Question

A polynomial f(x) has the given zeros of 7-1, and-3

Answer 1

The polynomial f(x) in expanded form is $f(x) = x^3 - 3x^2 - 25x - 21$

The polynomial f(x)

The polynomial zeros are given as:

x = 7, x = -1 and x = -3

Rewrite as:

x - 7 = 0, x + 1 = 0 and x + 3 = 0

Multiply the zeros

f(x) = (x - 7)(x + 1)(x + 3)

Expand

$f(x) = (x - 7)(x^2 + 4x + 3)$

Further, expand

$f(x) = x^3 + 4x^2 - 7x^2 + 3x - 28x - 21$

Evaluate

$f(x) = x^3 - 3x^2 - 25x - 21$

Hence, the polynomial f(x) in expanded form is $f(x) = x^3 - 3x^2 - 25x - 21$

Rational function g(x)

We have:

$g(x) = \frac{f(x)}{x^2 - x- 2}$

This gives

$g(x) = \frac{x^3 - 3x^2 - 25x - 21}{x^2 - x- 2}$

Expand the numerator

$g(x) = \frac{x^3 - x^2 - 2x^2 - 2x + 2x - 25x + 4 - 25}{x^2 - x - 2}$

Rewrite as:

$g(x) = \frac{x^3 - x^2 - 2x - 2x^2 + 2x + 4 - 25x - 25}{x^2 - x - 2}$

Factorize

$g(x) = \frac{(x - 2)(x^2 - x- 2) - 25x - 25}{x^2 - x- 2}$

Evaluate the quotient

$g(x) = x - 2 - \frac{25x + 25}{x^2 - x- 2}$

The slant asymptote is the quotient i.e. x - 2

Hence, the slant asymptote of the function g(x) is x - 2

The discontinuities of g(x)

In (b), we have:

$g(x) = \frac{x^3 - 3x^2 - 25x - 21}{x^2 - x- 2}$

Set the denominator to 0

$x^2 - x - 2 = 0$

Expand

$x^2 + x - 2x - 2 = 0$

Factorize

x(x + 1) - 2(x + 1) = 0

Factor our x + 1

(x - 2)(x + 1) = 0

Solve for x

x = 2 or x = -1

2 is greater than -1.

So, the discontinuities and their types are:

• Essential discontinuity at x = 2
• Removable discontinuity at x = -1

Read more about polynomials at:

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