|
Question 183030This question is from textbook College Algebra
: Help with Find all rational zeros of the polynomial (using synthetic division)
22. p(x)= 
What I did: multiples of 1= +-1
multiples of 4= +-1 +-4 +-2
possible zeros: +-1 +-4 +-2 (after dividing constant over leading coeficient)
I used synthetic division to test which are zeros. +1 worked, +4 did not, +2 using the quotient of +1 (1 -1 -4 4) did not work gave me remainder of 8, but when I used the original coeficients (1 -2 -3 8 -4)it gave me a zero. I was told by my professor that either way it should work.
Any information is much appreciated, have at test coming up. Thank you
This question is from textbook College Algebra
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find all rational zeros of the polynomial (using synthetic division)
22. p(x)=
What I did: multiples of 1= +-1
multiples of 4= +-1 +-4 +-2
possible zeros: +-1 +-4 +-2 (after dividing constant over leading coeficient)
I used synthetic division to test which are zeros. +1 worked, +4 did not, +2 using the quotient of +1 (1 -1 -4 4) did not work gave me remainder of 8, but when I used the original coeficients (1 -2 -3 8 -4)it gave me a zero. I was told by my professor that either way it should work.
---------------------------------------------------
Since the coefficients add up to zero, x = 1 is a zero.
1)....1....-2....-3....8....-4
.......1....-1....-4...4..|..0
Since the quotient coefficients add up to zero. x = 1 is AGAIN a zero.
1)....1....-1....-4....4
.......1.....0....-4...|..0
Now you have a quadratic which says x^2-4 = 0
Factor to get (x-2)(x+2)=0
x = 2 or x = -2
-------------------------
The 4 zeros are 1,1,2,-2
============================
Cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! Any rational zero can be found through this equation
 where p and q are the factors of the last and first coefficients
So let's list the factors of -4 (the last coefficient):
Now let's list the factors of 1 (the first coefficient):
Now let's divide each factor of the last coefficient by each factor of the first coefficient
Now simplify
These are all the distinct rational zeros of the function that could occur (ie some of these values are NOT zeros, but could be)
===========================================================================
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| 1 | | | 1 | -2 | -3 | 8 | -4 | | | | | 1 | -1 | -4 | 4 | | | 1 | -1 | -4 | 4 | 0 |
Since the remainder  (the right most entry in the last row) is equal to zero, this means that is a zero of
------------------------------------------------------
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of
------------------------------------------------------
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| 4 | | | 1 | -2 | -3 | 8 | -4 | | | | | 4 | 8 | 20 | 112 | | | 1 | 2 | 5 | 28 | 108 |
Since the remainder  (the right most entry in the last row) is not equal to zero, this means that is not a zero of
------------------------------------------------------
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| -1 | | | 1 | -2 | -3 | 8 | -4 | | | | | -1 | 3 | 0 | -8 | | | 1 | -3 | 0 | 8 | -12 |
Since the remainder  (the right most entry in the last row) is not equal to zero, this means that is not a zero of
------------------------------------------------------
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| -2 | | | 1 | -2 | -3 | 8 | -4 | | | | | -2 | 8 | -10 | 4 | | | 1 | -4 | 5 | -2 | 0 |
Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of
------------------------------------------------------
Let's see if the possible zero  is really a root for the function
So let's make the synthetic division table for the function given the possible zero :
| -4 | | | 1 | -2 | -3 | 8 | -4 | | | | | -4 | 24 | -84 | 304 | | | 1 | -6 | 21 | -76 | 300 |
Since the remainder  (the right most entry in the last row) is not equal to zero, this means that is not a zero of
======================================================================
Answer:
So the rational zeros of  are: 1,2,-2
In other words, if we plug in  ,  or  into , we'll get 0 as a result (try it out if you aren't sure)
Note: the zero 1 has a multiplicity of 2 (ie it is counted twice)
| |
|
| |
