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Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{3}+x^{2}+9 x+9$$

Short Answer

Expert verified
The zeros are \(x = -1, 3i, -3i\) with multiplicities of 1.

Step by step solution

01

Factor by Grouping

First, let's look for a way to group the terms in the polynomial such that we can factor them more easily. Consider the polynomial \( P(x) = x^3 + x^2 + 9x + 9 \). We can group the terms as follows: \((x^3 + x^2) + (9x + 9)\).
02

Factor Out Common Factors

Within each group from Step 1, we factor out the greatest common factor. From the first group \((x^3 + x^2)\), factor out \(x^2\): \(x^2(x + 1)\). From the second group \((9x + 9)\), factor out \(9\): \(9(x + 1)\). This gives us: \(x^2(x + 1) + 9(x + 1)\).
03

Factor Out Common Binomial

Notice that \((x + 1)\) is a common factor in both terms from Step 2. We factor out \((x + 1)\): \((x + 1)(x^2 + 9)\).
04

Find Zeros of the Polynomial

To find the zeros of the polynomial, set each factor equal to zero: \((x + 1) = 0\) and \(x^2 + 9 = 0\). Solving \(x + 1 = 0\) gives \(x = -1\). Solving \(x^2 + 9 = 0\) gives \(x^2 = -9\) or \(x = \pm 3i\).
05

Determine the Multiplicity of Each Zero

The zero \(x = -1\) comes from the factor \((x + 1)\), which is linear, so it has a multiplicity of 1. The zeros \(x = 3i\) and \(x = -3i\) come from the factor \(x^2 + 9\), which is quadratic, so each of these zeros also has a multiplicity of 1.

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Key Concepts

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

Finding Zeros of Polynomials
Zeros of a polynomial, sometimes called roots or solutions, are the values of the variable that make the polynomial equal to zero. To find the zeros of a polynomial, you can factor the polynomial and set each factor to zero. This is a crucial step because it simplifies the polynomial into a product of simpler expressions, making it easier to solve.

For example, consider the polynomial \(P(x) = x^{3} + x^{2} + 9x + 9\). By grouping terms and factoring, as shown in the original solution, the polynomial can be rewritten as \((x + 1)(x^2 + 9)\). By setting each factor equal to zero, \(x + 1 = 0\) gives the zero \(x = -1\) and \(x^2 + 9 = 0\) gives zeros \(x = 3i\) and \(x = -3i\).

Once factored, solving for zeros involves straightforward algebraic manipulation. Remember, zeros might be real or complex, depending on the polynomial's equation.
Multiplicity of Zeros
Multiplicity refers to how many times a particular zero occurs in a polynomial. If a zero is repeated, it has a multiplicity greater than one. When a factor is raised to a power, the number in the exponent indicates the multiplicity of the zero.

In the polynomial \(P(x) = (x + 1)(x^2 + 9)\), the factor \((x + 1)\) implies a zero at \(x = -1\) with a multiplicity of 1, because it appears only once. Similarly, \(x^2 + 9\) results in complex zeros \(x = 3i\) and \(x = -3i\), each with a multiplicity of 1. This means both complex zeros occur only once in the solution.

Understanding multiplicities is important because they affect the shape and behavior of the polynomial's graph. A zero with even multiplicity will touch the x-axis and turn back, while a zero with odd multiplicity will cross through the axis.
Complex Numbers
Complex numbers include real numbers and imaginary numbers, usually expressed in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. The imaginary unit \(i\) is defined as \(\sqrt{-1}\). Therefore, any expression involving the square root of a negative number can be rewritten in terms of \(i\).

In our polynomial example, solving \(x^2 + 9 = 0\), leads to a step where \(x^2 = -9\). Taking the square root of both sides, we find \(x = \pm 3i\), which are complex solutions. This concept arises often in polynomial equations that do not intersect the real number line.

Complex numbers are foundational in advanced mathematics and engineering. They allow the extension of solutions to equations where real numbers are insufficient.

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