Find the function $f(x)$. Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?
I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.
 A: Let's try to determine the degree of the polynomial $f(x).$
If $\ f(x)\ $ has degree $0$, then $f(x) = a_0\ $ which implies $\ f(x+2)-f(x) = 0 \neq 8x-2.$
If $\ f(x)\ $ has degree $1$, then $f(x) = a_0 + a_1x\ $ which implies $\ f(x+2)-f(x) = 2a_1 \neq 8x-2.$
If $\ f(x)\ $ has degree $2$, then $f(x) = a_0 + a_1x + a_2x^2\ $ which, after some cancelling, implies $\ f(x+2)-f(x) = 4a_2x + (2a_1+4a_2).$ For this to equal $8x-2,\ $ we must have $a_2 = 2\ $ and $a_1 = -5.$ So $f(x) = a_0 - 5x + 2x^2.\ $ The last condition, $f(0) = 5\ $ implies that $a_0 = 5.$ So $f(x) = 5 - 5x + 2x^2\ $ is certainly a solution - the only solution of degree 2, since we got there by deduction.
Are there any other solutions (of degree $\geq 3$)?
Well if the degree of $f(x) \geq 3,\ $ then $$f(x+2)-f(x) = a_0 + a_1(x+2) + a_2(x+2)^2 + ... + a_n(x+2)^n - (a_0 + a_1x+a_2x^2+...+a_nx^n),\ $$ where $n\geq3.\ $ From the binomial expansion, the $x^{n-1}\ $ coefficient of $f(x+2)-f(x)\ $ is equal to $a_{n-1} + 2na_n - a_{n-1} = 2na_n.\ a_n \neq 0\ $ and $n \neq 0,\ $ therefore the $x^{n-1}\ $ term in $f(x+2)-f(x)\ $ has at degree of at least $2\ $, and so $f(x+2)-f(x)\neq 8x-2.$ So no, there are no other solutions.
A: Suppose that $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots$ where $n$ is the degree of the polynomial. What is the order of the polynomial $f(x+2)-f(x)$ in this case?
You should then be able to use the equality to ascertain $n$. At this point your task is much easier - using the fact that $f(0)=5$ and the equality you should be able to completely determine $f$.
A: We have two conditions, as the result is $(8x-2)$.
To be solvable we need to use a quadratic with two parameters
$$f(x)=ax^2+bx+5$$
so we have
$$f(x+2)-f(x)=4 a x+(4 a+2 b)\equiv 8x-2$$
setting
$$
\begin{cases}
4a=8\\
4a+2b=-2\\
\end{cases}
$$
we have $a=2,b=-5$ so finally
$$f(x)=2 x^2-5 x+5$$
A: You stated, that $f$ is a polynomial function, hence we can write $f$ as $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ plugging that into your given equation we get
$$f(x+2)-f(x)=a_n(x+2)^n+a_{n-1}(x+2)^{n-1}+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$
using the binomial formula, we can extend this to
$$a_n(x^n+2nx^{n-1}+...+2^{n-1}nx+2^n)+a_{n-1}(x^{n-1}+2(n-1)x^{n-2}+...+2^{n-2}(n-1)x+2^{n-1})+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$
We see, that the $a_nx^n$ cancel each other out; if we look at the coefficient of $x^{n-1}$ we have $2na_n+a_{n-1}-a_{n-1}=2na_n$. This is nonzero, as $a_n$ as the leading coefficient can't be zero.
Thus your polynomial is of degree $2$. That means $f(x)=a_2x^2+a_1x+a_0$. Since $f(0)=5$, we know that $a_0=5$. Plugging what we know so far back into our original equation we get:
$$f(x+2)-f(x)=a_2(x+2)^2+a_1(x+2)+5-a_2x^2-a_1x-5=4a_2x+4a_2+2a_1=8x-2$$
which yields $a_2=2$ and thus $a_1=-5$ giving us the complete polynomial $f(x)=2x^2-5x+5$.
We can plug that back in to verify our solution.
