Question

Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=9 x^{3}-15 x^{2}+11 x-5$$

Short answer

The zeros of the function \(f(x) = 9x^3 - 15x^2 + 11x - 5\) are \(x = 1\), \(x = 1/3\) and \(x = 5/3\). The polynomial can be written as the product of linear factors: \(f(x) = 9(x - 1)(x - 1/3)(x - 5/3)\).

Step-by-step solution

Identify the polynomial

Identify the polynomial function given in the problem, which is \(f(x)=9x^3 - 15x^2 + 11x - 5\).

Set the function to zero

To find the zeros of the function, set \(f(x)\) to zero. The resulting equation becomes \(9x^3 - 15x^2 + 11x - 5 = 0\).

Solve the equation

Solving this equation might be challenging. Apply methods such as factoring, the quadratic formula, or synthetic division, depending on what is most suitable. Since the coefficients don't have any obvious common factors and the cubic function is not easily factored, synthetic division by guessing a root might be the most applicable approach. By trial and error or by using the Rational Root Theorem, one possible solution (roots) is \(x = 1\). Apply synthetic division using this root.

Implement Synthetic Division

Write 1 in the leftmost column, and under the dividend (the numbers before x, which are 9, -15, 11, and -5), write the coefficients of the polynomial. After implementing Synthetic Division, the reduced polynomial is \(9x^2 - 6x + 5\).

Solve the Quadratic Equation

Set the remaining quadratic equation equal to zero and solve for x. \(9x^2 - 6x + 5 = 0\). This quadratic equation could be solved using the quadratic formula, \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\). Solving it gives \(x = 1/3\) and \(x = 5/3\).

Write the polynomial as a product of linear factors

Now the polynomial can be written as a product of linear factors. The linear factors are obtained by setting each of the solutions equal to x and then subtracting them from both sides. The factors are \(x - 1\), \(x - 1/3\) and \(x - 5/3\). Therefore, \(f(x)=9(x - 1)(x - 1/3)(x - 5/3)\).

Key concepts

Synthetic Division

Synthetic Division is a simplified way to divide a polynomial by a binomial of the form \((x - c)\). It is widely used because it requires fewer steps and is much quicker compared to the long division of polynomials.
To start with synthetic division, choose a value for \(c\) that could be a root of the polynomial. This is often determined using the Rational Root Theorem, which we'll discuss next. Place this \(c\) outside of the synthetic division table, which is represented as a horizontal line dividing two rows.

• Write the coefficients of the polynomial to the right of this divider, keeping them in order from the highest degree to the constant term.
• Drop the leading coefficient down to the answer row unchanged.
• Multiply this number by \(c\) and write the resulting value underneath the second coefficient.
• Add this to the second coefficient and write the result in the answer row.

Repeat the multiply and add steps for all coefficients. The final row provides the coefficients of the quotient. If there is a remainder, it will be the last number in this row.
For our example, checking the value \(x = 1\) was confirmed through trial and error, and using synthetic division with \(x = 1\) reduced the cubic polynomial to a quadratic.

Rational Root Theorem

The Rational Root Theorem is a handy tool for finding potential rational zeros (or roots) of a polynomial. The theorem states that any rational root of the polynomial \(P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0\) must be of the form \(\frac{p}{q}\), where:

• \(p\) is a factor of the constant term \(a_0\).
• \(q\) is a factor of the leading coefficient \(a_n\).

This list of possible rational roots can be plugged into the polynomial to see which, if any, are actual roots.
In our polynomial \(f(x) = 9x^3 - 15x^2 + 11x - 5\), the constant term is -5, and the leading coefficient is 9. The factors of -5 are \(\pm1, \pm5\), and the factors of 9 are \(\pm1, \pm3, \pm9\). This means potential rational roots could be any combination of these fractions.
By inputting these values, \(x = 1\) was determined to be a root, thereby justifying the use of synthetic division with this particular value.

Quadratic Formula

The Quadratic Formula is a universal method for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is particularly useful when the quadratic equation does not factor easily.
The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The term \(b^2 - 4ac\) is known as the discriminant. It indicates the nature of the roots.
• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root.
• If it is negative, there are no real roots, but two complex ones instead.

In the reduced polynomial from our exercise, \(9x^2 - 6x + 5\), applying the quadratic formula results in finding two additional roots \(x = \frac{1}{3}\) and \(x = \frac{5}{3}\).
These roots, together with the one found through synthetic division, allow us to express the original polynomial as a product of linear factors.