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## वर्ग 10 (Foundation)

### Course: वर्ग 10 (Foundation)>Unit 1

Lesson 5: Standard identities

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

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

## Want to join the conversation?

• 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 , Sal mentions FOIL. What is it?
• How do I know when I have my final answer?
• 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 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
• 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.
• 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!