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The polynomial f(x) has degree 3. If f(0) = 4, f(1)= -2 , f(2) = 0, and f(3) = 31, then what are the x -intercepts of the graph of f(x)?

 Apr 23, 2022
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We have the form

ax^3 + bx^2 + cx + d

 

Because    f(0) = 4

  then d  =4

 

And because   f(2)  = 0.....then 2 is one root  =one x int

 

And we have these equations

 

a(1)^3  + b(1)^2  + c(1)  +  4  = -2

a(2)^3  + b(2)^2  + c(2)  +  4  =  0

a(3)^3  + b(3)^2  + c(3)  + 4  =  31

 

Simplifying  these we have

 

a + b + c  =   -6

8a + 4b + 2c  = -4

27a + 9b + 3c  =  27

 

Solving these   produces

 

a =7/2    b = -13/2   c  =-3    and d = 4

 

So  our polynomial     =  (7/2)x^3  - (13/2)x^2  - 3x  + 4

 

Since   2  is a root, we can use some synthetic division to  whittle  the cubic down to a  quadratic

 

 

2 [     7/2          -13/2             -3             4   ]

                           7                 1            -4

      _________________________________

         7/2            1/2            -2             0

 

So   the remaining quadratic    =   (7/2)x^2  + (1/2)x  -  2

 

Set this  to 0   and multiply through by 2 to get

 

7x^2  + x   -  4   =  0

 

Using the Q Formula  the other two roots are

 

-1 ±  sqrt [ 1 - 4 (7) (-4) ]                         -1  ± sqrt [ 113 ]

____________________  =                   _____________       x   ≈   .688   and     x  ≈   -.83  

          2 *  7                                                     14

 

These are the other two x ints

 

cool cool cool

 Apr 23, 2022
edited by CPhill  Apr 23, 2022
edited by CPhill  Apr 23, 2022

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