SOLUTION: please help me solve this division polynomial. I need to set up the equation and solve (2x^3-14x^2+19x+20)/(x-4)

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Question 88876: please help me solve this division polynomial. I need to set up the equation and solve
(2x^3-14x^2+19x+20)/(x-4)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Lets use synthetic division to divide

Start with the given polynomial %282x%5E3+-+14x%5E2+%2B+19x+%2B+20%29%2F%28x-4%29

First lets find our test zero:

x-4=0 Set the denominator x-4 equal to zero
x=4 Solve for x.

so our test zero is 4


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.
4|2-141920
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
4|2-141920
|
2

Multiply 4 by 2 and place the product (which is 8) right underneath the second coefficient (which is -14)
4|2-141920
|8
2

Add 8 and -14 to get -6. Place the sum right underneath 8.
4|2-141920
|8
2-6

Multiply 4 by -6 and place the product (which is -24) right underneath the third coefficient (which is 19)
4|2-141920
|8-24
2-6

Add -24 and 19 to get -5. Place the sum right underneath -24.
4|2-141920
|8-24
2-6-5

Multiply 4 by -5 and place the product (which is -20) right underneath the fourth coefficient (which is 20)
4|2-141920
|8-24-20
2-6-5

Add -20 and 20 to get 0. Place the sum right underneath -20.
4|2-141920
|8-24-20
2-6-50

Since the last column adds to zero, we have a remainder of zero. This means x-4 is a factor of 2x%5E3+-+14x%5E2+%2B+19x+%2B+20

Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,-6,-5) form the quotient

2x%5E2+-+6x+-+5


So %282x%5E3+-+14x%5E2+%2B+19x+%2B+20%29%2F%28x-4%29=2x%5E2+-+6x+-+5