Algebra (all content)
- Special products of the form (x+a)(x-a)
- Squaring binomials of the form (x+a)²
- Multiply difference of squares
- Special products of the form (ax+b)(ax-b)
- Squaring binomials of the form (ax+b)²
- Special products of binomials: two variables
- More examples of special products
- Polynomial special products: perfect square
- Squaring a binomial (old)
- Binomial special products review
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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- Just wondering, what is a coefficient?(7 votes)
A coefficient is the number beside the variable
For instance in
the coefficient is 5 and the variable is x(28 votes)
- Wouldn't it be 4x^2+16x-16x-64??(4 votes)
- Well, you'd have to cancel out the positive 16x with the negative 16x.
So you'd be left with 4x^2 - 64.(7 votes)
- At0:08, Sal mentions FOIL. What is it?(3 votes)
- 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?(3 votes)
- I broke khan academy the video won't go into full screen can you please help(3 votes)
- 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.(2 votes)
- 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) ?(0 votes)
- That's a close guess, but it's a bit more complicated than multiplying both of the terms.
For the first and last terms, we can add the exponents because the bases are the same.
There are 2 -(d^3x)(b^3y), so we can make both -2(d^3x)(b^3y).
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!(8 votes)
- can some one please factor 4x^2 - 64 which is the result that sal get(1 vote)
- At2:20how 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?(1 vote)
- 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(3 votes)
- so im having a problem with a assignment problem of mine its (3x + n + 4)(3x - 4n - 1) i cant figure it out but im also really bad at math... Can someone please help?(1 vote)
- At this point, you should be familiar with FOIL for multiplying 2 binomials. The basic process of FOIL is expanded as the polynomials have more term.
If we had: (3x +4 )(3x - 1), we take the "3x" and we multiply it with both the "3x" and the "-1" in the 2nd binomial. Then, we move tot he 4, and we repeat that process, multiplying the 4 with both the "3x" and the "-1" in the 2nd binomial.
With your 2 trinomials:
1) We take the 1st "3x" and we multiply it with every term in the 2nd trinomial:
(3x + n + 4)(3x - 4n - 1) = 3x(3x) + 3x(-4n) + 3x(-1) ...
(3x + n + 4)(3x - 4n - 1) = 9x^2 - 12nx - 3x ...
2) We repeat the process, multiplying the "n" with every term in the 2nd trinomial:
(3x + n + 4)(3x - 4n - 1) = 9x^2 - 12nx - 3x + n(3x) + n(-4n) + n(-1) ...
(3x + n + 4)(3x - 4n - 1) = 9x^2 - 12nx - 3x + 3nx -4n^2 - n ...
3) We repeat the process again, multiplying the "4" with every term in the 2nd trinomial:
(3x + n + 4)(3x - 4n - 1) = 9x^2 - 12nx - 3x + 3nx -4n^2 - n + 4(3x) + 4(-4n) + 4(-1)
(3x + n + 4)(3x - 4n - 1) = 9x^2 - 12nx - 3x + 3nx - 4n^2 - n + 12x - 16n - 4
4) Last, combine any like terms:
9x^2 + (- 12nx + 3nx) + (- 3x + 12x) - 4n^2 + (- n - 16n) - 4
9x^2 - 9nx + 9x - 4n^2 - 17n - 4
And, we're done. Hope this helps.(4 votes)
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.