# Cubing a simple thing

I am trying to expand $\quad (x + 2)^3$

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.

The following is correct, though not fully expanded:

$$(x^2 + 4x + 4)(x + 2) = (x+2)^2(x+2) = (x+2)^3$$

Consider writing this as $(x + 2)(x^2 + 4x + 4)$, and then distribute (multiply) each term in the first factor, with each term of the second factor.

\begin{align} (\color{blue}{\bf x + 2})(x^2 + 4x + 4) & = \color{blue}{\bf x}(x^2 + 4x + 4) + \color{blue}{\bf 2}(x^2 + 4x + 4) \\ \\ & = (x^3 + {\bf 4x^2} + \color{blue}{\bf 4x}) + ({\bf 2x^2 }+ \color{blue}{\bf 8x} + 8) \\ \\ & = x^3 + {\bf 6x^2} + \color{blue}{\bf 12x} + 8\\ \\ \end{align}

• I still get it wrong, I end up with $2x^3 + 8x^2 + 8x$ – toby yeats May 18 '13 at 2:35
• Can you see you have $x^3 + (4 + 2)x^2 + (4+8)x + 8$? – Namaste May 18 '13 at 2:38
• Is this making more sense now? – Namaste May 18 '13 at 2:45
• I see what I did wrong now, I multiplied by 2 the part I already multiplied by x when I should have multiplied seperated and added. – toby yeats May 18 '13 at 2:48
• Just practice a lot: it will become automatic in no time! ;-) – Namaste May 18 '13 at 2:55

Why is $(x^{2} + 4x + 4)(x+2)$ wrong?

$(x^{2} + 4x + 4)(x+2) = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8 = x^{3} + 6x^{2} + 12x + 8$

And it is the right answer. If you have problem remembering the power formulas for binomials, just use pascal's triangle to calculate the coefficients:

\begin{align*} (a + b)^{0}\to &1\\ (a + b)^{1}\to &1 &1\\ (a + b)^{2}\to &1 &2 &&1\\ (a + b)^{3}\to &1 &3 &&3 &&1\\ (a + b)^{4}\to &1 &4 &&6 &&4 &&1\\ \end{align*} (and so on)

The pattern here is that every number is the sum of the two adjacent number from the previous line. These numbers are the coefficients of the nth power. The pattern of the exponents is: $$\sum_{i=0}^{n} a^{n-i} b^{i}$$ Therefore: $$(x + 2)^{3} = (1)(x^{3})(2^{0}) + (3)(x^{2})(2^{1}) + (3)(x^{1})(2^{2}) + (1)(x^{0})(2^{3}) = x^{3} + 6x^{2} + 12x + 8$$ By the way:

$(x^{2}+4x+4)(x^{2}+4x+4) = (x + 2)^{2} (x + 2)^{2} = (x+2)^{4} \neq (x+3)^{3}$

• You know, you're the first person that explained the binomial theorem in a way I understand it... though I've only looked at textbooks... +1! – Stephen J May 18 '13 at 7:21
• I'm glad you liked it =) – Felipe Gavilan May 18 '13 at 14:41

Given: $(x+2)^3$ we can rewrite $(x+2)^3$ as: $(x+2)(x+2)(x+2)$

When expanding 3 terms we must first calculate $(x+2)(x+2)$ before we can multiply by the third $(x+2)$.

= $(x+2)(x+2)\implies x^2+2x+2x+4\implies x^2+4x+4$

= $(x+2)(x^2+4x+4)\implies x(x^2+4x+4)$ + $2(x^2+4x+4)$

= $x^3+4x^2+4x+2x^2+8x+8$

= $x^3+6x^2+12x+8$

• nitrous2, why did you edit it back? Do you see that you forgot a plus when you say $(x+2)(x^2+4x+4)\implies x(x^2+4x+4)$ $2(x^2 + 4x + 4)$? $(x+2)(x^2+4x+4) = x(x^2+4x+4)\color{red}{+}2(x^2 + 4x + 4)$ – Stahl May 18 '13 at 2:48
• Fixed, thanks for noticing. – nitrous2 May 18 '13 at 3:00

You had a correct step, when you had $(x^2 + 4x+4)(x+2)$. Now just distribute.

$$(x^2 + 4x + 4)x + (x^2 + 4x + 4)2 = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 = x^3 + 6x^2 + 12x + 8$$

$$(x^2 + 4x + 4)(x + 2)$$

was correct, because $(x + 2)^2 = (x^2 + 4x + 4)$. So it follows that $(x + 2)^3 = (x^2 + 4x + 4)(x + 2)$.

Let's try just multiplying it out:

To multiply $(x^2 + 4x + 4)$ by $(x+2)$, we'll multiply every term in $(x^2 + 4x + 4)$ by every term in $(x+2)$.

That's easy to do, and easy to visualise, because $(x^2 + 4x + 4)\cdot (x + 2)$ is the same as $(x \cdot(x^2 + 4x + 4)) + (2 \cdot (x^2 + 4x + 4)).$

$$(x \cdot(x^2 + 4x + 4)) = x^3 + 4x^2 + 4x$$

$$(2 \cdot (x^2 + 4x + 4)) = 2x^2 + 8x + 8$$

$$x^3 + \color{red}{4x^2} + \color{blue}{4x} + \color{red}{2x^2} + \color{blue}{8x} + 8 = x^3 + 6x^2 + 12x + 8$$

There is a more intuitive way to do this that works like the FOIL method.

Write $(x+2)^3$ as $(x+2)(x+2)(x+2)$.

When we multiply these we look at all the ways we can pick one element from each of the $(x+2)$s, multiply each possibility and add them all up.

What do these look like? Well, from each $(x+2)$ you can pick either an $x$ or a $2$. Since we pick one from each, we will get three things multiplied together.

If we pick an $x$ from each we will get an $x\cdot x\cdot x$ which equals $x^3$. We can only do this one way so we get $x^3$.

What if we pick an $x$ from only two of the $(x+2)$s? Then we have to pick a $2$ from the third. How many ways are there to do this? Well, we can pick an $x$ from the first two and a number from the third; we can pick an $x$ from the first and the third and a number from the second or we can pick an $x$ from the second and the third and a number from the first. This gives us $3(2x^2)=6x^2$. So far we have $x^3+6x^2$.

What if we pick only one $x$ and a $2$ from the other two? What do we get and how many ways are there to do this? We get $2\cdot 2\cdot x=4x$ and there are three ways to do this because we can pick the $x$ from either of the three. So we get $3(4x)=12x$ which gives us $x^3+6x^2+12x$ all together.

Are there any other possibilities? Yes, we can pick a number from each and there is only one way to do that so we get $2\cdot 2\cdot 2=8$ and finally we arrive at $(x+2)^3=x^3+6x^2+12x+8$.

This approach is easier than it seems if you practice a few and it works for $(x+a)^n$ for any positive n.

• I have never seen that, seems like it would get very complicated with powers over 3 though. – toby yeats May 18 '13 at 2:59
• @Jordan It is the intuition behind the binomial theorem. If you practice a few you will like it. – John Douma May 18 '13 at 3:00