SOLUTION: Find the number of times r is a root of P(x)=0. P(x)= x^4+4x^3-16x-16; r=-2 i found the depressed eruation twice but the numbers were all messed up. please help and explain.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Find the number of times r is a root of P(x)=0. P(x)= x^4+4x^3-16x-16; r=-2 i found the depressed eruation twice but the numbers were all messed up. please help and explain.      Log On


   



Question 191251This question is from textbook algebra and trigonometry structure and method book 2
: Find the number of times r is a root of P(x)=0.
P(x)= x^4+4x^3-16x-16; r=-2
i found the depressed eruation twice but the numbers were all messed up. please help and explain.
This question is from textbook algebra and trigonometry structure and method book 2

Found 3 solutions by Edwin McCravy, jim_thompson5910, solver91311:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find the number of times r is a root of P(x)=0.
P(x)= x^4+4x^3-16x-16; r=-2
i found the depressed eruation twice but the
numbers were all messed up. please help and
explain.

You have to insert a place holder %22+%22%2B0x%5E2
and start with

P%28x%29=+x%5E4%2B4x%5E3%2B0x%5E2-16x-16

-2 | 1  4  0 -16 -16
   |   -2 -4   8  16
    ----------------
     1  2 -4  -8   0

-2 | 1  2 -4  -8
   |   -2  0   8 
    ------------
     1  0 -4   0


-2 | 1  0 -4
   |   -2  4 
    --------
     1 -2  0


-2 | 1 -2
   |   -2
    -----
     1 -4

We get a 0 remainder 3 times, but
not the 4th time, so -2 is a root
3 times.

Edwin

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, perform synthetic division where -2 is the test zero (let me know if you need help with synthetic division)

-2 | 1   4   0   -16   -16
   |    -2  -4     8    16
   ------------------------
     1   2  -4    -8     0



Since the last number in the bottom row is zero, this means that the remainder is 0. So -2 is a root of P%28x%29=+x%5E4%2B4x%5E3-16x-16

The first 4 numbers form the depressed polynomial x%5E3%2B2x%5E2-4x-8. This means that x%5E4%2B4x%5E3-16x-16=%28x%2B2%29%28x%5E3%2B2x%5E2-4x-8%29
=================================


Now perform synthetic division on x%5E3%2B2x%5E2-4x-8 using the same test zero:

-2 | 1   2  -4    -8   
   |    -2   0     8   
   ------------------
     1   0  -4     0   


Notice how the last number in the bottom row is 0. So -2 is a root of x%5E3%2B2x%5E2-4x-8. So far, r=-2 is a root of multiplicity 2 (ie -2 is a root twice).


The first 3 numbers in the bottom row form the new polynomial x%5E2-4. This tells us that x%5E3%2B2x%5E2-4x-8=%28x%2B2%29%28x%5E2-4%29


So

=================================


Now perform synthetic division on the polynomial x%5E2-4
-2 | 1   0  -4     
   |    -2   4    
   ------------
     1  -2   0




So -2 is a root of x%5E2-4. So this means that r=-2 is a root of multiplicity 3 (ie -2 is a root three times).


The first two numbers in the bottom row form the new polynomial: x-2


Now because -2 is NOT a root of x-2, this means that we can stop looking for more roots of -2.


So

Or in other words, x%5E4%2B4x%5E3-16x-16=%28x%2B2%29%28x%2B2%29%28x%2B2%29%28x-2%29


Notice how the factor x%2B2 is repeated 3 times, this supports our conclusion that r=-2 is a root of multiplicity 3

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If you found the depressed equation twice, you should be at an easy to solve quadratic.

Check your work based on the following results when I performed the polynomial long division:





And

, hence the roots are

-2 (with a multiplicity of 3) and 2.

I used polynomial long division, but synthetic division would work just as well. A common error when performing these division operations is failing to explictly specify all orders of the variable in the dividend. In other words, you have to divide into:



or if you are using synthetic division, your first line of coefficients must look like:



Another possible source of error would be if you were dividing by x - 2. Remember, if the root is -2, then the factor is x + 2, and that should have been your divisor.

John