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?
+33616 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.
