For what values of a ​​the function $y=x^6+ax^3-2x^3-2x^2+1$ is even I want to know for what valuyes this function is even
I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!
 A: Let us start by evaluationg $f(x)$ and $f(-x)$.
$f(x) = x^6 + ax^3 - 2x^3 -2x^2 +1$
and
$f(-x) = x^6 - ax^3 + 2x^3 -2x^2 +1$
If $f(x)$ should equal $f(-x)$ then $a = $...
A: $$f(x)=f(-x)\\
x^6+ax^3-2x^3-2x^2+1=x^6-ax^3+2x^3-2x^2+1\\
(a-2)x^3=(2-a)x^3
$$
So for what values of $a$ is this equality satisfied?
$$a-2=2-a\iff 2a=4\iff a=2$$
A: Hint: Show that a polynomial is even if and only if the coefficients of all the odd powers are 0.
This is a complete classification of even polynomial functions, which is more general than the question you're asking, but a good concept for you to be aware of.
A: $f(-x)=(-x)^6 + a(-x)^3 -2(-x)^3 -2(-x)^2 + 1 = x^6 -ax^3 + 2x^3 - 2x^2 +1 $
$= x^6 +(-a+2)x^3 -2x^2 +1$
So for the two polynomials to be equal we need $2-a=a-2$ Hence, $2a=4, a=2$
A: Having even values is different to being an even function. For example, the function $\operatorname{f} : x \mapsto x$ has many even values, e.g. $2 \mapsto 2$ and $4 \mapsto 4$. However, $\operatorname{f}$ is not an even function since $\operatorname{f}(-x) \neq x$ for all $x \neq 0$.
In the example you give, the function cannot be even if it contains odd powers of $x$. Consider 
\begin{array}{ccc}
\operatorname{f}(-x) &\equiv& (-x)^6 + a(-x)^3 - 2(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& (-x)^6 + (a-2)(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& x^6 +(2-a)x^3 -2x^2 + 1
\end{array}
Hence, $\operatorname{f}(x) \equiv \operatorname{f}(-x)$ if and only if $a-2=2-a$, i.e. $a=2$. Notice that $a=2$ if and only if the $x^3$, i.e. the odd powered term, vanishes.
