Problem 38 - Algebra and Trigonometry

Question

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=6 x^{3}+11 x^{2}-3 x-2 $$

Step-by-Step Solution

Answer

The rational zeros of the polynomial are \( x = -1, -\frac{1}{2}, -\frac{2}{3} \). The factored form is \( P(x) = (x + 1)(2x + 1)(3x + 2) \).

Step 1: Use the Rational Root Theorem

According to the Rational Root Theorem, any rational zero, \( \frac{p}{q} \), of the polynomial \( P(x) \) is such that \( p \) is a factor of the constant term (-2) and \( q \) is a factor of the leading coefficient (6). The factors of -2 are \( \pm 1, \pm 2 \), and the factors of 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \). Therefore, the possible rational zeros are \( \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6} \).

Step 2: Test Possible Rational Zeros

We test each possible rational zero using synthetic division or direct substitution into the polynomial. After testing, we discover that \( x = -1 \) is a zero of the polynomial because \( P(-1) = 0 \).

Step 3: Synthetic Division

Perform synthetic division using \( x = -1 \) to divide the polynomial \( P(x) = 6x^3 + 11x^2 - 3x - 2 \). Divide the coefficients [6, 11, -3, -2]. The result is a quotient of \( 6x^2 + 5x + 2 \) with a remainder of 0, confirming \( x + 1 \) as a factor.

Step 4: Factor the Quadratic

Now factor the quadratic \( 6x^2 + 5x + 2 \). We look for two numbers that multiply to \( 6 \times 2 = 12 \) and add to 5. These numbers are 3 and 2. Factor the quadratic as \( (2x + 1)(3x + 2) \).

Step 5: Write the Polynomial in Factored Form

The original polynomial \( P(x) \) can be expressed in factored form as \( (x + 1)(2x + 1)(3x + 2) \). All these factors are confirmed from our synthetic division and factoring practices.

Key Concepts

Polynomial Factoring

At its core, polynomial factoring is the process of breaking down a polynomial into simpler factors that, when multiplied together, give back the original polynomial. This is similar to factoring numbers, like how 12 can be factored into 3 and 4. Factoring polynomials has several purposes, one being to find the zeros or roots of the polynomial.

For a polynomial like \( P(x) = 6x^3 + 11x^2 - 3x - 2 \), factoring means expressing it as a product of its simpler components. These components might be linear factors like \( x + 1 \) or quadratic factors. Factoring helps us understand more about the polynomial, including its behavior and graph.

This task often begins by using the Rational Root Theorem to identify potential rational roots, but we'll explore that in more detail in sections ahead.

Synthetic Division

Synthetic division is a quick method to divide polynomials when you know one of the roots. It’s particularly handy because it’s faster and more streamlined than the traditional polynomial division, especially when dealing with a potential zero. In synthetic division, we work with the coefficients of the polynomial rather than the whole expression itself.

Let's say you suspect \( x = -1 \) is a root of \( P(x) = 6x^3 + 11x^2 - 3x - 2 \). You'd set up the synthetic division by listing the coefficients, which are \([6, 11, -3, -2]\). The root goes outside the division bracket. The division process is an iteration of bringing down, multiplying, adding, and repeating, leading to a quotient and possibly a remainder.

Here, the result was a quotient of \( 6x^2 + 5x + 2 \) and a remainder of 0. This confirms that \( x + 1 \) is a factor of the polynomial.

Rational Zeros

Finding rational zeros involves determining which fractions could possibly be zeros of the polynomial. The Rational Root Theorem provides a systematic way to find these values. For a polynomial \( P(x) \) with integer coefficients, any rational zero, \( \frac{p}{q} \), is such that:

• \( p \) is a factor of the constant term (in our example, \(-2\)),
• \( q \) is a factor of the leading coefficient (here, 6).

From these factors, you list all possible values of \( \frac{p}{q} \). For our example, possible rational zeros are \( \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6} \).

Testing each using substitution or synthetic division identifies actual zeros, like \( x = -1 \) in this case.

Quadratic Factoring

Once you've used synthetic division to divide the polynomial and reduce it, you might need to factor any remaining quadratic expressions. Quadratic factoring involves rewriting the quadratic as a product of two binomials.

For the quadratic \( 6x^2 + 5x + 2 \), we need numbers that multiply to 12 (the product of the leading coefficient 6 and the constant term 2) and add to 5 (the middle coefficient). Here, these numbers are 3 and 2.

With these numbers, the quadratic can be factored into \((2x + 1)(3x + 2)\). By putting it all together from earlier steps and synthetic division results, the original polynomial \( P(x) = 6x^3 + 11x^2 - 3x - 2 \) is fully factored into \((x + 1)(2x + 1)(3x + 2)\).