SOLUTION: can you please help me with: (((x(3)+2x(2)+3x-6 over x-1)))

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Question 152642This question is from textbook Algebra structure and method book 1
: can you please help me with: (((x(3)+2x(2)+3x-6 over x-1))) This question is from textbook Algebra structure and method book 1

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

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|123-6
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
1|123-6
|
1

Multiply 1 by 1 and place the product (which is 1) right underneath the second coefficient (which is 2)
1|123-6
|1
1

Add 1 and 2 to get 3. Place the sum right underneath 1.
1|123-6
|1
13

Multiply 1 by 3 and place the product (which is 3) right underneath the third coefficient (which is 3)
1|123-6
|13
13

Add 3 and 3 to get 6. Place the sum right underneath 3.
1|123-6
|13
136

Multiply 1 by 6 and place the product (which is 6) right underneath the fourth coefficient (which is -6)
1|123-6
|136
136

Add 6 and -6 to get 0. Place the sum right underneath 6.
1|123-6
|136
1360

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (1,3,6) form the quotient




So where

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