|
Question 138661: Find the zeros of 
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! First graph the function
 Graph of 
From the graph we can see that the function has a zero at  . So our test zero is -6. So our first factor is 
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.
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
Multiply -6 by 1 and place the product (which is -6) right underneath the second coefficient (which is -2)
Add -6 and -2 to get -8. Place the sum right underneath -6.
Multiply -6 by -8 and place the product (which is 48) right underneath the third coefficient (which is -19)
Add 48 and -19 to get 29. Place the sum right underneath 48.
| -6 | | | 1 | -2 | -19 | 122 | -312 | | | | | -6 | 48 | | | | | | 1 | -8 | 29 | | |
Multiply -6 by 29 and place the product (which is -174) right underneath the fourth coefficient (which is 122)
| -6 | | | 1 | -2 | -19 | 122 | -312 | | | | | -6 | 48 | -174 | | | | | 1 | -8 | 29 | | |
Add -174 and 122 to get -52. Place the sum right underneath -174.
| -6 | | | 1 | -2 | -19 | 122 | -312 | | | | | -6 | 48 | -174 | | | | | 1 | -8 | 29 | -52 | |
Multiply -6 by -52 and place the product (which is 312) right underneath the fifth coefficient (which is -312)
| -6 | | | 1 | -2 | -19 | 122 | -312 | | | | | -6 | 48 | -174 | 312 | | | | 1 | -8 | 29 | -52 | |
Add 312 and -312 to get 0. Place the sum right underneath 312.
| -6 | | | 1 | -2 | -19 | 122 | -312 | | | | | -6 | 48 | -174 | 312 | | | | 1 | -8 | 29 | -52 | 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 4 coefficients (1,-8,29,-52) form the quotient
So 
You can use this online polynomial division calculator to check your work
Basically factors to 
Now lets break down further
Also from the graph, we can see that the function has another zero at  . So our test zero is 4. So our second factor is 
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.
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
Multiply 4 by 1 and place the product (which is 4) right underneath the second coefficient (which is -8)
Add 4 and -8 to get -4. Place the sum right underneath 4.
Multiply 4 by -4 and place the product (which is -16) right underneath the third coefficient (which is 29)
Add -16 and 29 to get 13. Place the sum right underneath -16.
Multiply 4 by 13 and place the product (which is 52) right underneath the fourth coefficient (which is -52)
Add 52 and -52 to get 0. Place the sum right underneath 52.
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,-4,13) form the quotient
So 
You can use this online polynomial division calculator to check your work
Basically factors to 
Now lets break down further
Let's use the quadratic formula to solve for x:
Starting with the general quadratic
the general solution using the quadratic equation is:
So lets solve  ( notice  ,  , and  )
 Plug in a=1, b=-4, and c=13
 Negate -4 to get 4
 Square -4 to get 16 (note: remember when you square -4, you must square the negative as well. This is because  .)
 Multiply  to get 
 Combine like terms in the radicand (everything under the square root)
 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)
 Multiply 2 and 1 to get 2
After simplifying, the quadratic has roots of
 or 
-------------------------------------------
Answer:
So the zeros of the original polynomial are:
, , and
|
|
|
| |
