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## Multiplying and factoring expressions

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## Example 1: Factoring quadratic expressions

Factoring trinomials with a leading 1 coefficient
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## Example 1: Factoring quadratic expressions

Discussion and questions for this video
What happens at 3:35???? I got really confused at that part.
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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)
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where is the other t+3 gone can anyone explain please
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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!
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When doing the KhanAcademy lessons for Factoring Polynomials 1 (https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/quad_factoring/e/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,

(x−6)(x−4)

Can anybody tell me where I went wrong? I just don't get it. They both solve the same way.
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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.

(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.
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
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I have a question?

HOW DO YOU SOLVE THIS!

Example:
3/2x+4=-9

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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.
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at 2:27 he mentions grouping, what is grouping
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Taking out separate thing and putting them into groups
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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
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sifuentesj,
-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

I hope that helps make it click for you.
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.
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Julie,

has some problems that fit your request. If you use the "i'd like a hint" button, it will show you how to work each problem step by step.
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What if at 3:40 you had t(t+3) + 5(t-3). How do you deal with one of the values being negative?
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Is there an easier way of doing this?
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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)
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2xto the forth +2xto the third-xsquared-x
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Sal is an awesome teacher explaining it in short times and no questions needed for his methods
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(x - 1)(x + 3) = 4 what is the next step
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Expand the polynomial, to get x^2 + 2x - 3 = 4
Set the equation equal to zero, and then factor.
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so the final answer, the thing you would write down and box is "(t+3)(t+5)" right?
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Yes; all you're doing when you're factoring a polynomial is breaking it down into its respective binomials, or the terms you multiplied together to create the polynomial.
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After watching this video, I still cant figure out how to solve my problem. t squared plus 7t+12.
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Sometimes you may not be able to group them easily and that is why there are other methods. BUT this one is easy and it can be solved by grouping method.
t^2 + 7t + 12 =0
=> t^2 + 4t + 3t + 12 =0
=>t(t+4)+3(t+4) = 0
=>(t+4)(t + 3) = 0
=> t+4 = 0 OR t+3 = 0
=> t = - 4 or t = - 3
I hope this helps !
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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)
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what's the advantage of factoring. If you simplify you go the other way right? Remove parentheses. Just wondering.
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Am I performing a correct arithmetic operation if I factor like following:
(x+4)(4+4)=8x+32
I haven't seen anything like it, so I got a bit confused.
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What do I do if the exponent with the first monomial is a 3?
Ex: 6x^3-13x^2-28x
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For that problem in particular, you can factor out an x from all the terms so that you have x(6x^2-13x-28), and doing the factoring we've been learning in the last few videos, we get x(3x+4)(2x-7).
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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?
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You can view the x variable in the quadratic as having an implied 1 in front of it. In other words you could rewrite it as

x^2+1x-90

so now all that is left is finding factors of -90 whose sum is 1. 10 and -9 work.
The factored version of the quadratic would be

(x+10)(x-9)

or if you factor it through grouping the steps would be

x^2+10x-9x-90

x(x+10)-9(x+10)

(x+10)(x-9)
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how do you solve 4x2 - 6x - 30 ???
I've tried factoring it but can't seenm to find an answer
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I think it is
2(2x^2 - 3x - 15) ?
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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.
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4x^2+4x+13
Cause i always see examples that i can solve very quickly, but i haven't see an example what i can't factor by grouping. Is that possible or i just can't see the factors in my example?
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My question for my my friend, is where does your name come from? Ive never heard it before and it is interesting...
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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.
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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?

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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
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I understand everything except at 3:28, when t(t+3) + 5(t+3) equals (t+3)(t+5), I do not know how Sal got this, plz someone explain to me....

By the way Sal is awesome!!! LOL
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Thanks Jules :)
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what is the factor trinomial for y2+y-20
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(x+5)(x-4)
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What's coefficient?
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The number in front of the variable. An example is 9x where 9 is the coefficient and x is the variable.
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i don't think that's factorable.
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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???
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The expression you gave is not factorable. Since the roots of that expression are not integers, you cannot easily factor it using only integers.
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ok, what if there is a variable in the middle like X to the second,minus x, minus 20?
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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.
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At 3:45, how does Mr Khan go from (tsq.+3t)+(5t+15) to (t+3)*(t+5)?
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(t^2+3t)+(5t+15)
t(t+3)+5(t+3)
(t+3)*(t+5)
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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.
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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.
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Do you have videos that show you how to factor this type of polynomial ---> x3-8 or 10x3-25
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Does the method shown in the Factoring quadratic expression video only work if the term with an exponent is squared? Could it work if it was an equation with an exponent of 4?
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i can't use what sal is doing in the problem in my homework..

2a^2-5a-4^2
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There are a couple of ways to approach this, but factoring always starts by:

1) Look for a GCF (greatest common factor)
2a^2-5a-16 There are no common factors.

2) Look for a common pattern (a^2-b^2)=(a+b)(a-b); a^2+2ab+b^2=(a+b)^2; a^2-2ab+b^2=(a-b)^2
These are three patterns you just have to brute force memorize called difference of squares and perfect square patterns.
None of these patterns work.

3) Use a generic factoring method (ac method, slide and divide, grouping, educated guess and check, etc.)
ac method: 2•-16=-32, so look for factor pairs that multiply to equal -32 and add to equal -5
1+-32=-31
2+-16=-14
4+-8=-4
There are no other factor pairs of -32 so this will not factor evenly (aka it is 'prime')

4) For more advanced factoring of quadratics you can use the quadratic formula to get irrational and imaginary factors. a*x^2+b*x+c will factor as a(x-r1)(x-r2) where r1=(-b+√(b^2-4ac))/(2a) and r2=(-b-√(b^2-4ac))/(2a)

[-(-5)±√((-5)^2-4(2)(-16))]/[2(2)]
[5±√(25+128)]/4
[5±√153]/4
[5±√(3•3•17)]/4
[5±3√17]/4

2(a+[5+3√17]/4)(a-[5-3√17]/4)
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How do you factor a problem such as m^2+m-20 ?
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m^2 -5m + 4m -20 =0
m(m-5)+4(m-5)=0
m=-4,5
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thank u.u have helped me very much.can you do simultaneous equations please?
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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
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Is there any way this system of factoring can be incorrect?
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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??
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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)
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My teacher in the past two years dont help and i been watching you for one hour and i alraedy get it? I need u as my teacher
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is the a and b step really necessary i feel that there is a easier way to factoring if so please tell me
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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.
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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!
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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. http://www.youtube.com/watch?v=uFenBVfPxio
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What would I do if I had a c that couldn't be divided by my b?
Like in b^2+*b=7?
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I need help determining when to factor and when to apply the quadratic equation. For instance, in the problem:
Solve for x:
2x^2+6x−140=0
My first inclination was to apply the quadratic formula, which gave me x= (-3+or- 2(70)^1/2)/2
And that's OK, but not pretty.
But when I entered it, it was wrong. So I tried again and got the same answer. So I asked for a hint and they started in factoring. And I ran with the whole factoring thing and got the nice, pretty answer x=-10 or x=7.
Why did the two methods get me different answers?
And how can I tell when looking at a problem which one to use?
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these vids seem a little backwards in their order
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oops, sorry only just realised they were different :-/
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y2+y-20
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(x+5)(x-4)
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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??
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It factors to (x-6)(4x+19)
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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.
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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
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)
(t+3)
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Is 6x^2-13x+5 = (3x-5)(2x+1)?
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No. Use FOIL First Inside Outside Last to do it backwards

3x(2x) -10x +3x - 5 =
6x^2 -7x -5
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how do i get the coefficient, and the numbers,variables, to have a similar factor?
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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.
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how would you factorise 2x squared +3x -7 = 0
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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.
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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.
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i need help i don't understand how my instructor got the answer
x^2+xz/3; use x=3, and z=5
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Is there a simpler way to do this?
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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!
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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)
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Why is B there in (t+b)?
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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.
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why don't just do the magic x method?
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say the sum is 4 x^3 y^3 - 12 x^4 y^5 + 16 x^6 y^6
for the multiplication and addition part im unable to get the right numbers
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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:
81p^4-18p^2q^2+q^4
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or
(9p^2-q^2)(9p^2-q^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)
or
(9p^2 - q^2)^2
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What is a coefficient?
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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.
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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!
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Can x^2+24+143 be factored, having trouble with the operation.
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You can save some time by noticing that you only have to check for divisibility up to the square root of the number. For example, when checking factors of 143, we only have to go up to 11 because sqrt(143)~=11.9. All of the factors of 143 that are above 11.9 will be flushed out when you test the lower numbers.

In other words, you can stop checking factors when the result you get is smaller then the number you divided by.
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At 1:58, Sal says you can use this method if the coefficient is 1.
What if the coefficient of x term raised to the second power, is greater than one?
Like: 2x^2 + 2x + 1
Can you use this same factoring method?
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You can do something similar.
2X^2 +10X + 8.
Multiply A and C. 2*8 = 16.
Find factors of AC (16) that add to equal B (10). These are 8 and 2.
Divide each factor by A. 8/2 and 2/2.
Have each coefficient of X as 1
(X+ )(X+ )
(X+8/2)(X+2/2)
Multiply this by A
2(X+8/2)(X+2/2). This reduces to
2(x+4)(x+1).
The same thing that happens when A is 1, but multiplying and dividing by 1 don't change your values.
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What is a polynomial? (at 0:13)
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It is an expression that has two or more terms, in this case there are 3 terms.
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A good shortcut would be finding the factors of the last monomial or variable which you can add or subtract and the answer would be the numeral coefficient in the middle.
That sounded more complicated
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I think he was just showing the basis of the concept.
Yeah, the way you explained was how I was taught too.
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I was factoring polynomials 1 and the problem was x2(exponent)+3x-54 and I got +9 and -6 and it said it was wrong what did I do wrong?
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x^2+3x-54=0
x^2+(9x-6x)-54=0
(x^2+9x)-(6x+54)=0
x(x+9)-6(x+9)=0
(x+9)(x-6)=0
=>x+9=0 and x-6=0
=>x=-9 and x=6
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How do you do the problem a^3-9a+3a^2-27? How do you factor it. I'm really confused
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Continuing from Kate Ballard's explanation, the first quantity can be factored more using the difference of two squares. The final answer should be (a-3)(a+3)(a+3) and this can be changed to (a-3)(a+3)^2.
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Whats going on? Where did Sal get a and b from? can someone explain this to me?
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The a and b are what he uses to represent the numbers and can be used to represent the the coefficients of terms in any quadratic expression
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Sal is factoring whole numbers. What about fractions? Ex: x^2 - 4/3x + 4/9
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The factors of 4/9 are: 1, 4/9 or 2/3,2/3.
because in the original expression the middle term is negative the 4/9 is positive we are looking for two negative factors that add up to -4/3.

(x-2/3)(x-2/3)
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how do I factor by grouping polynomials to the third power? ie.
12n to the third -28n squared -15n + 35?
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You factor by grouping.
12n^3-28n^2-15n+35
You essentially create two binomials through grouping, so you have:
(12n^3-28n^2) and (-15n+35)
Then, you reduce each binomial by taking out a term (i.e. dividing by 5 or 5n or something) so the binomials left int the parentheses are equivalent:
4n^2(3n-7) -5(3n-7)
So in the equation above, 4n^2 was taken out of the first binomial by dividing, and -5 was taken out of the second binomial through division as well. Then, you can reduce this even further by taking out, or factoring out the 3n-7 and leaving 4n^2-5:
(3n-7)(4n^2-5)
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sigh... I'm so terrible at grasping concepts... I will give you a problem i dont understand..
If i factor by grouping: 7w squared+14w+wb+2b... how does it turn out to be (w+2)(7w+b)????
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(w+2)(7w+b)= w times 7w which gives you 7w squared plus w times b which gives you wb plus 2 timea 7w plus 2 times b.this gives you 7w squared+wb+14w+2b.when u put them in order from greatest to least you get 7w squared+14w+wb+2b.Hope this is helpful!
Comment
Report a mistake in the video
Example:

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

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