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I've mentioned in previous questions that I have a hard time with simplifying algebraic equations.

For this equation, I assumed we would first put the equation into easier to read form:

$(k+1)(k+1)(k+1)$

Then apply FOIL to the first two:

EDIT: $(k^2+2k+1)(k+1)$

Although this is where I'm stuck. Any help is appreciated.

With all your help my answer is:

$k^3+3k^2+3k+1$

Thanks.

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Look into the binomial theorem. –  Max Oct 30 '12 at 23:24
    
Yes, I say that in my question. –  Unknown Oct 30 '12 at 23:24
    
You don't mention the binomial theorem anywhere in your question. –  Max Oct 30 '12 at 23:25
1  
@Max: I think the OP was referring to my comment, which I had deleted. –  wj32 Oct 30 '12 at 23:26
1  
There's a nice post on MathOverflow about FOIL, in "Most harmful heuristic?" Unfortunately, MO is down right now: mathoverflow.net/questions/2358/most-harmful-heuristic –  wj32 Oct 30 '12 at 23:27

6 Answers 6

up vote 4 down vote accepted

You just multiply each pair of terms: $$(k^2+2k+2)(k+1)=k^2\cdot k + k^2\cdot 1 + 2k\cdot k + 2k\cdot 1 + 2 \cdot k + 2 \cdot 1 $$

Of course you can also use the binomial theorem, but I'd try to master the basics first.

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Ah, that's simple and easy, thank you. –  Unknown Oct 30 '12 at 23:27

You can write $(k^2+2k+1)(k+1)=k(k^2+2k+1)+1(k^2+2k+1)$ then distribute to get $k(k^2+2k+1)+1(k^2+2k+1)=k^3+2k^2+k+k^2+2k+1$ then collect like terms to get $k^3+3k^2+3k+1$. This is more general than FOIL, which only handles the product of two items of two terms each.

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Use the distributive law (after correcting your first multiplication):

$$(k^2+2k+1)(\color{red}{k}+\color{blue}{1})=(k^2+2k+1)\cdot\color{red}{k}+(k^2+2k+1)\cdot\color{blue}{1}\;.$$

Now $(k^2+2k+1)\cdot k=k^3+2k^2+k$, and $(k^2+2k+1)\cdot 1=k^2+2k+1$, so all that’s left is to add the two polynomials:

$$\begin{align*} (k^3+2k^2+k)+(k^2+2k+1)&=k^3+(2k^2+k^2)+(k+2k)+1\\ &=k^3+3k^2+3k+1\;. \end{align*}$$

You can also organize this calculation like a pencil-and-paper multiplication:

$$\begin{array}{r} &&k^2&+&2k&+&1\\ &&&&k&+&1\\ \hline &&k^2&+&2k&+&1\\ k^3&+&2k^2&+&k\\ \hline k^2&+&3k^2&+&3k&+&1 \end{array}$$

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I'll add \color into my memory banks. –  muntoo Oct 31 '12 at 4:15

You can apply a similar procedure and distribute the two binomial terms over the tri-nomial:

$$ \begin{align*}(k^2 + 2k + 1) (k+1) &= (k^2 + 2k^\hphantom{2} + 1) k + (k^2 + 2k + 1) 1\\ &= \hphantom{(}k^3 + 2k^2 + k\hphantom{1)} + \hphantom{(}k^2 + 2k + 1 \\ &= k^3 + 3k^2 + 3k + 1 \end{align*}$$

Note:

  • You have a mistake in the question: $(k+1)^2 = (k^2 + 2k + 1)$
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Final line isn't correct... –  Epictetus Oct 30 '12 at 23:29
    
@Epictetus: Yeah, I naively assumed that the beginning was correct in the problem. –  Peter Grill Oct 30 '12 at 23:34
    
Please don't call the expansion of a trinomial times a binomial FOIL! –  The Chaz 2.0 Oct 30 '12 at 23:44
    
@TheChaz: Corrected. Let me know if this is better. –  Peter Grill Oct 31 '12 at 0:03

You might want to learn how this triangle works:

$$ 1\\ 1\,1\\ 1\,2\,1\\ 1\,3\,3\,1\\ 1\,4\,6\,4\,1\\ 1\,5\,10\,10\,5\,1 $$ The $n^{th}$ line (first one is zero) gives the coefficient of $x^{k}y^{n-k}$ in $(x+y)^n$

Since you had $(k+1)^3$, the coefficient are gonna be $1,3,3,1$ and now we have to figure out how to work the $k$.

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Multiplication is distributive over addition, so:

$$(k^2+2k+2)(k+1)=(k^2+2k+2)(k)+ (k^2+2k+2)(1) $$ $$=(k^3+2k^2+2k)+(k^2+2k+2)$$ Now, sum up like terms: $$=k^3+3k^2+4k+2$$

Hope that helps.

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