The function that we are to start with,

The "beginning" finction: 

y=(x-4)(x+1)(x-8)

has zeros 

4, -1, 8

We put them in order smallest to largest

-1, 4, 8

The function that we are to transform it into,

the "final" function, 

y=2(x-1)(x+4)(x-5),  

has zeros 1, -4, and 5

We put them in order:

-4, 1, 5

We compare them:

beginning function's zeros:  -1, 4, 8
final function's zeros:      -4, 1, 5

We observe that 

moving 3 units left from -1 gives -4
moving 3 units left from  4 gives  1
moving 3 units left from  8  gives 5

So shifting the beginning graph left by 3 units
takes care of moving the zeros of the beginning function
to the zeros of the final function.

Also we notice that the final function has a 2 factor, 
which involves a stretch by a factor of 2.

So yes we can create the graph of

y=2(x-1)(x+4)(x-5) 

by shifting the graph of

y=(x-4)(x+1)(x-8)

3 units left and stretching it by a
factor of 2.

To see it done, we start with the graph of the beginning
function 

y=(x-4)(x+1)(x-8), which is the red graph below:



We shift it left 3 units by replacing each x by x+3

y=(x-4)(x+1)(x-8)
y=(x+3-4)(x+3+1)(x+3-8)

Simplified,
the intermediate function's equation is: 

y=(x-1)(x+4)(x-5)

This intermediate function is graphed below (in green). It is the
beginning graph (in red) shifted 3 units left.  



Now we only need to stretch the intermediate green graph vertically 
by a factor of 2 to have the graph of the final function:

To stretch the intermediate green graph vertically by a factor of 2,
we multiply the right side by 2, and get this equation, which is
equivalent to the final function:

y = 2(x-1)(x+4)(x-5) 

Think of it as if the graph were drawn on a rubber sheet and we 
took hold of the top and bottom and stretched it double. The green
graph would become the blue graph below:

.

Edwin