Problem 24 - Algebra 2

Question

Find all of the zeros of each function. \(h(x)=6 x^{3}+11 x^{2}-3 x-2\)

Step-by-Step Solution

Answer

The zeros are \(-1, \frac{-5 + \sqrt{73}}{12}, \frac{-5 - \sqrt{73}}{12}\).

Step 1: Identify Possible Rational Zeros

The possible rational zeros can be found using the Rational Root Theorem, which suggests that the possible rational zeros are the factors of the constant term divided by the factors of the leading coefficient. For the function \( h(x) = 6x^3 + 11x^2 - 3x - 2 \), the factors of \(-2\) (constant term) are \(\pm 1, \pm 2\) and the factors of \(6\) (leading coefficient) are \(\pm 1, \pm 2, \pm 3, \pm 6\). 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 Zeros Using Synthetic Division

Use synthetic division to test potential zeros from the list determined in Step 1. Begin with \(x = -1\). Using synthetic division, if \(x = -1\) is a zero, the remainder should be 0. Applying synthetic division, the result is \(6x^2 + 5x - 2\) with a remainder of 0, proving \(-1\) is a zero of \(h(x)\).

Step 3: Solve Reduced Quadratic Equation

Once \(x = -1\) is verified as a zero, the polynomial can be factored to \((x + 1)(6x^2 + 5x - 2)\). We need to solve the quadratic \(6x^2 + 5x - 2 = 0\) either by factoring, completing the square, or the quadratic formula. Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].For \(a = 6\), \(b = 5\), and \(c = -2\), plug in these values:\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} \].

Step 4: Simplify Using Quadratic Formula

Simplify the quadratic formula:\[ x = \frac{-5 \pm \sqrt{25 + 48}}{12} \]\[ x = \frac{-5 \pm \sqrt{73}}{12} \].This yields two real zeros: \(x = \frac{-5 + \sqrt{73}}{12}\) and \(x = \frac{-5 - \sqrt{73}}{12}\).

Key Concepts

Rational Root Theorem

The rational root theorem is a helpful tool in finding possible rational roots or zeros of a polynomial function. By applying this theorem, you can predict potential rational zeros of a polynomial function of the form \[ ax^n + bx^{n-1} + ... + k = 0 \]Here's the key idea:

• Consider the constant term, \(k\), and the leading coefficient, \(a\).
• The possible rational zeros are given by \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(k\), and \(q\) is a factor of the leading coefficient \(a\).

Let's take the function \(h(x) = 6x^3 + 11x^2 - 3x - 2\) as an example. Here, the constant term is \(-2\) with potential factors \(\pm 1, \pm 2\), and the leading coefficient is \(6\) with potential factors \(\pm 1, \pm 2, \pm 3, \pm 6\). Thus, 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}\}. This list gives you a set of potential values to test as zeros of the function, starting your journey into determining the actual zeros.

Synthetic Division

Synthetic division is a streamlined form of polynomial division, much faster and simpler than traditional long division. It's particularly beneficial for testing possible zeros that you've identified with the Rational Root Theorem.To use synthetic division:

• Grab the coefficients of the polynomial you want to divide.
• Choose a potential zero from your list and place it in the synthetic division setup.

As an example, for \(h(x) = 6x^3 + 11x^2 - 3x - 2\), let's test \(x = -1\):

• Write the coefficients: \(6, 11, -3, -2\)
• The process involves bringing down the leading coefficient and performing operations across the row that involve your test zero \(-1\).

If your final remainder is zero, you've found a zero of the polynomial. With \(x = -1\), the polynomial reduces to \(6x^2 + 5x - 2\), confirming that \(-1\) is indeed a zero because the remainder here is zero.

Quadratic Formula

When you're left with a quadratic equation after simplifying a polynomial, as we did with \(6x^2 + 5x - 2\), the quadratic formula is a handy tool. It's general and applicable when factoring is not straightforward.The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Just plug in the coefficients from your quadratic equation.For \(6x^2 + 5x - 2 = 0\):

• Set \(a = 6\), \(b = 5\), \(c = -2\).
• Insert these into the formula to find your zeros: \[x = \frac{-5 \pm \sqrt{(5)^2 - 4(6)(-2)}}{2(6)}\]
• Simplifying gives the solutions: \[x = \frac{-5 \pm \sqrt{73}}{12}\]

These steps give you the two real solutions, which are more zeros of your function.

Zeros of Functions

The zeros of a function are the \(x\)-values for which the function equals zero, typically found where the graph intersects the \(x\)-axis. Discovering all zeros of a polynomial involves steps using the rational root theorem, synthetic division, and possibly the quadratic formula, as we've done.For the function \(h(x) = 6x^3 + 11x^2 - 3x - 2\), the zeros we identified are:

• \(x = -1\)
• \(x = \frac{-5 + \sqrt{73}}{12}\)
• \(x = \frac{-5 - \sqrt{73}}{12}\)

Each zero represents a point on the graph of \(h(x)\) where the graph crosses or touches the \(x\)-axis. Understanding zeros is crucial because they reveal valuable information about the behavior and properties of the function itself.