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

Guest Jul 30, 2018
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You can write the polynomial as:

 

f(x) = x(ax2+bx+c)     This form immediately gives f(0) = 0.

 

Now use the other conditions to find constants a, b and c:

 

15 = -1( (-1)2a + (-1)b +c)   or  15 = -a + b - c   ...(1)

 

-5 = 1(12a + 1b + c)  or  -5 = a + b + c    ...(2)

 

12 = 2(22a + 2b + c)  or  12 = 8a + 4b + 2c   or  6 = 4a + 2b + c...(3)

 

Add (1) and (2) to get:  10 = 2b  or  b = 5

 

Substitute this into (2) and (3) and rearrange:

-10 = a + c   ...(2b)

-4 = 4a + c  ...(3b)

 

Subtract (2b) from (3b)

6 = 3a   or  a = 2

 

Substitute this into (2b)

-10 = 2 + c  or  c = -12

 

Now we have:  f(x) = x(2x2 + 5x - 12) 

This factors as f(x) = x(2x - 3)(x + 4)

 

Hence the roots are: 0, 3/2, -4

 

You should check this by substituting the values -1, 0, 1 and 2 for x into the function to see they give the values specified in the question.

Alan  Jul 31, 2018

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