There are many ways to attack this problem, but here is one way to
look at it, showing the solution for a similar problem. Although they
explanation appears long, it really isn?t; the explanation is just
spread out, with all the steps shown.
(Note that exponents, such as x to the power 2 or ?x-squared? is
written as ?x^2? for this response).
Remember that you can move terms in and out of parenthesis, as long as
you do the same thing to each term within the parentheses.
So, you can take:
x^2 + x
and rewrite it as
x^2 + x = x(x+1).
So, what happens when we have the slightly more complex
(x+1)(x+1)
We can rewrite it as the first term multiplied by the whole second
expression, and then the second term multiplied by the whole second
expression, or
(x+1)(x+1) = (x)(x+1) + (1)(x+1)
or
(x+1)(x+1) = x^2 + x + x + 1
which, by combining terms, is the same as
(x+1)(x+1) = x^2 + 2x + 1.
Your question is slightly more complex. So, what if the expression were
(x+1)(x+1)(x+1)
The approach would be to split the problem up and tackle two of them
first, then multiply by the third-- it doesn?t matter which two you
choose to do first, since the order of multiplication doesn?t change
the result.
So, we can think of the expression as
(x+1)(x+1)(x+1) = first multiply and expand (x+1)(x+1), then
multiply the whole thing by (x+1)
We already know that (x+1)(x+1) = x^2 + 2x +1, so we can rewrite this as
(x^2 + 2x +1) (x+1)
We can then take each term within one of the parenthesis, and multiply
it by each term within the other parenthesis, or:
(x^2 + 2x +1) (x) + (x^2 + 2x +1)(1)
multiplying out?
= x^3 + 2x^2 + x + x^2 + 2x + 1
Combining terms gives
x^3 + 3x^2 + 3x + 1
Your problem is slightly nastier, since I used the same expression
within each parenthesis, and ?1? is the easiest non-zero number to do
these with. But, the approach is the same whether it is
(x+1)(x+1)(x+1) or (x+2)(x+3)(x+4)
Good luck. |
