Question

Use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor \(f\) over the real numbers. $$ f(x)=3 x^{3}+6 x^{2}-15 x-30 $$

Short answer

The real zeros are \(-2, \sqrt{5}, -\sqrt{5}\). The polynomial factors over the real numbers as \(3(x+2)(x-\sqrt{5})(x+\sqrt{5})\).

Step-by-step solution

Identify the polynomial's coefficients

First, note the polynomial provided: \[ f(x) = 3x^3 + 6x^2 - 15x - 30 \] The coefficients are: 3 (leading coefficient) and -30 (constant term).

List possible rational zeros

Apply the Rational Zeros Theorem, which states that any rational zero of the polynomial function will be a fraction \( \frac{p}{q} \) where p and q are factors of the constant term and leading coefficient, respectively. Factors of -30 (constant term): \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 \)Factors of 3 (leading coefficient): \( \pm 1, \pm 3 \)Possible rational zeros: \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3}, \pm \frac{15}{3} \)

Test possible rational zeros

Use synthetic division or direct substitution to test the possible rational zeros. A zero will result in a remainder of zero. Testing \(x = -2\):\[ f(-2) = 3(-2)^3 + 6(-2)^2 - 15(-2) - 30 = -24 + 24 + 30 - 30 = 0\]So, \(x = -2\) is a zero.

Factor out the found zero

Since \(x = -2\) is a zero, \(x + 2\) is a factor. Use synthetic division to factorize \[ f(x) = 3x^3 + 6x^2 - 15x - 30 \]by \(x + 2\).Set up the synthetic division:\[\begin{array}{r|rrrr} -2 & 3 & 6 & -15 & -30 \ \ \footnotesize{} & {} & -6 & 0 & 30 \ \ \text{} & 3 & 0 & -15 & 0 \ \ \text{} & \text{} & {} & 0 \ \text{} & 3x^2 & -0x & -15 ewline & \text{} & 3x^2 - 15ewline & \text{} ewline & \text{} So, \[ f(x) = (x + 2)(3x^2 - 15) \]\]

Factor the quadratic polynomial

Factoring the remaining quadratic polynomial: \[ 3x^2 - 15 = 3(x^2 - 5) = 3(x - \sqrt{5})(x + \sqrt{5})\]So, the fully factored form is:\[ f(x) = 3(x+2)(x-\sqrt{5})(x+\sqrt{5})\]

List all real zeros

The real zeros found are: \(x = -2\), \(x = \sqrt{5}\), and \(x = -\sqrt{5}\).

Key concepts

Polynomial Functions

A polynomial function is an expression that consists of variables and coefficients. The variables in a polynomial are raised to whole number exponents, and the coefficients are real numbers. Here's a typical form of a polynomial function:

• For example, the polynomial function given in the exercise is

\[ f(x) = 3x^3 + 6x^2 - 15x - 30\] The highest exponent in the polynomial determines its degree; in this case, it is degree 3 because the highest power of \(x\) is \(x^3\). Polynomial functions can have various types of solutions, including real zeros. Finding the real zeros is crucial because it helps in factoring the polynomial, making it easier to understand and utilize in different applications.

Synthetic Division

Synthetic division is a simplified form of polynomial division, specifically used to divide polynomials by binomials of the form \(x - c\). It is a quicker and more straightforward method compared to long division. Here's a basic rundown on how synthetic division works:

• First, write down the coefficients of the polynomial.
• Use a test zero \(c\) (in this case, from the Rational Zeros Theorem) and place it to the left.
• Bring down the leading coefficient to the bottom row.
• Multiply \(c\) by the value just written in the bottom row and add the result to the next coefficient.

In the exercise, synthetic division is used after identifying \(x = -2\) as a zero:\[\begin{array}{r|rrrr}-2 & 3 & 6 & -15 & -30 \ & & -6 & & \ & 3 & 0 &-15 & 0 \end{array}\]\ The result shows the new polynomial as \(3x^2 - 15\), indicating that \(x + 2\) is a factor of the original polynomial.

Factoring Quadratic Polynomials

After reducing the polynomial using synthetic division and finding one real zero, the next step is to factorize the remaining quadratic polynomial. Quadratics are typically in the form \(ax^2 + b\)x \(+ c\). In this exercise, the quadratic obtained is: \(f(x) = 3(x^2 - 5)\).

To factor it further:

• Factor out the common factor 3:

\(3(x^2 - 5)\) Now, look at \(x^2 - 5\), which can be factored further using the difference of squares formula: \(a \) / )(a - b)(a + b):
\(x^2 - 5 = (x + )( x - )(\sqrt{5} \). This gives the fully factored form: \(f(x) = 3(x + 2)(x + \sqrt{5})(x - \sqrt{5})\).These factors reveal the real zeros, which are the solutions to the polynomial function. For this exercise, the real zeros are:

• \(x = - 2 \)
• \(x = - \sqrt {5}\)
• \(x = \sqrt {5}\)

Understanding how to factorize quadratics is essential for solving and simplifying polynomial functions.