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### Course: Get ready for Algebra 2 > Unit 1

Lesson 5: Special products of binomials- Special products of the form (x+a)(x-a)
- Squaring binomials of the form (x+a)²
- Multiply difference of squares
- Multiply perfect squares of binomials
- Special products of the form (ax+b)(ax-b)
- Squaring binomials of the form (ax+b)²
- Binomial special products review

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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?(23 votes)
- 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!(44 votes)

- when will we ever use this IRL please It's a genuine question(12 votes)
- Finance, Medicine, Engineering, many many science fields, list goes on..(37 votes)

- Can you distribute it like: x(x-3) 3(x-3) instead of x(x+3) -3(x+3)(10 votes)
- Almost... You would need to have: x(x-3) + 3(x-3)

Your version is missing the plus sign in the middle.

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.(21 votes)

- It is common to type x squared or x to the power 2 with a 'carrot.'

x^2?(7 votes)- Yes... that is a common way of typing exponents when you can't use superscripts.(11 votes)

- doesn't (x+10)(x-10)=xsquared -100?(8 votes)
- yup (x+10)(x-10) does equal x^2 - 100(6 votes)

- what are the solutions to x(x-a)-b(a+b)=0? Please explain in detail. TY.(3 votes)
- 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.(12 votes)

- 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.(7 votes)

- 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.(6 votes)

- 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".(5 votes)

- How do you graph polynomial equations like we did earlier, and why not do it now?(3 votes)
- I assume he will eventually.(1 vote)

## Video transcript

- [Voiceover] Let's see
if we can figure out what x plus three times x minus three is, and I encourage you to pause the video and see if you can work this out. Well, one way to tackle
it is the way that we've always tackled it when
we multiply binomials, is just apply the
distributive property twice. So first we can take this
entire yellow x plus three and multiply times each
of these two terms. So first we can multiply it times this x. So that's going to be
x times x plus three. And then we are going to multiply it times, we can
say, this negative three. So we could write minus three times, now that's going to be multiplied by x plus three again. And then we apply the
distributive property one more time. Where we take this magenta x and we distribute it across this x plus three so x times x is x squared, x times three is three x, and then we do it on this side. Negative three times x is negative three x and negative three times three is negative nine. And what does this simplify to? Well, we're gonna get x squared, and we have three x and minus three x so these two characters cancel out, and we are just left with x squared minus nine. And you might see a little pattern here, notice I added three and
then I subtracted three and I got this, I got the x squared and then if you take three and multiply it by negative three, you are going to get a negative nine. And notice, the middle terms canceled out. And one thing you might ask is, well, will that always be the case, if we add a number and we subtract that same number like that? And we could try it out. Let's talk in general terms. So if we, instead of doing x plus three times x minus three, we could write this same thing as, instead of three, let's just say you have x plus a times x minus a. And I encourage you to pause this video and work it all out, just assume a is some number, like three or some other number, and apply the distributive property
twice and see what you get. Well, let's work through it. So first we can distribute
this yellow x plus a onto the x and the negative a. So x plus a times x, or we could say x times x plus a, that's going to be x times x plus a, and then we're going to have minus a, or this negative
a, times x plus a. So minus, and then we're gonna have this minus a times x, plus a. Notice, all I did is I distributed this yellow, I distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a. And now we can apply the
distributive property again. X times x is x squared, x times a is ax, and then we get negative a times x is negative ax, and then negative a times a, is negative a squared. And notice, regardless of my choice of a, I'm going to have ax and then minus ax. So this is always going to cancel out. It didn't just work for
the case when a was three. For any a, if I have a times x and then I subtract a times x, that's
just going to cancel out. So this is just going to cancel out, and what are we going to be left with? We are going to be left with x squared minus a squared. And you can view this as a special case. When you have something, x plus something, times x minus that same something, it's going to be x squared minus that something squared. And this is a good one to know in general. And we could use it to quickly figure out the products of other binomials that fit this pattern here. So if I were to say, quick, what is x plus 10, times x minus 10? Well, you could say, all right this fits the pattern, it's x plus a times x minus a, so it's going to be x
squared minus a squared. If a is 10, a squared is going to be 100. So you can do it really quick once you recognize the pattern.