SOLUTION: Can you please explain to me how to go about dividing the expression below? {{{(x^3+x^2+x-3)/ (x-1)}}} Thank you.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Can you please explain to me how to go about dividing the expression below? {{{(x^3+x^2+x-3)/ (x-1)}}} Thank you.      Log On


   

Question 201752: Can you please explain to me how to go about dividing the expression below?

%28x%5E3%2Bx%5E2%2Bx-3%29%2F+%28x-1%29

Thank you.
Found 2 solutions by jim_thompson5910, RAY100:
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%5E3+%2B+x%5E2+%2B+x+-+3%29%2F%28x-1%29

First lets find our test zero:

x-1=0 Set the denominator x-1 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|111-3
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
1|111-3
|
1

Multiply 1 by 1 and place the product (which is 1) right underneath the second coefficient (which is 1)
1|111-3
|1
1

Add 1 and 1 to get 2. Place the sum right underneath 1.
1|111-3
|1
12

Multiply 1 by 2 and place the product (which is 2) right underneath the third coefficient (which is 1)
1|111-3
|12
12

Add 2 and 1 to get 3. Place the sum right underneath 2.
1|111-3
|12
123

Multiply 1 by 3 and place the product (which is 3) right underneath the fourth coefficient (which is -3)
1|111-3
|123
123

Add 3 and -3 to get 0. Place the sum right underneath 3.
1|111-3
|123
1230

Since the last column adds to zero, we have a remainder of zero. This means x-1 is a factor of x%5E3+%2B+x%5E2+%2B+x+-+3

Now lets look at the bottom row of coefficients:

The first 3 coefficients (1,2,3) form the quotient

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


So %28x%5E3+%2B+x%5E2+%2B+x+-+3%29%2F%28x-1%29=x%5E2+%2B+2x+%2B+3

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
Before we divide polynomials, let's review division in general.
.
The algorithm is,,,divide,,multiple,,substract,,and bring down,,,then repeat
.
,,,,,,,,,,,,,,x^2 +2x +3
,,,,,,,____________________
,,x-1,,[ x^3 +x^2 +x -3
,,,,,,,,,x^3 -x^2
,,,,,,,,,________
,,,,,,,,,,,,,2x^2 +x
,,,,,,,,,,,,,2x^2 -2x
,,,,,,,,,,,,,________
,,,,,,,,,,,,,,,,,,3x -3
,,,,,,,,,,,,,,,,,,3x -3
,,,,,,,,,,,,,,,,,,______
,,,,,,,,,,,,,,,,,,,,,,0
.
to check, multiply (x^2 +2x +3) * (x-1) = x^3 +x^2 +x -3,,,,,ok