Until you get used to them it seems quite tricky to multiply out brackets.

Using color to show what is happening.

Given: #y=color(blue)((x-3))color(brown)((x-4))#

You can split the multiplication up into parts like this:

#y=color(blue)(xcolor(brown)((x-4))-4color(brown)((x-4)) ..........(1)#

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#color(blue)("Using numbers to explain what is happening")#

This is a bit like: #3xx4 =12#

Now mater how we split the 3 we will always end up with 12

#color(brown)((1+2)color(blue)(xx4)#

=#color(brown)((1color(blue)(xx4))+(2color(blue)(xx4))#

#color(blue)(4+4+4)color(white)(.)= 12#

It will even work if you write 3 as 4-1

#color(brown)((4-1))color(blue)(xx4)#

#color(brown)((4color(blue)(xx4))-(1color(blue)(xx4)) # the last bracket gives you #color(green)(-4)#

#color(blue)(4+4+4+4color(green)( -4)) =12#

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#color(red)("Back to your question")#

So equation (1) becomes:

#y=(x^2-4x)color(red)(-)(color(green)(+4x-16)).........(2)#

As there is no multiplication left we can removing the brackets

#y=x^2-4xcolor(green)(color(blue)(bb-)4xcolor(blue)(+)16)......................(3)#

Notice the way the #color(red)("minus sign")color(white)(.) underline("outside")# the #underline("last")# bracket in equation (2) changes all the #color(blue)("signs inside")# it when the bracket is removed. Giving equation (3)

#y=x^2-8x+16#