Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then: 
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
 A: $\left(2x^2+x\right)^2=\left(2x^2+x\right)\left(2x^2+x\right)=4x^4+4x^3+x^2$
A: Hint:
$$
(A+B)^2= (A+B)(A+B)=A^2+AB+BA+B^2=A^2+B^2+2AB
$$
Use: $A=2x^2$ and $B=x$ and you find the right result.
A: In general, $(a + b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$. So, in your case,
$$(2x^2 + x)^2 = (2x^2)^2 + 2(2x^2\cdot x) + x^2 = 4x^4 + 4x^3 + x^2$$
A: The commutativity property states that:


*

*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}+\color{green}{b} = \color{green}{b}+\color{red}{a}$

*For all $\color{red}{a},\color{green}{b}$ we have $\color{red}{a}\times\color{green}{b} = \color{green}{b}\times\color{red}{a}$


distributivity property of multiplication over addition states that:


*

*For all $\color{red}{a},\color{green}{b},\color{blue}{c}$ you have: $(\color{red}{a}+\color{green}{b})\times \color{blue}{c} = \color{red}{a}\times \color{blue}{c} + \color{green}{b}\times \color{blue}{c}$


$(2x^2 + x)^2 = (\color{red}{2x^2}+\color{green}{x})\color{blue}{(2x^2+x)}$
For the moment, let us refer to the blue parenthesis as a single piece, and use the distributivity property above:
$=\color{red}{2x^2}\color{blue}{(2x^2+x)}+\color{green}{x}\color{blue}{(2x^2+x)}$
Now, using commutativity and distributivity, and reassigning colors, we see that this is:
$=(\color{red}{2x^2}+\color{green}{x})\color{blue}{2x^2} + (\color{red}{2x^2}+\color{green}{x})\color{blue}{x} = \color{red}{2x^2}\times \color{blue}{2x^2}+\color{green}{x}\times\color{blue}{2x^2}+\color{red}{2x^2}\times\color{blue}{x}+\color{green}{x}\times\color{blue}{x}$
$=4x^4+2x^3+2x^3+x^2=4x^4+4x^3+x^2$
A: Think of it like this.
$a^{2}$ means $a \times a$.
More generally, $a^{n}$ means $\underbrace{a \times a \times a \times \dots \times a}_{n \text{ times}}$.
So $2^{3}$ means $2 \times 2 \times 2$. That's $2 \times 2$, $3$ times.
Therefore $(2x^{2} + x)^{2}$ means $(2x^{2} + x) \times (2x^{2} + x)$.
Now, using your number, we see that:
$(2 \times 3^{2}) + 3 = 18 + 3 = 21$
And $21 \times 21$ is very different to $(2 \times 3^{4}) + 3^{2} = 162 + 9 = 171$.
What you have learned the hard way is that exponentiation is not distributive over addition. Which means that $(a + b)^{2} \neq a^{2} + b^{2}$. Why is this? Because $(a + b)^{2}$ means:
Multiply $(a + b)$ by $(a + b)$
Not multiply $a$ by $a$ and multiply $b$ by $b$ and add them together.
For a more simple example than yours, let's look at $(3 + 2)^{2}$
$(3 + 2)^{2} = (3 + 2) \times (3 + 2) = 5 \times 5 = 25$
But:
$(3 + 2)^{2} \neq 3 \times 3 + 2 \times 2 = 9 + 4 = 13$
N.B.: Here we are talking about the general case. There is certainly one specific example of where $(a + b)^{2} = a^{2} + b^{2}$ but this is basically just by coincidence.
