# How to expand binomials?

I'm working on a few proofs and am missing how this algebra works....

So, how does one expand $(k+1)^3\,$? Can I use FOIL? What does it expand to?

And how to expand $(k+1)^5\,$?

Thanks!

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look up Pascal's Triangle, and Binomial Theorem. Also no, you can't use FOIL, at very least you must use distributive property. – Joseph Skelton Jan 22 '13 at 20:59

Check out the entry Binomial Theorem in Wikipedia.

Putting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\; (k+1)^5, \;$ and $\,(k + 1)^n\,$ for any non-negative integer $n$.

FOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$

One can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$

But for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.

I'll include an animation and image of "Pascal's Triangle" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:

$\quad\quad\quad\quad$ $\quad\quad\quad\quad$

$$\text{Each number in the triangle is the sum of the two directly above it.}$$

To see how this "plays out" in the expansion of $(x + 1)^n,\;0 \le n \le 6$:

$$(x + 1)^0 = \color{blue}{\bf{1}}$$ $$(x + 1)^1 = \color{blue}{\bf{1}}\cdot x +\color{blue}{\bf{1}}$$ $$(x + 1)^2 = \color{blue}{\bf{1}}\cdot x^2 + \color{blue}{\bf{2}}x + \color{blue}{\bf{1}}$$ $$(x+1)^3 = \color{blue}{\bf{1}}\cdot x^3 + \color{blue}{\bf{3}}x^2 + \color{blue}{\bf{3}}x + \color{blue}{\bf{1}}$$ $$(x+1)^4 = \color{blue}{\bf{1}}\cdot x^4 + \color{blue}{\bf{4}} x^3+ \color{blue}{\bf{6}}x^2 + \color{blue}{\bf{4}}x +\color{blue}{\bf{1}}$$ $$(x+1)^5 = \color{blue}{\bf{1}}\cdot x^5 + \color{blue}{\bf{5}}x^4 + \color{blue}{\bf{10}} x^3 + \color{blue}{\bf{10}} x^2 + \color{blue}{\bf{5}}x + \color{blue}{\bf{1}}$$ $$(x + 1)^6 = \color{blue}{\bf{1}}\cdot x^6 + \color{blue}{\bf{6}}x^5 +\color{blue}{\bf{15}}x^4 + \color{blue}{\bf{20}}x^3 +\color{blue}{\bf{15}}x^2 + \color{blue}{\bf{6}}x + \color{blue}{\bf{1}}$$ $${\bf{\vdots}}$$

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Image source: Wikipedia, found at link to "Pascal's Triangle". – amWhy Jan 22 '13 at 22:17
Nice animated GIF! The answer is good, too :-) (+1) – robjohn Jan 22 '13 at 22:53
Thanks, @robjohn! It says more concisely, in animation, what would take a bit of explaining to write! – amWhy Jan 22 '13 at 22:56
@amWhy: It is unbelievable!! Great!! I cannot say anything. Nothing such this can describe the problem soooo well. +10! – Babak S. Jan 23 '13 at 15:50
This is incredible! Perfect explanation. Thank you! – user56763 Jan 23 '13 at 16:44

Before you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that $$(k+1)^3=(k+1)(k+1)(k+1)$$ We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have $$(k+1)(k+1)=k^2+2k+1$$ We're two-thirds of the way to the answer. Now we have $$(k+1)^3=(k+1)(k^2+2k+1)$$ and by the distributive property, namely that $(a+b)c=ac+bc$, we have \begin{align} (k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\ &=(k^3+2k^2+k)+(k^2+2k+1)\\ &=k^3+3k^2+3k+1 \end{align} This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.

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Thanks, makes sense. – user56763 Jan 23 '13 at 16:45
@ user56763: From what Rick Decker posted you can see that $(k+1)^3 = (k+1)(k+1)(k+1)$ can be expanded in the following way. Write down all possible 3-fold products of the form $ABC,$ where $A$ is a choice of one of the terms in the first $(k+1)$ factor, $B$ is a choice of one of the terms in the second $(k+1)$ factor, and $C$ is a choice of one of the terms in the third $(k+1)$ factor. Then add the 8 products. Thus, among the things you'll add will be $(k)(k)(1)$ and $(1)(k)(k)$ and $(k)(1)(k).$ Note that there are 3 of these with exactly two $k$'s and exactly one $1,$ so you get $3k^2$ ... – Dave L. Renfro Jan 28 '13 at 19:30

You can use "FOIL" twice. You should get $$k^3+3k^2+3k+1.$$ More generally, $$(a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3.$$

Please don't vacilate between lower-case $k$ and capital $K$ in mathematical notation. Pick one and stick to it. Mathematical notation is case sensitive. Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.

But it would take a while to use FOIL to get $$(a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9.$$ That's one reason to be aware of the binomial theorem, which explains the pattern.

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What you're looking for is the binomial theorem, where y = 1.

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This is very scant. Something more would be good: some explanation or at least a link to an article. Of course, an answer should be more than simply a link. Look at the other answers for examples. – robjohn Jan 22 '13 at 22:56
Yeah, I was going to, but was having trouble typing out the math for a more exhaustive answer. – MITjanitor Jan 23 '13 at 2:10