SOLUTION: solve the following in synthetic division 2x^2+10x+12/x+3 2x^4+x^3-5x^2-9/x+2 x^3-3x^2+12/x+2 3x^4-4x^3+5x^2-7x-5/x-2

Algebra ->  Equations -> SOLUTION: solve the following in synthetic division 2x^2+10x+12/x+3 2x^4+x^3-5x^2-9/x+2 x^3-3x^2+12/x+2 3x^4-4x^3+5x^2-7x-5/x-2      Log On


   



Question 254793: solve the following in synthetic division
2x^2+10x+12/x+3
2x^4+x^3-5x^2-9/x+2
x^3-3x^2+12/x+2
3x^4-4x^3+5x^2-7x-5/x-2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you started.

# 1


Let's simplify this expression using synthetic division


Start with the given expression %282x%5E2+%2B+10x+%2B+12%29%2F%28x%2B3%29

First lets find our test zero:

x%2B3=0 Set the denominator x%2B3 equal to zero

x=-3 Solve for x.

so our test zero is -3


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.
-3|21012
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
-3|21012
|
2

Multiply -3 by 2 and place the product (which is -6) right underneath the second coefficient (which is 10)
-3|21012
|-6
2

Add -6 and 10 to get 4. Place the sum right underneath -6.
-3|21012
|-6
24

Multiply -3 by 4 and place the product (which is -12) right underneath the third coefficient (which is 12)
-3|21012
|-6-12
24

Add -12 and 12 to get 0. Place the sum right underneath -12.
-3|21012
|-6-12
240

Since the last column adds to zero, we have a remainder of zero. This means x%2B3 is a factor of 2x%5E2+%2B+10x+%2B+12

Now lets look at the bottom row of coefficients:

The first 2 coefficients (2,4) form the quotient

2x+%2B+4


So %282x%5E2+%2B+10x+%2B+12%29%2F%28x%2B3%29=2x+%2B+4

You can use this online polynomial division calculator to check your work.