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?
 A: Your doing is correct, you just need to expand the expressions inside the parenthesis:
\begin{align}((x-1) - (x-5))((x-1)+(x-5))&=(x-1-x+5)(x-1+x-5)\\ & =4(2x-6)\\ & =8(x-3)\end{align}
A: 
$${ \left( x-1 \right)  }^{ 2 }-{ \left( x-5 \right)  }^{ 2 }=\left( x-1-x+5 \right) \left( x-1+x-5 \right) =4\left( 2x-6 \right) =8\left( x-3 \right)  $$

A: using the formula $$a^2-b^2=(a-b)(a+b)$$ we obtain
$$(x-1)^2-(x-5)^2=(x-1-x+5)(x-1+x-5)=4(2x-6)=8(x-3)$$
A: $$\begin{array}{rl} & \begin{bmatrix} -1\\ 1\end{bmatrix} \begin{bmatrix} -1\\ 1\end{bmatrix}^T - \begin{bmatrix} -5\\ 1\end{bmatrix} \begin{bmatrix} -5\\ 1\end{bmatrix}^T = \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix} - \begin{bmatrix} 25 & -5\\ -5 & 1\end{bmatrix}\\\\ &= \begin{bmatrix} -24 & 4\\ 4 & 0\end{bmatrix} = 4 \begin{bmatrix} -6 & 1\\ 1 & 0\end{bmatrix} \equiv 4 \begin{bmatrix} -6 & 2\\ 0 & 0\end{bmatrix} = 8 \begin{bmatrix} -3 & 1\\ 0 & 0\end{bmatrix} = \begin{bmatrix} 8\\ 0\end{bmatrix} \begin{bmatrix} -3\\ 1\end{bmatrix}^T\end{array}$$
Hence,
$$(x-1)^2 - (x-5)^2 = 8 (x-3)$$
