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The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

 Aug 26, 2019
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Since f(0)  = 0,,,,then x  = 0  is one root

 

So.....we have the  form    ax^3 + bx^2 + cx + d   where d  = 0

 

So  we know that

 

a(-1)^3 + b(-1)^2 + c(-1)  = 15  ⇒    -a +b - c = 15     (1)

a (1)^3  + b(1)^2 + c(1)  = -5   ⇒      a+ b + c = -5   (2)

a(2)^3 + b(2)^2 + c(2)  = 12  ⇒         8a + 4b  + 2c = 12    (3)

 

Add (1) and (2)  and we have that 2b = 10  ⇒   b  = 5

 

Multiply (1) by 8  and add to (3)   and we have that

 

12b - 6c  = 132     sub 5 in for b  and we have

12(5) - 6c  = 132

60 - 6c  = 132     subtract 60 from both sides

-6c  =  72   divider both sides by -6

c  = -12

 

And using (2)  we can find  a  as

 

a + 5  - 12  = -5

a - 7  =  -5    add 7 to both sides

a  = 2

 

So.....the polynomial  is     2x^3  + 5x^2 - 12x

 

Set to 0  and factor

 

x (2x^2 + 5x - 12)  = 0

 

x (2x  - 3) ( x + 4)  = 0      set each factor to 0   and solve for x

 

x  = 0         2x - 3  =  0        x + 4  = 0

 

x = 0       x  = 3/2          x  = -4       these  are the x intercepts

 

Here is a graph  : https://www.desmos.com/calculator/o040mctwqm

 

 

cool cool cool

 Aug 26, 2019

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