Remainder of $\frac{3x^{2019}+5x^{1019}-7x+4}{x^2-1}$ I don't understand how I should go about solving the following question:

Find the remainder when polynomial $f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.

I tried to use the factor theorem, but I never encountered a problem with a divisor which, in this case, is $(x+1)(x-1)$, so I simply found $f$ of both roots, so $f(1)$ and $f(-1)$.
Allegedly the remainder is a linear polynomial in the form $ax+b$ but I fail to see how they derived that fact. Apparently they made use of simultaneous equations, but I'm not sure how or why.
Any help would be appreciated!

Degree of the remainder less then degree of quotient then remainder is linear , $R(x)=ax+b$
$f(x)=3x^{2019}+5x^{1019}-7x+4=Q(x)(x-1)(x+1)+ax+b$
$f(1)=a+b=5$
$f(-1)=-a+b=3$
then $a=1,b=4$
$R(x)=x+4$
OR
To find remainder when
$f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.
By using remainder theorem
$x^2-1=0$
then
$x^2=1$
$f(x)=3x^{2018+1}+5x^{1018+1}-7x+4=3((x^2))^{1009}\times x +5((x^2))^{509}\times x -7x+4=3\times 1 \times x +5\times x\times 1\times x-7x+4=x+4$
Hence
$R(x)=x+4$
