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I've seen

Binomial Theorem Question (Expansion of Three Terms)

Binomial Theorem with Three Terms

Expanding Equation with Binomial Theorem

but I'm not such a math expert, I need things explained in simple terms.

Basically I've heard that the solve

(x + y)^

it's essentially

(x + y)(x + y)

and then multiply the first terms, (leftmost), then first term of one with last term of the other (outer) then second term of one with first term of other (inner) then the two last terms of each (right most), and add them up, so

x * x + x * y + y * x + y * y =

x^2 + 2xy + y^2

that is pretty much all I know, now if I want to solve a more complicated binomial, with three or more terms, for example

(x + y + z)(x + y + z)

would I use a similar method? Meaning do I start with the left most terms, then x * y, then, what? then do I do x * z and then move on to the next right term, y, and do y * x + y * y + y * z, and then do the same for z, meaning z * x + z * y + z * z ? Am I missing something here, or is that it?

B''H Bi'ezras -- Boruch Hashem
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1 Answers1

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You do not need any theorems or results other than the distributive property: $a(b+c)=ab+ac$. Applying this once, we have $(x+y)(x+y)=x(x+y)+y(x+y)$. Here, the left $x+y$ is playing the role of $a$, and the right $x$ and $y$ are playing the roles of $b$ and $c$ respectively. Then we can use the same property again to see $x(x+y)=x^2+xy$ and $y(x+y)=yx+y^2$. Since $xy=yx$, we have

\begin{align} (x+y)(x+y)&=x(x+y)+y(x+y)\\ &=x^2+xy+yx+y^2\\ &=x^2+xy+xy+y^2\\ &=x^2+2xy+y^2. \end{align}

For three terms we can do the same thing:

\begin{align} (x+y+z)(x+y+z)&=x(x+y+z)+y(x+y+z)+z(x+y+z)\\ &=x^2+xy+xz+xy+y^2+yz+xz+yz+z^2\\ &=x^2+y^2+z^2+2xy+2xz+2yz. \end{align}

pancini
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  • ok thanks, I'm just trying to understand how in (x + y)(x + y), did the x get on the outside of one and y on the other, like you have x(x + y) + y(x + y), its probably very simply but I'm just drawing a blank as to how that was reached, also is the final result pretty much the same as multiplying the left most term by the left most term, then the left most term by the middle term, then adding it to the left most term times the right most term, etc.? – B''H Bi'ezras -- Boruch Hashem Jul 07 '20 at 05:27
  • The point is that $a(x+y)=ax+ay$ for any $a$. This includes when $a$ itself is a sum of many things. So $(x+y)(x+y)=(x+y)x+(x+y)y$, which is the same as $x(x+y)+y(x+y)$ (because we can multiply in any oder). – pancini Jul 07 '20 at 05:31
  • OK, and is the method mentioned at the end of the last comment the correct one, basically? – B''H Bi'ezras -- Boruch Hashem Jul 07 '20 at 05:42
  • Yes; in practice you just need to multiply every possible combination of terms from the left and the right and add them up. So $(a+b+c+d)(a+b+c+d)$ would have $16$ terms (before simplifying). – pancini Jul 07 '20 at 05:44
  • also would the same paply for multipaction, i.e. (x * y)^2 would be (x * y)(x * y) = (x ^ 2 * xy * xy * y^2) = (x ^2 * 2xy * y^2)? – B''H Bi'ezras -- Boruch Hashem Jul 07 '20 at 06:01