Will somebody please help me with this Pre-Calc question.

Question

A polynomial f (x) has the given zeros of 6, –1, and –3.

Part A: Using the Factor Theorem, determine the polynomial f (x) in expanded form. Show all necessary calculations.

Part B: Divide the polynomial f (x) by (x² – x – 2) to create a rational function g(x) in simplest factored form. Determine g(x) and find its slant asymptote.

Part C: List all locations and types of discontinuities of the function g(x).

Answer 1

Part A. Because the zeros are 6, -1, -3, the polynomial has the factors (x-6)(x+1)(x+3). To get the expanded form, you multiply those out.

(x-6)(x+1)(x+3)

(x-6)(x²+4x+3)

x³+4x²+3x-6x²-24x-18

x³-2x²-21x-18

So f(x) = x³-2x²-21x-18

Part B. Divide f(x) by (x² - x - 2). The easiest way to do this is to get the factored form (x² - x - 2) = (x+1)(x-2)

And divide using using the factored forms: g(x) = (x-6)(x+1)(x+3) / (x+1)(x-2).

To get the simplest factored form, you can divide out the common factor (x+1) and get g(x) = (x-6)(x+3) / (x-2)

In expanded form, this is g(x) = x²-3x-18 / x-2

To get the slant asymptote, you need to divide these using polynomial division.

x - 1

x - 2 | x² - 3x - 18

x² - 2x

-x - 18

-x + 2

-20

The slant asymptote is y = x-1

Part C.

The discontinuities of a rational function are the zeros of the denominator. It's easiest to see this when g(x) is in factored form:

g(x) = (x-6)(x+1)(x+3) / (x+1)(x-2)

The zeros of the denominator are -1 and 2. So, the discontinuities of g(x) are x = -1 and x = 2.

Because we can simplify g(x) by dividing out the (x+1), g(x) = (x-6)(x+3) / (x-2), then the type of discontinuity at x = -1 is "removable". (We removed it, right?)

But the other discontinuity at x = 2 is still there. It is a "infinite" discontinuity. (Because at x = 2, g(x) goes to infinity.)

I think that's everything.

Answer 2

zeros 6, -1, -3

x=6

x-6=0

x=-1

x +1=0

x=-3

x+3 =0

factors are (x-6)(x+1)(x+3) multiplied together = (x^2-5x -6)(x+3)

f(x)= x^3 -2x^2 -21x -18

x^-x -2 + (x-2)(x+1)

divide (x-6)(x+1)(x+3) by (x-2)(x+1) and the x+1 factors cancel leaving

(x-6)(x+3)/(x-2)= g(x) = (x^2 -3x -18)/(x-2)=

= x-1 wiith a remainder of -20

g(x) has a discontinuity where the denominator x-2 = 0, x+1=0 where x=2, -1

g(x) has a vertical asymptote x=2, slant asymptote y= x+1, as x gets very large g(x) approaches (x^2+x)/x = x+1