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# Polynomial problem

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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

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