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)?
+130071 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
_________________________________
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
