Answer 1

We have: 
\[40 + 3x - x^2 \]
\[ \Rightarrow - ( x^2 - 3x - 40) \]
\[\text{ To factorise }( x^2 - 3x - 40),\text{ we will find two numbers p and q such that }p + q = - 3\text{ and pq }= - 40 . \]
Now,
\[5 + ( - 8) = - 3 \]
and 
\[5 \times ( - 8) = - 40\]
\[\text{ Splitting the middle term }- 3x \text{ in the given quadratic as }5x - 8x,\text{ we get: }\]
\[40 + 3x - x^2 = - ( x^2 - 3x - 40)\]
\[ = - ( x^2 + 5x - 8x - 40)\]
\[ = - [( x^2 + 5x) - (8x + 40)]\]
\[ = - [x(x + 5) - 8(x + 5)]\]
\[ = - (x - 8)(x + 5)\]
\[ = (x + 5)( - x + 8)\]