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

#1**0 **

You can write the polynomial as:

f(x) = x(ax^{2}+bx+c) This form immediately gives f(0) = 0.

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

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

-5 = 1(1^{2}a + 1b + c) or -5 = a + b + c ...(2)

12 = 2(2^{2}a + 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(2x^{2} + 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