why at 0:50 it is (x+a)(x+b)?????
'a' and 'b' are just placeholders. It shows you the pattern. Both 'a' and 'b' will be replaced with numbers in an actual problem. 'x' is also a placeholder, but in this type of a problem, it is left as a placeholder. If you multiply out (x+a)(x+b) you will get x^2 + (a+b)x + (a*b)x. If you were to replace 'a' and 'b' with your actual numbers the outcome will look like his initial problems: (x+1)(x+2) = x * x + 1x + 2x + 1 + 2 = x^2 + 3x + 3
I don't understand this question.... "draw a divided rectangle whose area is represented by each expression.". X(x+7). - label the length and area of each section. Then write an equivalent expression in expanded form.
Okay I have a lot of trouble with math, so I am just going to ask this a question. How on earth does -x^2+18x-72=-12?
You can solve using the *Completeing the Square* method: http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/solving-quadratic-equations-by-completing-the-square

Or you can use the *Quadratic Formula*: http://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_formula_tutorial/v/using-the-quadratic-formula

The first method is easier to figure out how to do when you can't remember the quadratci formula. It is a really good idea to understand the first method.

Good luck!
How do you figure out a question that has the answer but you have to figure out x's value?
For example: x^2 - 3x -18 = 0
Is there an easier or faster way to figure out what x equals besides guess and check?
To factor the polynomial more easily look at the videos on completing the square or the quadratic formula.
At 16:04 for the expression -(x^2-18x+72), couldn't Sal take out a common factor of -9 to change the expression to -9(x^2-2x+8) ?
There is no 9 in the x^2 term. He would have to change it to -9(1/9 x^2 - 2x + 8) if he did that.
What if the coefficient of x-squared is more than 1? Then can we solve something like 6p(square)-7p-20? Or 12x(square)+10x-12? Thank you :D
you have to divide all the terms by the co-efficient of x-squared to get the x-squared term to 1 and then factor.
I recently learned that quadratic equations can be viewed as the greatest approximation or the 'limit' an area of an object actually has. We all are taught to break down (x^2 + 9x + 9) into (x+3) and (x+3) and the equation of that line represents a parabola when graphed.
Instead of seeing it as a line, imagine we are trying to find the area of a post it or sticky note.

We would measure both sides, and the lengths would be somewhat close to 3 by 3, but that is not EXACTLY right. Every measurement we make is only good to a certain number of significant digits. If we had a more precise ruler and a microscope that could zoom in to help us measure beyond every 1/16 inch, we might find that the sides of our sticky note is actually slighter longer than 3 inches, but so small that we have no way of correctly calculating it. The 2 x's in (3+x)(3+x) can be used to represent the unknown additional lengths we are not able to measure.

We might form a hypothesis that each side is actually 3.05 inches long, so we could then put .05 as the x in (x^2 + 9x + 9) to give us a greater approximation of the area it has. This would cause the area to be 9.4525 sq.in. instead of just 9 sq. in.

While this makes an little difference on the area of a piece of paper, when calculating the areas of counties, cities, and states, you better believe they take into account for every half inch that was unaccounted for. Over hundreds and thousands of miles, a quarter inch can add a lot of area.

I am still trying to understand how equations that break down into (x-8)(x-3) would be thought of in terms of area when lengths cannot be negative, but I thought maybe some other students here might come up with ideas of what the quadratic equation can be used to represent that I have not visualized yet.. so please share your thoughts and ideas. I appreciate them.
How do you factor an equation such as 2x^2+8x+8? Or anything with a higher or lower coefficient than 1 for x^2?
(2x + 4)(x+2)
You factor it the exact same as if there wasn't a coefficient >1, it's just a little trickier.
Report a mistake in the video
Example:

At 2:33, Sal said "single bonds" but meant "covalent bonds."

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