If $a>0$ and $(x+1)(x+2)(x+3)(x+4)+1=(ax^2+bx+c)^2$, then find the value of $a+b+c$

Question

Q3/30 If $a>0$ and $(x+1)(x+2)(x+3)(x+4)+1=(ax^2+bx+c)^2$, then find the value of $a+b+c$.

Answer 1

If you tried setting $x=1,$ then you'd have all you want.

Answer 2

$\begin{split} (x+1)(x+2)(x+3)(x+4)+1 & =[(x+1)(x+4)][(x+2)(x+3)]+1\\ &=[(x^2+5x+5)-1][(x^2+5x+5)+1]+1\\ &=(x^2+5x+5)^2\end{split}$

$\therefore$The answer equals $1+5+5=11$