SOLUTION: How would I solve this? (x-2)^3 Thanks :)

Algebra ->  Exponents -> SOLUTION: How would I solve this? (x-2)^3 Thanks :)      Log On


   



Question 243608: How would I solve this? (x-2)^3 Thanks :)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!


%28x-2%29%5E3 Start with the given expression

To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle:
1   

1   1   

1   2   1   

1   3   3   1   




Looking at the row that starts with 1,3, etc, we can see that this row has the numbers:

1, 3, 3, and 1

These numbers will be the coefficients of our expansion. So to expand %28x-2%29%5E3, simply follow this procedure:
Write the first coefficient. Multiply that coefficient with the first binomial term x and then the second binomial term -2. Repeat this until all of the coefficients have been written.

Once that has been done, add up the terms like this:


Notice how the coefficients are in front of each term.



However, we're not done yet.


Looking at the first term 1%28x%29%28-2%29, raise x to the 3rd power and raise -2 to the 0th power.

Looking at the second term 3%28x%29%28-2%29 raise x to the 2nd power and raise -2 to the 1st power.

Continue this until you reach the final term.


Notice how the exponents of x are stepping down and the exponents of -2 are stepping up.


So the fully expanded expression should now look like this:





1%28x%5E3%29%281%29%2B3%28x%5E2%29%28-2%29%2B3%28x%5E1%29%284%29%2B1%28x%5E0%29%28-8%29 Distribute the exponents


1%28x%5E3%29%2B3%28-2x%5E2%29%2B3%284x%29%2B1%28-8%29 Multiply


x%5E3-6x%5E2%2B12x-8 Multiply the terms with their coefficients


So %28x-2%29%5E3 expands and simplifies to x%5E3-6x%5E2%2B12x-8.


In other words, %28x-2%29%5E3=x%5E3-6x%5E2%2B12x-8