SOLUTION: A partial solution set is given. Find the complete solution set. x^3+3x^2-13x-15=0;{-1}

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Question 167735: A partial solution set is given. Find the complete solution set.
x^3+3x^2-13x-15=0;{-1}
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Note: since -1 is a given solution, this means that is a given factor

Since -1 is a given zero, we can use with synthetic division where -1 is the test zero.




Set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the polynomial to the right of the test zero.
-1|13-13-15
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1|13-13-15
|
1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 3)
-1|13-13-15
|-1
1

Add -1 and 3 to get 2. Place the sum right underneath -1.
-1|13-13-15
|-1
12

Multiply -1 by 2 and place the product (which is -2) right underneath the third coefficient (which is -13)
-1|13-13-15
|-1-2
12

Add -2 and -13 to get -15. Place the sum right underneath -2.
-1|13-13-15
|-1-2
12-15

Multiply -1 by -15 and place the product (which is 15) right underneath the fourth coefficient (which is -15)
-1|13-13-15
|-1-215
12-15

Add 15 and -15 to get 0. Place the sum right underneath 15.
-1|13-13-15
|-1-215
12-150

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,-15) form the quotient




So


Basically factors to


Now simply solve to find the remaining two solutions (I'll let you do that)
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