Question

I will give 50 points

Answer 1

Answer:

f(x) = x³+2x²-16x-32

Step-by-step explanation:

f(x) = 0 when x = -4, -2, 4

So these are the roots

Factors are (x+4)(x+2)(x-4)

f(x) = [(x+4)(x-4)](x+2)

f(x) = (x²-16)(x+2)

f(x) = x³+2x²-16x-32

Answer 2

Answer:

f(x) = 1x^3 + 2x^2  -16x - 32

Step-by-step explanation:

Looking at the first row of the table, I see f(x) = -27 when x = -5.  We know that -27 is the cube of -3.  So, try adding 2 to x = -5, obtaining -3.  The result (-5 + 2)^3 equals -27, as we wanted.

Unfortunately the proposed function f(x) = (x + 2)^3 does not "work" for other x values.

The governing polynomial involves four coefficients:  f(x) = ax^3 + bx^2 + cx + d.

We can pick any four x values and the corresponding y-values and come up with four versions of the polynomial f(x) = ax^3 + bx^2 + cx + d:

(1)  Choose x = -4 and f(x) = 0 from the second line in the table.  Then

     0 = a(-4)^3 + b(-4)^2 + c(-4) + d, or -64a + 16b - 4c + d = 0, and

(2) Choose x = 0 and y  = -32.  Then -32 = f(x) = a(0)^3 + b(0)^2 + c(0) + d.  Then we have -32 = 0 + 0 + 0 + d, or d = -32.

(3) Choose x = 4 and y = 0.  Then we have 0 = 64a + 16b + 4c - 32.

(4) Choose x = 5 and y = 63.  Then 63 = 125a + 25b + 5c - 32

Now we have four equations in four unknowns {a, b, c, d}.

Let's simplify these four equations, as follows:

(2) We have already seen that d = -32.  Therefore,

(3) becomes 64a + 16b + 4c = 32

(4) becomes 125a + 25b + 5c = 95.  We need one more equation.

(5) will be based on (2, -48):  -48 = 2^3a + 2^2b + 2c - 32, or

                                                 -16 = 8a + 4b + 2c

Now we have a system of linear equations to solve:

  8a + 4b + 2c = -16

125a +25b+5c = 95

 64a  + 16b + 4c = 32

Here I used the MATRIX functions of my old TI-83Plus calculator to find the values of {a, b, c}.  They are:

a=1, b=2 and c= -16

and so the general 3rd order polynomial becomes

f(x) = 1x^3 + 2x^2  -16x - 32