SOLUTION: 1. A polynomial function is shown below. f(x) = x^6 + 12x^5 + 43x^4 + 22x^3 - 138x^2 - 280x -200 a. Find the real zeros of f (x). b. Express f (x) in terms of linear fa

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 1. A polynomial function is shown below. f(x) = x^6 + 12x^5 + 43x^4 + 22x^3 - 138x^2 - 280x -200 a. Find the real zeros of f (x). b. Express f (x) in terms of linear fa      Log On


   



Question 1034287: 1. A polynomial function is shown below.
f(x) = x^6 + 12x^5 + 43x^4 + 22x^3 - 138x^2 - 280x -200
a. Find the real zeros of f (x).
b. Express f (x) in terms of linear factors.
c. Find all of the zeros of f (x).
(3 points)
''.

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!

Rational Roots Theorem with synthetic division indicate possible zeros to check are the positive and negative
of 1,2,4,5,8,10,25,50,100. A graphing tool might help to decide which of the possibilities to try first.
Here are some of the productive synthetic-division root checkings. The first dividend of coefficients to start is
 1  12  43  22  -138  -280  -200
; the roots being checked are shown to the left, in the "divisor" location.

 -2 |    1    12    43    22    -138    -280   -200
    |
    |          -2   -20   -46     48     180    200
    |____________________________________________________
         1    10    23    -24     -90    -100    0

 -5 |    1    10    23    -24    -90    -100
    |
    |         -5    -25    10    70      100
    |____________________________________________________
        1     5    -2     -14    -20      0

 -5 |     1     5    -2     -14    -20   
    |
    |          -5    0       10     20
    |____________________________________
          1    0      -2     -4     0

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

This last quotient, having remainder 0, indicates the polynomial factor x%5E2%2B2x%2B2
and the zeros are found using general solution formula of a quadratic equation.

Zeros are %28-2%2B-+sqrt%282%5E2-4%2A2%29%29%2F2

%28-2%2B-+sqrt%284-8%29%29%2F2

%28-2%2B-+sqrt%28-4%29%29%2F2

%28-2%2B-+2%2Asqrt%28-1%29%29%2F2

highlight%28-1%2B-+i%29

---------------------------------------------------------
To summarize all of these zeros-finding results,
REAL RATIONAL ZEROS:
-5 of multiplicity two;
-2
+2
COMPLEX ZEROS:
-1-i and -1+i
----------------------------------------------------------

To begin in forming the factorized form of your function, start with this, and then simplify the complex part if you want.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
  The roots are 
        -5  of multiplicity 2
        -2  
         2
         -1 + i
         -1 - i

  f(x) = (x+5)*(x+5)*(x+2)*(x-2)*(x-(-1+i))*(x-(-1-i)) = 

        %28x%2B5%29%5E2%2A%28x%2B2%29%2A%28x-2%29%2A%28x%5E2+%2B+2x+%2B2%29