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
|
1
3
-13
-15
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1
|
1
3
-13
-15
|
1
Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 3)
-1
|
1
3
-13
-15
|
-1
1
Add -1 and 3 to get 2. Place the sum right underneath -1.
-1
|
1
3
-13
-15
|
-1
1
2
Multiply -1 by 2 and place the product (which is -2) right underneath the third coefficient (which is -13)
-1
|
1
3
-13
-15
|
-1
-2
1
2
Add -2 and -13 to get -15. Place the sum right underneath -2.
-1
|
1
3
-13
-15
|
-1
-2
1
2
-15
Multiply -1 by -15 and place the product (which is 15) right underneath the fourth coefficient (which is -15)
-1
|
1
3
-13
-15
|
-1
-2
15
1
2
-15
Add 15 and -15 to get 0. Place the sum right underneath 15.
-1
|
1
3
-13
-15
|
-1
-2
15
1
2
-15
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,-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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