SOLUTION: Preform the indicated divison x4+4x3+6x2+4x+1 ----------- ---- x+1 the numbers after the x such as 4x3 is 4x to the third power

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Question 149911: Preform the indicated divison

x4+4x3+6x2+4x+1
----------------
x+1

the numbers after the x such as 4x3 is 4x to the third power

Found 2 solutions by jim_thompson5910, vleith:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Let's simplify this expression using synthetic division


Start with the given expression %28x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1%29%2F%28x%2B1%29

First lets find our test zero:

x%2B1=0 Set the denominator x%2B1 equal to zero

x=-1 Solve for x.

so our test zero is -1


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
-1|14641
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1|14641
|
1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 4)
-1|14641
|-1
1

Add -1 and 4 to get 3. Place the sum right underneath -1.
-1|14641
|-1
13

Multiply -1 by 3 and place the product (which is -3) right underneath the third coefficient (which is 6)
-1|14641
|-1-3
13

Add -3 and 6 to get 3. Place the sum right underneath -3.
-1|14641
|-1-3
133

Multiply -1 by 3 and place the product (which is -3) right underneath the fourth coefficient (which is 4)
-1|14641
|-1-3-3
133

Add -3 and 4 to get 1. Place the sum right underneath -3.
-1|14641
|-1-3-3
1331

Multiply -1 by 1 and place the product (which is -1) right underneath the fifth coefficient (which is 1)
-1|14641
|-1-3-3-1
1331

Add -1 and 1 to get 0. Place the sum right underneath -1.
-1|14641
|-1-3-3-1
13310


Since the last column adds to zero, we have a remainder of zero. This means x%2B1 is a factor of x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1

Now lets look at the bottom row of coefficients:

The first 4 coefficients (1,3,3,1) form the quotient

x%5E3+%2B+3x%5E2+%2B+3x+%2B+1


So %28x%5E4+%2B+4x%5E3+%2B+6x%5E2+%2B+4x+%2B+1%29%2F%28x%2B1%29=x%5E3+%2B+3x%5E2+%2B+3x+%2B+1

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Use this --> http://www.calc101.com/webMathematica/long-divide.jsp#topdoit