Introduction to special products of binomials | Algebra I | Khan Academy

Introduction to special products of binomials | Algebra I | Khan Academy

And now I want to do a bunch
of examples dealing with probably the two most typical
types of polynomial multiplication that you’ll see,
definitely, in algebra. And the first is just
squaring a binomial. So if I have x plus 9 squared, I
know that your temptation is going to say, oh, isn’t that
x squared plus 9 squared? And I’ll say, no, it isn’t. You have to resist every
temptation on the planet to do this. It is not x squared
plus 9 squared. Remember, x plus 9 squared,
this is equal to x plus 9, times x plus 9. This is a multiplication of this
binomial times itself. You always need to
remember that. It’s very tempting to think that
it’s just x squared plus 9 squared, but no, you have
to expand it out. And now that we’ve expanded it
out, we can use some of the skills we learned in the last
video to actually multiply it. And just to show you that we can
do it in the way that we multiplied the trinomial last
time, let’s multiply x plus 9, times x plus a magenta 9. And I’m doing it this way just
to show you when I’m multiplying by this
9 versus this x. But let’s just do it. So we go 9 times 9 is 81. Put it in the constants’
place. 9 times x is 9x. Then we have– go switch to this
x term– we have a yellow x. x times 9x is 9x. Put it in the first
degree space. x times x is x squared. And then we add everything up. And we get x squared
plus 18x plus 81. So this is equal to x squared
plus 18x plus 81. Now you might see a little bit
of a pattern here, and I’ll actually make the pattern
explicit in a second. But when you square a binomial,
what happened? You have x squared. You have this x times this
x, gives you x squared. You have the 9 times
the 9, which is 81. And then you have this term
here which is 18x. How did we get that 18x? Well, we multiplied this x times
9 to get 9x, and then we multiplied this 9 times
x to get another 9x. And then we added the two
right here to get 18x. So in general, whenever you have
a squared binomial– let me do it this way. I’ll do it in very general
terms. Let’s say we have a plus b squared. Let me multiply it this way
again, just to give you the hang of it. This is equal to a plus b, times
a plus– I’ll do a green b right there. So we have to b times
b is b squared. Let’s just assume that this
is a constant term. I’ll put it in the b squared
right there. I’m assuming this is constant. So this would be a constant,
this would be analogous to our 81. a is a variable that we–
actually let me change that up even better. Let me make this into x plus b
squared, and we’re assuming b is a constant. So it would be x plus b, times x
plus a green b, right there. So assuming b’s a constant,
b times b is b squared. b times x is bx. And then we’ll do
the magenta x. x times b is bx. And then x times
x is x squared. So when you add everything,
you’re left with x squared plus 2bx, plus b squared. So what you see is, the end
product, what you have when you have x plus b squared, is
x squared, plus 2 times the product of x and b,
plus b squared. So given that pattern, let’s
do a bunch more of these. And I’m going to do
it the fast way. So 3x minus 7 squared. Let’s just remember
what I told you. Just don’t remember it, in the
back of your mind, you should know why it makes sense. If I were to multiply this
out, do the distributive property twice, you know you’ll
get the same answer. So this is going to be equal
to 3x squared, plus 2 times 3x, times negative 7. Right? We know that it’s 2 times each
the product of these terms, plus negative 7 squared. And if we use our product rules
here, 3x squared is the same thing as 9x squared. This right here, you’re going
to have a 2 times a 3, which is 6, times a negative 7,
which is negative 42x. And then a negative 7
squared is plus 49. That was the fast way. And just to make sure that I’m
not doing something bizarre, let me do it the slow
way for you. 3x minus 7, times 3x minus 7. Negative 7 times negative
7 is positive 49. Negative 7 times 3x
is negative 21x. 3x times negative 7
is negative 21x. 3x times 3x is 9 x squared. Scroll to the left
a little bit. Add everything. You’re left with 9x squared,
minus 42x, plus 49. So we did indeed get
the same answer. Let’s do one more, and we’ll
do it the fast way. So if we have 8x minus 3–
actually, let me do one which has more variables in it. Let’s say we had 4x squared plus
y squared, and we wanted to square that. Well, same idea. This is going to be equal to
this term squared, 4x squared, squared, plus 2 times the
product of both terms, 2 times 4x squared times y
squared, plus y squared, this term, squared. And what’s this going
to be equal to? This is going to be equal to
16– right, 4 squared is 16– x squared, squared, that’s 2
times 2, so it’s x to the fourth power. And then plus, 2 times
4 times 1, that’s 8x squared y squared. And then y squared, squared,
is y to the fourth. Now, we’ve been dealing with
squaring a binomial. The next example I want to show
you is when I take the product of a sum and
a difference. And this one actually comes
out pretty neat. So I’m going to do a very
general one for you. Let’s just do a plus
b, times a minus b. So what’s this going
to be equal to? This is going to be equal to a
times a– let me make these actually in different colors–
so a minus b, just like that. So it’s going to be this green
a times this magenta a, a times a, plus, or maybe I should
say minus, the green a times this b. I got the minus from
right there. And then we’re going to have the
green b, so plus the green b times the magenta a. I’m just multiplying every
term by every term. And then finally minus the green
b– that’s where the minus is coming from–
minus the green b times the magenta b. And what is this going
to be equal to? This is going to be equal
to a squared, and then this is minus ab. This could be rewritten as
plus ab, and then we have minus b squared. These right here cancel out,
minus ab plus ab, so you’re just left with a squared
minus b squared. Which is a really neat
result because it really simplifies things. So let’s use that notion to
do some multiplication. So if we say 2x minus
1, times 2x plus 1. Well, these are the
same thing. The 2x plus 1, you could view
this as, if you like, a plus b, and the 2x minus 1, you can
view it as a minus b, where this is a, and that b is 1. This is b. That is a. Just using this pattern that
we figured out just now. So what is this going
to be equal to? It’s going to be a squared, it’s
going to be 2x squared, minus b squared, minus
1 squared. 2x squared is 4x squared. 1 squared is just
1, so minus 1. So it’s going to be 4x
squared minus 1. Let’s do one more of these,
just to really hit the point home. I’ll just focus on
multiplication right now. If I have 5a minus 2b, and
I’m multiplying that times 5a plus 2b. And remember, this only applies
when I have at a product of a sum and
a difference. That’s the only time that
I can use this. And I’ve shown you why. And if you’re ever in doubt,
just multiply it out. It’ll take you a little
bit longer. And you’ll see the terms
canceling out. You can’t do this for just any
binomial multiplication. You saw that earlier in the
video, when we were multiplying, when we were
taking squares. So this is going to be, using
the pattern, it’s going to be 5a squared minus 2b squared,
which is equal to 25 a squared minus 4b squared. And, well, I’ll leave it there,
and I’ll see you in the next video.

58 thoughts on “Introduction to special products of binomials | Algebra I | Khan Academy

  1. Your use of colour is, in my mind, what stands out between your tutorials and those of others. This video is a good example of what works, though there could still be more colour used in a way that maintained the sensibility, and improved the understanding of students of the material.

  2. i wish more teachers taught like this. the colors are awesome. its been a long time since i've been in school but this video is great! i wish i had this in school!

    keep up the good work sir!

  3. Thank you so much. I am currently going to an online college, and I haven't taken Algebra in years, watching you go through the problems enables me to better understand what it is that i am doing. thanks again =)

  4. Or … 
    Step 1 : Square of its first term .
    Step 2 : Product of its terms.
    Step 3 : Square of its last term.

    Simple .

  5. How about this…. (3-4m)²  , (8-3x) (8+3x) , and (2t-1) (t+5) ?? 

    PS: Can you solve this.. pretty please.??

  6. Wow thank you soo much this was our next lesson and since now that I learned it in advance I probably wont have any problems with catching up in my class 😁😁 thanks so much

  7. you can also do this if the sign is plus:
    1st step: Square the first term
    2nd step: Multiply the 2nd term/last term by 2 and the multiply to the 1st term
    3rd step: is to square the 2nd term/last term

  8. My teacher teaches me this for 1 hour, and you teach me for only 10 minutes… And I understand your lesson more than my teacher 😂😂

Leave a Reply

Your email address will not be published. Required fields are marked *