SOLUTION: Multiply each of the polynomial by x+1 with solution: a. x-1 b. x^2-x+1 c. x^3-x^2+x-1 d. x^4-x^3+x^2-x+1 2. Look for the pattern in Question 1 and use it to multiply: (x+1)

Algebra ->  Exponents -> SOLUTION: Multiply each of the polynomial by x+1 with solution: a. x-1 b. x^2-x+1 c. x^3-x^2+x-1 d. x^4-x^3+x^2-x+1 2. Look for the pattern in Question 1 and use it to multiply: (x+1)       Log On


   

Question 885807: Multiply each of the polynomial by x+1 with solution:
a. x-1
b. x^2-x+1
c. x^3-x^2+x-1
d. x^4-x^3+x^2-x+1
2. Look for the pattern in Question 1 and use it to multiply:
(x+1) (x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1)
3. Predict what you think will be the product of (x+1) and (x^100-x^98+x^97...x^2-x+1) when simplified, can you explain why you're must be correct?
All should have a solution. Please help. :)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1)
%28x%2B1%29%2A%28x-1%29=x%2A%28x-1%29%2B1%2A%28x-1%29=x%5E2-x%2Bx-1=x%5E2-1


=?
Those two products would be too long to write in one line, so I will add them up vertically:

So, %28x%2B1%29%2A%28x%5E4-x%5E3%2Bx%5E2-x%2B1%29=X%5E6%2B1

2) In all the products above %28x%2B1%29 is one factor,
and the other factor is a polynomial with sign that alternate,
so all the middle products cancel out.
You are left with just the product of the first terms plus the product of the last terms.
So, %28x%2B1%29%2A%28x%5E8-x%5E7%2Bx%5E6-x%5E5%2Bx%5E4-x%5E3%2Bx%5E2-x%2B1%29=x%5E9%2B1
All the other products cancel out and you are left with the product of the x in %28x%2B1%29 times the x%5E8 ,
plus the product of the %22%2B1%22 and %22%2B1%22 at the end of both factors.

3) I believe what you meant to write was
%28x%2B1%29%2A%28x%5E100-x%5E99%2Bx%5E98-x%5E97%2B+%22...%22+%2Bx%5E2-x%2B1%29 , because if the signs alternate,
all the terms with even exponents will follow a %22%2B%22 sign.
%28x%2B1%29%2A%28x%5E100-x%5E99%2Bx%5E98-x%5E97%2B+%22...%22+%2Bx%5E2-x%2B1%29=x%5E101%2B1