$(ax+b)(bx+a)=10x^2+cx+10$ where $a$ and $b$ are positive integers. Find the two possible values of $c$

Question

I know this seems simple but i am really struggling please help me thank you

Answer 1

$$(ax+b)(bx+a)=10^2+cx+10 \\or\\ (ax+b)(bx+a)=x^2+cx+10 \\or\\(ax+b)(bx+a)=10x^2+cx+10$$I think 3rd is right $$so\\abx^2+(a^2+b^2)x+ab=10x^2+cx+10\\ab=10\\ ab=10 \to \begin{cases}a=1 & b = 10 \to c=a^2+b^2=1+100=101\\a=10 & b = 1 \to c=a^2+b^2=1+100=101 \\ a=2 & b=5 \to c=a^2+b^2=4+25=29\\ a=5 & b=2 \to c=a^2+b^2=4+25=29\end{cases}$$