Problem 34 - College Algebra with Corequisite Support

Question

For the following exercises, find the zeros and give the multiplicity of each. $$ f(x)=(2 x+1)^{3}\left(9 x^{2}-6 x+1\right) $$

Step-by-Step Solution

Answer

The zeros are \(x = -\frac{1}{2}\) with multiplicity 3 and \(x = \frac{1}{3}\) with multiplicity 2.

Step 1: Set the Function Equal to Zero

To find the zeros of the function \( f(x) = (2x + 1)^3 (9x^2 - 6x + 1) \), set it equal to zero: \[ (2x + 1)^3 (9x^2 - 6x + 1) = 0 \]. The product is zero if either \((2x + 1)^3 = 0\) or \(9x^2 - 6x + 1 = 0\).

Step 2: Solve for Zeros of \((2x + 1)^3 = 0\)

Solve \((2x + 1) = 0\) to find the zeros for this term. Rearrange to find: \[ 2x + 1 = 0 \] \[ 2x = -1 \] \[ x = -\frac{1}{2} \]. Since the power is 3, the multiplicity of this zero is 3.

Step 3: Solve the Quadratic Formula

Solve \(9x^2 - 6x + 1 = 0\) using the quadratic formula. For a quadratic equation \(ax^2 + bx + c = 0\), the solutions are \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \(a = 9\), \(b = -6\), and \(c = 1\).

Step 4: Calculate Discriminant and Solutions

Calculate the discriminant first: \(b^2 - 4ac = (-6)^2 - 4(9)(1) = 36 - 36 = 0\). The discriminant is zero, indicating one repeated real root. Now calculate the root:\[ x = \frac{-(-6) \pm \sqrt{0}}{2(9)} \] \[ x = \frac{6}{18} = \frac{1}{3} \]. This root has multiplicity 2.

Key Concepts

Quadratic Formula

When faced with a quadratic equation like \( ax^2 + bx + c = 0 \), the quadratic formula is incredibly useful. It provides a straightforward way to find the zeros of the equation, which are the values of \( x \) that make the equation true. The formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The terms \( a \), \( b \), and \( c \) represent the coefficients of the equation. In this particular exercise, we have a quadratic term in the function:

• \( 9x^2 - 6x + 1 = 0 \)

Breaking down the quadratic formula step by step can simplify the solution process:

• Calculate the discriminant: \( b^2 - 4ac \).
• Place values into the formula to find \( x \).

In this case, the discriminant determines the nature of the roots, whether they are real or complex, and how many there are.

Multiplicity of Zeros

The multiplicity of zeros refers to the number of times a particular zero appears in the function. It indicates how a graph touches or crosses the x-axis at zero.

• If a zero has multiplicity 1, the graph crosses the x-axis at that point.
• If a zero has an even multiplicity (like 2 or 4), the graph touches but does not cross the x-axis at that zero.
• If a zero has an odd multiplicity greater than 1, the graph crosses the axis and changes direction.

In the provided exercise, \( x = -\frac{1}{2} \) is a zero with a multiplicity of 3, meaning the graph will cross and twist slightly around this point. On the other hand, \( x = \frac{1}{3} \) is a zero with multiplicity 2, meaning the function's graph will just touch and not cross the x-axis here. Understanding the multiplicity helps with sketching the behavior of the polynomial graph.

Real Root Calculation

Calculating real roots is essential in determining where the function equals zero on the x-axis. For polynomials, these calculations often begin by solving for zeros of smaller factors, as we did by setting \((2x + 1)^3 = 0 \) and \(9x^2 - 6x + 1 = 0 \).

• For simple linear factors like \((2x + 1)\), rearrange and solve the equation \(2x + 1 = 0\) to find \( x = -\frac{1}{2} \).
• For quadratic factors like \(9x^2 - 6x + 1 \), use the quadratic formula. Compute the discriminant first to determine the nature of the roots.

In this exercise, the function \(9x^2 - 6x + 1\) has a discriminant of 0, indicating one real root. The calculation showed that \( x = \frac{1}{3} \) with multiplicity 2. Real root calculation is crucial in understanding the polynomial's full behavior and how it interacts with the x-axis.