Direct proof that$ Var(cX) = c^2Var(X)$

Question

I'm trying to prove that $Var(cX) = c^2Var(X)$ and this is what I have so far:

$Var(cX) = E((cX - \mu)^2)$

$ = E(c^2X^2 - 2cX\mu + \mu^2)$

$ = c^2E(X)-2c\mu E(X)+\mu^2$ by linearity of expectation

$ = c^2E(X)-2c\mu^2+\mu^2$ by $E(X) =\mu$

Here, I am stuck, I know that I have a $c^2$ now but I'm having trouble making the right side of $c^2$ equal to $Var(X)$.

EDIT: With the help of below commenters, this is the correct proof.

$Var(cX) = E((cX - c\mu)^2)$

$ = E(c^2(X^2-2x\mu+\mu^2))$

$ = c^2E(X^2-2x\mu+\mu^2)$ by linearity

$=c^2E((X -\mu)^2)$

$= c^2Var(X)$

Answer 1

The definition of variance is:

$$\mbox{Var}[X]=E[(X-E[X])^2].$$

Notice that the inner expectation depends on $X$. Hence if you sub in $cX$ youll be able to pull out $c$ from both terms.