Problem 66 - College Algebra

Question

For the following exercises, use the given information about the polynomial graph to write the equation. Double zero at \(x=-3\) and triple zero at \(x=0\) . Passes through the point \((1,32)\) .

Step-by-Step Solution

Answer

The polynomial is \(f(x) = 2(x + 3)^2 x^3\).

Step 1: Identify the Zeros and Multiplicities

We are given that the polynomial has a double zero at \(x = -3\) and a triple zero at \(x = 0\). This means the factors of the polynomial include \((x + 3)^2\) and \(x^3\). The general form of the polynomial is then \(f(x) = a \cdot (x + 3)^2 \cdot x^3\), where \(a\) is a constant we need to determine.

Step 2: Substitute the Point into the Polynomial

We know that the polynomial passes through the point \((1,32)\). Substitute \(x = 1\) and \(f(x) = 32\) into the polynomial equation: \[ f(x) = a \cdot (x + 3)^2 \cdot x^3 \Rightarrow 32 = a \cdot (1 + 3)^2 \cdot 1^3. \]

Step 3: Solve for the Constant 'a'

Simplify the equation from step 2: \[ 32 = a \cdot (4)^2 \cdot 1. \] This simplifies to \(32 = 16a\). Solve for \(a\) by dividing both sides of the equation by 16:\[ a = \frac{32}{16} = 2. \]

Step 4: Write the Polynomial Equation

Now that we have found \(a = 2\), substitute it back into the general polynomial equation to get the specific equation: \[ f(x) = 2 \cdot (x + 3)^2 \cdot x^3. \] Simplify this to \[ f(x) = 2x^3(x^2 + 6x + 9). \]

Key Concepts

Understanding Zeros in Polynomials

Zeros of a polynomial are values of the variable that make the polynomial equal to zero. In simpler terms, zeros are the x-values where the graph of the polynomial intersects the x-axis. They are crucial because they tell us where the polynomial "dips" below the x-axis, hits it, or just touches it lightly.

• If a polynomial has a single zero, the graph crosses the x-axis at that point.
• A double zero means the graph touches the x-axis and turns back.
• A triple zero would cause the graph to flatten out at the axis before crossing.

In our exercise, the double zero at \(x=-3\) tells us the graph "bounces" off the x-axis at this point. The triple zero at \(x=0\) means the graph flattens and goes through the x-axis. This behavior is crucial for plotting and understanding the nature of polynomial graphs.

The Role of Multiplicities

Multiplicities refer to the number of times a particular zero occurs in the polynomial. The multiplicity influences the shape and the behavior of the graph at that zero.

• A zero with odd multiplicity means the graph will pass through the x-axis at that zero.
• A zero with even multiplicity causes the graph to just touch the x-axis and turn back.

For example, if a polynomial has a zero of multiplicity 2, like \((x+3)^2\) in our problem, the graph will touch and then turn back at \(x=-3\). Similarly, a multiplicity of 3, such as \(x^3\), will make the graph pass through the x-axis but will appear more flattened at the point of zero. This is why understanding multiplicities helps predict how the graph behaves around its zeros, giving us a more comprehensive view.

Factored Form and Its Benefits

Writing a polynomial in its factored form means expressing it as a product of its zero-determined factors. This method is beneficial for easily identifying zeros and their multiplicities. The process converts the polynomial into a form that highlights its root characteristics.

• In our exercise, the factors are \((x+3)^2\) and \(x^3\), showing the zeros at \(x=-3\) and \(x=0\) respectively.
• The exponents on these factors (2 and 3) show the multiplicities of the corresponding zeros.

Factoring a polynomial simplifies solving and graphing. It allows one to quickly spot the zeros while also providing insight into the graph's general shape through the behavior dictated by those zeros and their multiplicities.

Finding Polynomial Equations from Graph Information

To find a polynomial equation from graph information, you need zeros, their multiplicities, and potentially a point that the graph passes through. Here are the key steps:

• Identify the zeros and set up the factors with their respective multiplicities. For example, a double zero at \(x=-3\) becomes \((x+3)^2\).
• Include a variable constant \(a\) which is solved using the given point that the graph passes through.
• Substitute the x-value of this point into your equation and solve for \(a\), ensuring the equation matches the point's y-value.

With \(a\) found, substitute it back into the polynomial to get the specific equation. In our exercise, \(a\) was determined to be 2 when the graph passed through \((1,32)\), leading to the polynomial equation \(f(x) = 2 \cdot (x + 3)^2 \cdot x^3\). These steps ensure that the final polynomial equation correctly reflects the behavior and position of its graphical representation.