How to factorise $(x-1)^2 - (x-5)^2$? My attempt: 
$a = (x-1)$ 
$c = (x-5)$
$a^2 - c^2$
which is equal to: 
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is : 
$8(x-3)$
Can you explain, please?

$$\begin{array}{rl} & \begin{bmatrix} -1\\ 1\end{bmatrix} \begin{bmatrix} -1\\ 1\end{bmatrix}^T - \begin{bmatrix} -5\\ 1\end{bmatrix} \begin{bmatrix} -5\\ 1\end{bmatrix}^T = \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix} - \begin{bmatrix} 25 & -5\\ -5 & 1\end{bmatrix}\\\\ &= \begin{bmatrix} -24 & 4\\ 4 & 0\end{bmatrix} = 4 \begin{bmatrix} -6 & 1\\ 1 & 0\end{bmatrix} \equiv 4 \begin{bmatrix} -6 & 2\\ 0 & 0\end{bmatrix} = 8 \begin{bmatrix} -3 & 1\\ 0 & 0\end{bmatrix} = \begin{bmatrix} 8\\ 0\end{bmatrix} \begin{bmatrix} -3\\ 1\end{bmatrix}^T\end{array}$$
Hence,
$$(x-1)^2 - (x-5)^2 = 8 (x-3)$$
