Problem 31
Question
For the following exercises, find the zeros and give the multiplicity of each. $$ f(x)=x^{2}(2 x+3)^{5}(x-4)^{2} $$
Step-by-Step Solution
Verified Answer
Zeros are at \( x = 0 \) (multiplicity 2), \( x = -\frac{3}{2} \) (multiplicity 5), \( x = 4 \) (multiplicity 2).
1Step 1: Identify the Factors
The function given is \( f(x)=x^{2}(2x+3)^{5}(x-4)^{2} \). The zeros of the function correspond to the values of \( x \) that make any factor of the product zero. The factors are \( x^{2} \), \( (2x+3)^{5} \), and \( (x-4)^{2} \). We will find the zeros by setting each factor to zero independently.
2Step 2: Find the Zero of \( x^2 \)
The factor \( x^2 = 0 \) when \( x = 0 \). This gives us a zero at \( x = 0 \) with multiplicity 2, since the exponent of \( x \) in this factor is 2.
3Step 3: Find the Zero of \( (2x+3)^5 \)
Set the factor \( (2x+3)^5 = 0 \) to find the zero. Solve for \( x \):\[2x + 3 = 0 \2x = -3 \x = -\frac{3}{2}\]This gives us a zero at \( x = -\frac{3}{2} \) with multiplicity 5, since the exponent of \( (2x+3) \) is 5.
4Step 4: Find the Zero of \( (x-4)^2 \)
Set the factor \( (x-4)^2 = 0 \) to find the zero. Solve for \( x \):\[x - 4 = 0 \x = 4\]This gives us a zero at \( x = 4 \) with multiplicity 2, since the exponent of \( (x-4) \) is 2.
5Step 5: Summarize Zeros and Their Multiplicities
The zeros of the function are:- \( x = 0 \) with multiplicity 2- \( x = -\frac{3}{2} \) with multiplicity 5- \( x = 4 \) with multiplicity 2
Key Concepts
Multiplicity of ZerosFactoring PolynomialsFinding Zeros of Functions
Multiplicity of Zeros
In polynomial functions, the **multiplicity of a zero** refers to the number of times a particular zero appears as a root of the polynomial. This determines the behavior of the graph at the zero's location. The multiplicity is directly related to the exponent of the factor that provides the zero. For instance, in the polynomial given by \[ f(x) = x^{2}(2x+3)^{5}(x-4)^{2} \]we have different zeros with distinct multiplicities:
- For \( x^2 \), the zero is at \( x = 0 \) with multiplicity 2, because the exponent of \( x \) in the factor is 2.
- The factor \( (2x+3)^5 \) results in a zero at \( x = -\frac{3}{2} \) with multiplicity 5, since the exponent is 5.
- Lastly, \( (x-4)^2 \) gives a zero at \( x = 4 \) with multiplicity 2, due to its exponent being 2.
Factoring Polynomials
**Factoring polynomials** involves breaking them down into their simplest parts, or factors, which are polynomials of lower degrees. This process is crucial for finding the zeros of the polynomial function because the factors set to zero give the roots. For the polynomial function \[ f(x) = x^{2}(2x+3)^{5}(x-4)^{2} \]we can observe that it's already factored into three distinct parts:
- \( x^2 \)
- \( (2x+3)^5 \)
- \( (x-4)^2 \)
Finding Zeros of Functions
The process of **finding zeros of functions** involves determining the values of \( x \) that make a polynomial equal to zero. These zeros are also known as roots or solutions of the polynomial function. In the case of \[ f(x) = x^{2}(2x+3)^{5}(x-4)^{2} \]we find its zeros by setting each factor to zero:
- For \( x^2 = 0 \), \( x = 0 \) is a zero.
- From \( (2x+3)^5 = 0 \), solving \( 2x + 3 = 0 \) gives \( x = -\frac{3}{2} \).
- For \( (x-4)^2 = 0 \), \( x = 4 \) is the zero.
Other exercises in this chapter
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