# SOLUTION: Help how do I solve (14y+8y^2+y^3+12)/6+y

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Question 133085: Help how do I solve (14y+8y^2+y^3+12)/6+y
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Let's simplify this expression using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for y.

so our test zero is -6

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.
 -6 | 1 8 14 12 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
 -6 | 1 8 14 12 | 1

Multiply -6 by 1 and place the product (which is -6) right underneath the second coefficient (which is 8)
 -6 | 1 8 14 12 | -6 1

Add -6 and 8 to get 2. Place the sum right underneath -6.
 -6 | 1 8 14 12 | -6 1 2

Multiply -6 by 2 and place the product (which is -12) right underneath the third coefficient (which is 14)
 -6 | 1 8 14 12 | -6 -12 1 2

Add -12 and 14 to get 2. Place the sum right underneath -12.
 -6 | 1 8 14 12 | -6 -12 1 2 2

Multiply -6 by 2 and place the product (which is -12) right underneath the fourth coefficient (which is 12)
 -6 | 1 8 14 12 | -6 -12 -12 1 2 2

Add -12 and 12 to get 0. Place the sum right underneath -12.
 -6 | 1 8 14 12 | -6 -12 -12 1 2 2 0

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 (1,2,2) form the quotient

So

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