Just wondering, what is a coefficient?

`A coefficient is the number beside the variable` For instance in 5x the coefficient is 5 and the variable is x

Wouldn't it be 4x^2+16x-16x-64??

Well, you'd have to cancel out the positive 16x with the negative 16x. So you'd be left with 4x^2 - 64.

At 0:08, Sal mentions FOIL. What is it?

Sal talks about it in this video: www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:multiply-binomial/v/multiplying-binomials

I broke khan academy the video won't go into full screen can you please help

As always, make sure the problem isn't fixed by closing out of Khan Academy and going back into it. You can also find the youtube version of the KA video by just searching the title of the video on youtube. You should be able to go into fullscreen there.

Wait Sal says at the end that the outcome of the 2x^2 - 8^2 is 4x^2 - 64 wouldn't it be 4x - 64 since "a" is 2?

Sal starts with (2x-8)(2x+8) "a" = 2x, not 2 (2x)^2 = 2x * 2x = 4x^2, not 4x. Hope this helps.

At 2:20 how come you do not get rid of the exponent after you square 2x^2? Is it because it still contains a variable? because after you squared (8)^2 = 64 ...no more exponent, but when you square (2x)^2, it turns into 4x^2?

In (2x)^2, both the 2 and the X have to be squared. We can calculate 2^2 = 4. But, we don't know the numeric value for X. So, we can't calculate X*X. We just write it in exponent form: x^2. (2x)^2 = (2x) (2x) = (2*2) (x*x) = 4x^2

can some one please factor 4x^2 - 64 which is the result that sal get

Of course! In factored form it'd be: 4(x + 4)(x - 4); hope that helps.

Is it possible to square an exponent? For example (d^3x-b^3y)^2 In other way of writing (d^3x-b^3y)(d^3x-b^3y) Would it be (d^6x^2+b^6y^2) ?

That's a close guess, but it's a bit more complicated than multiplying both of the terms. (d^3x-b^3y)(d^3x-b^3y) d^3x(d^3x-b^3y)-b^3y(d^3x-b^3y) (d^3x)(d^3x)-(d^3x)(b^3y)-(b^3y)(d^3x)+(b^3y)(b^3y) For the first and last terms, we can add the exponents because the bases are the same. d^6x-(d^3x)(b^3y)-(d^3x)(b^3y)+b^6y There are 2 -(d^3x)(b^3y), so we can make both -2(d^3x)(b^3y). d^6x-2(d^3x)(b^3y)+b^6y Thus, (d^3x-b^3y)^2 or (d^3x-b^3y)(d^3x-b^3y) is d^6x-2(d^3x)(b^3y)+b^6y. I hope this helped!

Main content

Get ready for Algebra 2

Course: Get ready for Algebra 2 > Unit 1

Lesson 5: Special products of binomials

Special products of the form (ax+b)(ax-b)

Name: Special products of the form (ax+b)(ax-b)
Uploaded: 2011-02-20T16:40:59Z
Description: Sal expands the difference of squares (2x+8)(2x-8) as 4x²-64.

CCSS.Math:

HSA.APR.A.1

Google Classroom

Sal expands the difference of squares (2x+8)(2x-8) as 4x²-64. Created by Sal Khan and Monterey Institute for Technology and Education.

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TheEpicOne
Posted 9 years ago. Direct link to TheEpicOne's post “Just wondering, what is a...”
Just wondering, what is a coefficient?
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(7 votes)
Answer
- Petrie (Peter S. Asiain III)
  Posted 9 years ago. Direct link to Petrie (Peter S. Asiain III)'s post “`A coefficient is the num...”
  A coefficient is the number beside the variable
  For instance in
  5x
  the coefficient is 5 and the variable is x
  Comment on Petrie (Peter S. Asiain III)'s post “`A coefficient is the num...”
  (32 votes)
Contreras-Norwood Luis
Posted 9 years ago. Direct link to Contreras-Norwood Luis's post “Wouldn't it be 4x^2+16x-1...”
Wouldn't it be 4x^2+16x-16x-64??
Button navigates to signup pageComment on Contreras-Norwood Luis's post “Wouldn't it be 4x^2+16x-1...”
(4 votes)
Answer
- Karla Bolton
  Posted 9 years ago. Direct link to Karla Bolton's post “Well, you'd have to cance...”
  Well, you'd have to cancel out the positive 16x with the negative 16x.
  So you'd be left with 4x^2 - 64.
  Comment on Karla Bolton's post “Well, you'd have to cance...”
  (7 votes)
lms106
Posted 4 years ago. Direct link to lms106's post “At 0:08, Sal mentions FOI...”
At
0:08
, Sal mentions FOIL. What is it?
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(3 votes)
Answer
- Poydoo
  Posted 3 years ago. Direct link to Poydoo's post “Sal talks about it in thi...”
  Sal talks about it in this video:
  https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:multiply-binomial/v/multiplying-binomials
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  (4 votes)
jlampkin410
Posted 6 years ago. Direct link to jlampkin410's post “How do I know when I have...”
How do I know when I have my final answer?
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Glenn Crosby
Posted 4 years ago. Direct link to Glenn Crosby's post “Wait Sal says at the end ...”
Wait Sal says at the end that the outcome of the 2x^2 - 8^2 is 4x^2 - 64 wouldn't it be 4x - 64 since "a" is 2?
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(2 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “Sal starts with (2x-8)(2x...”
  Sal starts with (2x-8)(2x+8)
  "a" = 2x, not 2
  (2x)^2 = 2x * 2x = 4x^2, not 4x.
  Hope this helps.
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  (5 votes)
Colin Louis Blake
Posted 7 years ago. Direct link to Colin Louis Blake's post “At 2:20 how come you do n...”
At
2:20
how come you do not get rid of the exponent after you square 2x^2? Is it because it still contains a variable? because after you squared (8)^2 = 64 ...no more exponent, but when you square (2x)^2, it turns into 4x^2?
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(2 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “In (2x)^2, both the 2 and...”
  In (2x)^2, both the 2 and the X have to be squared. We can calculate 2^2 = 4. But, we don't know the numeric value for X. So, we can't calculate X*X. We just write it in exponent form: x^2.
  (2x)^2 = (2x) (2x) = (2*2) (x*x) = 4x^2
  Comment on Kim Seidel's post “In (2x)^2, both the 2 and...”
  (4 votes)
RedBlackandBlue
Posted 3 years ago. Direct link to RedBlackandBlue's post “I broke khan academy the ...”
I broke khan academy the video won't go into full screen can you please help
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(3 votes)
Answer
- Hecretary Bird
  Posted 3 years ago. Direct link to Hecretary Bird's post “As always, make sure the ...”
  As always, make sure the problem isn't fixed by closing out of Khan Academy and going back into it.
  You can also find the youtube version of the KA video by just searching the title of the video on youtube. You should be able to go into fullscreen there.
  Comment on Hecretary Bird's post “As always, make sure the ...”
  (2 votes)
Chaguz Antonio Mangubat Oca
Posted 10 years ago. Direct link to Chaguz Antonio Mangubat Oca's post “Is it possible to square ...”
Is it possible to square an exponent? For example (d^3x-b^3y)^2
In other way of writing (d^3x-b^3y)(d^3x-b^3y)
Would it be (d^6x^2+b^6y^2) ?
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(0 votes)
Answer
- Noble Mushtak
  Posted 10 years ago. Direct link to Noble Mushtak's post “That's a close guess, but...”
  That's a close guess, but it's a bit more complicated than multiplying both of the terms.
  
  (d^3x-b^3y)(d^3x-b^3y)
  d^3x(d^3x-b^3y)-b^3y(d^3x-b^3y)
  (d^3x)(d^3x)-(d^3x)(b^3y)-(b^3y)(d^3x)+(b^3y)(b^3y)
  For the first and last terms, we can add the exponents because the bases are the same.
  d^6x-(d^3x)(b^3y)-(d^3x)(b^3y)+b^6y
  There are 2 -(d^3x)(b^3y), so we can make both -2(d^3x)(b^3y).
  d^6x-2(d^3x)(b^3y)+b^6y
  
  Thus, (d^3x-b^3y)^2 or (d^3x-b^3y)(d^3x-b^3y) is d^6x-2(d^3x)(b^3y)+b^6y.
  
  I hope this helped!
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  (8 votes)
abdishakur ali
Posted 8 years ago. Direct link to abdishakur ali's post “can some one please facto...”
can some one please factor 4x^2 - 64 which is the result that sal get
Button navigates to signup pageComment on abdishakur ali's post “can some one please facto...”
(1 vote)
Answer
- Karla Bolton
  Posted 8 years ago. Direct link to Karla Bolton's post “Of course! In factored fo...”
  Of course!
  In factored form it'd be: 4(x + 4)(x - 4); hope that helps.
  Button navigates to signup page
  (5 votes)
jonathanr48
Posted 4 months ago. Direct link to jonathanr48's post “The answer is 4x^2-64”
The answer is 4x^2-64
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(2 votes)
Answer

Video transcript

Find the product 2x plus 8 times 2x minus 8. So we're multiplying two binomials. So you could use FOIL, you could just straight up use the distributive property here. But the whole point of this problem, I'm guessing, is to see whether you recognize a pattern here. This is of the form a plus b times a minus b, where here a is 2x and b is 8. We have 2x plus 8 and then 2x minus 8. a plus b, a minus b. What I want to do is I'm just going to multiply this out for us. And then just see what happens. Whenever you have this pattern, what the product actually looks like. So if you were to multiply this out, we can distribute the a plus b. We could distribute this whole thing. Distribute the whole a plus b on the a and then distribute it on the b. And I could have done this with this problem right here, and it would have taken us less time to just solve it. But I want to find out the general pattern here. So a plus b times a. So we have a times a plus b, that's this times this. And then a plus b times negative b, that's negative b times a plus b. So I've done distributive property once, now I could do it again. I can distribute the a onto the a and this b and it gives me a squared. a times a is a squared, plus a times b, which is ab. And now I can do it with the negative b. Negative b times a is negative ab or negative ba, same thing. And negative b times b is negative b squared. Now, what does this simplify to? Well, I have an ab, and then I'm subtracting an ab. So these two guys cancel out and I am just left with a squared minus b squared. So the general pattern, and this is a good one to just kind of know super fast, is that a plus b times a minus b is always going to be a squared minus b squared. So we have an a plus b times an a minus b. So this product is going to be a squared. So it's going to be 2x squared minus b squared minus 8 squared. 2x squared, that's the same thing as 2 squared times x squared, or 4x squared. And from that, we're subtracting 8 squared. So it's going to be 4x squared minus 64.