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Chapter 3: Problem 23

Find the zeros of \(f(x),\) and state the multiplicity of each zero. $$f(x)=\left(x^{2}+x-12\right)^{3}\left(x^{2}-9\right)^{2}$$

Short Answer

Expert verified
The zeros are \(x = 3\) (multiplicity 5), \(x = -4\) (multiplicity 3), and \(x = -3\) (multiplicity 2).

Step by step solution

01

Identify Each Factor

First, check the expression to see how it can be factored further if necessary.\[ f(x) = ig(x^2 + x - 12\big)^3 ig(x^2 - 9\big)^2 \] To find the zeros, individually factor each polynomial inside the expression.
02

Factor the Quadratics

Let's factor each quadratic expression.Start with \(x^2 + x - 12\):- Find two numbers that multiply to \(-12\) and add up to \(1\). These numbers are \(4\) and \(-3\).- So, \(x^2 + x - 12 = (x - 3)(x + 4)\).Now, factor \(x^2 - 9\) using the difference of squares:- \(x^2 - 9 = (x - 3)(x + 3)\).
03

Rewrite the Function with Factors

Substitute the factors back into the main function:\[ f(x) = ig((x - 3)(x + 4)\big)^3 ig((x - 3)(x + 3)\big)^2 \] Expand the representation:- \((x - 3)\) appears in both expressions with multiplicities 3 and 2, which makes it have a total multiplicity of 5.- \((x + 4)\) appears with a multiplicity of 3.- \((x + 3)\) appears with a multiplicity of 2.
04

Identify the Zeros and Their Multiplicities

Each factor of the polynomial gives potential zeros. Set each factor equal to zero:- \(x - 3 = 0\) gives \(x = 3\) with multiplicity 5.- \(x + 4 = 0\) gives \(x = -4\) with multiplicity 3.- \(x + 3 = 0\) gives \(x = -3\) with multiplicity 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multiplicity of Zeros
The concept of multiplicity of zeros refers to how many times a particular zero occurs in a polynomial function. Essentially, it's about counting repeated roots.
If a polynomial can be expressed in the form
  • If \( (x-a)^n \) is a factor, then \( a \) is a zero of multiplicity \( n \).
In our exercise, the polynomial is \( f(x) = (x^2 + x - 12)^3 (x^2 - 9)^2 \).
After factoring:
  • The zeros are: \( x = 3 \), \( x = -4 \), and \( x = -3 \).
  • The factor \( (x-3) \) appears twice across two different factors with powers \( 3 \) and \( 2 \), so its total multiplicity is \( 5 \).
  • Similarly,\( (x + 4) \)appears once with a multiplicity of \( 3 \).
  • \( (x + 3) \)appears once with a multiplicity of \( 2 \).
Understanding multiplicities helps indicate the nature of each zero on the graph. When a zero has a higher multiplicity, the graph will "bounce" differently at that point.
Factoring Quadratics
Factoring quadratics is the process of breaking down a quadratic polynomial into simpler multiplication components.
The general form for a quadratic is \(ax^2 + bx + c\), where you look for two numbers that multiply to \(c\) and add to \(b\).
To factor:
  • Find two numbers that meet these criteria.
  • Rewrite \(b\) as a sum or difference of these two numbers.
  • Factor by grouping, if necessary.
For \(x^2 + x - 12\):
  • Numbers \(4\) and \(-3\) meet the requirements (multiply to \(-12\) and add to \(1\)).
  • The factorization is \((x - 3)(x + 4)\).
Factoring quadratics is essential for identifying polynomial zeros.
Difference of Squares
The concept of difference of squares involves expressions of the form \(a^2 - b^2\). This makes it straightforward to factor.
These expressions can always be rewritten as:
  • \((a - b)(a + b)\)
For example, in the exercise, observe \(x^2 - 9\) which is a difference of squares because \(9\) is \(3^2\).
So, it can be factored as:
  • \((x - 3)(x + 3)\)
Using the difference of squares makes it possible to break down difficult expressions into simpler, solvable components.
It's a vital technique when solving polynomial equations by factoring, as it quickly reveals zeros.

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