3(x–2)2 – 3(x–2) – 6
To see this better, I will color the parentheses red:
3(x-2)2 – 3(x-2) – 6
Look carefully at each of these and you will notice
just how similar they are. Especially observe the
red parts:
3(x-2)2 – 3(x-2) – 6 and 3y2 - 3y - 6
Do you see that the only difference between them is
their red parts? The expression on the right is
an ordinary factoring job, and we do it this way:
3y2 - 3y - 6
3(y2 - y - 2)
3(y - 2)(y + 1)
So we do the exact same thing with your problem:
3(x-2)2 – 3(x-2) – 6 ---> 3y2 - 3y - 6
3[(x-2)2 - (x-2) - 2] ---> 3(y2 - y - 2)
3[(x-2) - 2][(x-2) + 1] ---> 3(y - 2)(y + 1)
Do you see that what's done on the left is just like
what's done on the right except that on the right we
are doing it with just a single letter y, whereas on
the left we are doing those same steps only with
something a bit more complicated-looking, namely (x-2).
We're also using brackets [ ] on your problem because
the red expression already contains parentheses ( ).
This is a skill of algebra that you must develop --
Viewing a big chunk of algebra like (x-2) just as you
view a small chunk of algebra, such as y.
There is only a couple things more that we need to do with
your problem than we had to do with the one on the right.
They are to remove the parentheses inside the brackets,
collect terms, and change the brackets to parentheses.
3[(x-2) - 2][(x-2) + 1]
3[ x-2 - 2][x-2 + 1]
3[x-4][x-1]
3(x-4)(x-1)
Edwin
