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# Special products of the form (x+a)(x-a)

Sal introduces difference of squares expressions. For example, (x+3)(x-3) is expanded as x²-9.

## Want to join the conversation?

• cant you just used the FOIL method?
• Another thing is that the FOIL method is generally not useful when multiplying something other than two binomials (that is, multiplying either three or more polynomials or multiplying polynomials with three or more terms).

However, multiple applications of the distributive property work for finding any product of any polynomials. This ultimately means multiplying each term of each polynomial (using all the polynomials to be multiplied) and then adding the products.

Have a blessed, wonderful day!
• Can you distribute it like: x(x-3) 3(x-3) instead of x(x+3) -3(x+3)
• Almost... You would need to have: x(x-3) + 3(x-3)
Without it, you have an expression that is "3x (x-3)^2" which will create a completely different result that what you want.
Hope this helps.
• when will we ever use this IRL please It's a genuine question
• Finance, Medicine, Engineering, many many science fields, list goes on..
• It is common to type x squared or x to the power 2 with a 'carrot.'
x^2?
• Yes... that is a common way of typing exponents when you can't use superscripts.
• what are the solutions to x(x-a)-b(a+b)=0? Please explain in detail. TY.
• In its current form, it appears not to be factorable because for the 2 terms, the contents of the parentheses don't match. So, there is no common factor.

Distribute to eliminate the parentheses: x^2-ax-ab-b^2=0
Use commutative property to rearrange the terms. This may take a while to find the right sequence. I did:
x^2-b^2-ax-ab=0

Then group by pairs of terms and try to factor.
x^2-b^2 = (x-b)(x+b)
-ax-ab = -a(x+b)
So, not it looks like: (x-b)(x+b)-a(x+b)=0

Now the 2 terms have a matching factor of (x+b). Factor it out to get:
(x+b)(x-b-a) = 0

You can then use the zero product rule to split the factors and solve for the desired variable.
x+b=0 and x-b-a=0

Note: You didn't specify which variable you are solving for. So, I can't go further. I assume you were told to solve for "x" or solve for "b" as these occur in both factors.

You should be able to take it from here.
Comment back if you have questions.
• Why did Sal distribute (x-3) out to (x+3) and not vise versa?
(1 vote)
• Actually, Sal did distribute the (x+3). If you notice, the (x+3) is what is getting multiplied with the "x" and the "-3" after the distribution. If he had distributed the (x-3), the result would have been:
x(x-3) + 3(x-3)

Note: either method is acceptable because of the commutative property of multiplication.

Hope this helps.
• how do you solve (x+1)^3? My answer keeps besoming x^3+3x+1, but the book I use says the right answer is x^3+3x^2+3x+1.

How do I get to the 3x^2?
(1 vote)
• It looks like you tried to apply the technique for squaring a binomial to cube the binomial. It doesn't work to cube a binomial.

Think about how you cube any number...
If you were to do: 5^3, you would do 5*5*5 = 25*5 = 125
You need to do the same process.
-- Multiply 2 factors: (x+1)(x+1) = x^2+2x+1
-- Then, take the result and multiply it with the 3rd factor
(x+1)(x^2+2x+1) = x^3+2x^2+x+x^2+2x+1 = x^3+3x^2+3x+1

Hope this helps.
• Why is it ax instead of xa?
(1 vote)
• You could use either. But, the convention is to write the variables in alphabetic order which is "ax".
• Hello,
Why x (x+3) -3 (x+3) is equal to x^ +3x-3x-9?
I got confused a little bit about why the "3" have "x" on it?
Like how the "3" got "x" on it?
Like this "3x"?

But I understand a lot of this, except "3x"!
(1 vote)
• () why did we pick (x+a)to work with and not (x-a)
(1 vote)
• Probably because x+a is the first factor and x-a is the second factor