Chapter 5: Problem 36
For the following exercises, find the zeros and give the multiplicity of each. $$ f(x)=x\left(4 x^{2}-12 x+9\right)\left(x^{2}+8 x+16\right) $$
Short Answer
Expert verified
Zeros are \( x = 0 \) (multiplicity 1), \( x = \frac{3}{2} \) (multiplicity 2), \( x = -4 \) (multiplicity 2).
Step by step solution
01
Identify the Factors
First, look at the expression given: \( f(x) = x(4x^2 - 12x + 9)(x^2 + 8x + 16) \). The function is already factored into three parts: \( x \), \( 4x^2 - 12x + 9 \), and \( x^2 + 8x + 16 \).
02
Find Zeros of Each Factor
To find the zeros, set each factor equal to zero and solve for \( x \).1. For \( x \): \( x = 0 \).2. For \( 4x^2 - 12x + 9 \): Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -12 \), and \( c = 9 \). \( b^2 - 4ac = (-12)^2 - 4 \cdot 4 \cdot 9 = 144 - 144 = 0 \) There is one real solution since the discriminant is 0: \( x = \frac{-(-12)}{2 \cdot 4} = \frac{12}{8} = \frac{3}{2} \).3. For \( x^2 + 8x + 16 \): Also, use the quadratic formula with \( a = 1 \), \( b = 8 \), \( c = 16 \). \( b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 16 = 64 - 64 = 0 \) One real solution for \( x \): \( x = \frac{-8}{2} = -4 \).
03
Determine Multiplicities
Multiplicity refers to the number of times a particular zero appears in the factorization:1. The zero \( x = 0 \) comes from the factor \( x \), so it has a multiplicity of 1.2. The zero \( x = \frac{3}{2} \) is the root of the perfect square trinomial \( (2x - 3)^2 = 4x^2 - 12x + 9 \), so its multiplicity is 2.3. The zero \( x = -4 \) is the root of the perfect square trinomial \( (x + 4)^2 = x^2 + 8x + 16 \), so its multiplicity is 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool for finding the solutions, or "zeros," of a quadratic equation. A quadratic equation is generally written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). If you're scratching your head about how to find a zero, think of solving an equation to find where the graph of the quadratic touches the x-axis. This touchpoint, also called a 'root', can be found using this straightforward formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Discriminant
When dealing with the quadratic formula, the discriminant is a crucial part. It is the expression under the square root: \( b^2 - 4ac \). The discriminant helps determine the nature of the roots of the quadratic equation without solving the entire formula.
- If \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real solutions.
- If \( b^2 - 4ac = 0 \), it has exactly one real solution (a repeated root).
- If \( b^2 - 4ac < 0 \), there are no real solutions — the roots are complex numbers.
Multiplicity of a Root
Once you find the roots of a polynomial, it's important to determine the "multiplicity" of each root. Multiplicity refers to how many times a particular root appears in the factorization of the polynomial. This can affect the graph's shape at the root location.
- A root with multiplicity 1 means the graph simply crosses the x-axis at that point.
- If the root has an even multiplicity (like 2 or 4), the graph "bounces" off the x-axis at the root.
- An odd multiplicity greater than 1 (like 3, 5, etc.) indicates that the curve flattens at the x-axis and then crosses.