Problem 8 $1-12$ . A polynomial $P$ is... [FREE SOLUTION]

Chapter 4: Problem 8

$1-12$ . A polynomial $P$ is given. (a) Find all zeros of $P$ , real and complex. (b) Factor $P$ completely. $$ P(x)=x^{4}+6 x^{2}+9 $$

Short Answer

Expert verified

Zeros: $x = \pm i\sqrt{3}$; Factored: $(x^2 + 3)^2$.

Step by step solution

Convert Polynomial to Quadratic Form

Notice that the polynomial $ P(x) = x^4 + 6x^2 + 9 $ can be rewritten using a substitution. Let $ z = x^2 $. Then the polynomial becomes $ z^2 + 6z + 9 $.

Factor Quadratic Expression

The quadratic $ z^2 + 6z + 9 $ is in standard form. It factors as $ (z + 3)(z + 3) $ or $ (z + 3)^2 $ because $ 3 $ multiplied by $ 3 $ gives $ 9 $, and $ 3 + 3 $ gives $ 6 $.

Replace Substitution Variable

We substitute back $ z = x^2 $ into the factorization $ (z+3)^2 $. Hence, we get $ (x^2 + 3)^2 $. This implies $ (x^2 + 3)(x^2 + 3) $.

Solve for Zeros of Components

To find zeros, set $ x^2 + 3 = 0 $. Solving for $ x $, we have $ x^2 = -3 $. Since no real number squared equals $-3$, we take the square root to find $ x = \pm i\sqrt{3} $.

Write Final Answers

The polynomial $ x^4 + 6x^2 + 9 $ has complex zeros $ x = i\sqrt{3} $ and $ x = -i\sqrt{3} $. The factored form of the polynomial is $ (x^2 + 3)^2 $.

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Key Concepts

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

Factorization of Polynomials

Factorization of polynomials is a fundamental concept in algebra. It involves expressing a polynomial as a product of its simpler factors, which can often make solving equations more straightforward. For example, consider the polynomial given in the problem: \[ x^4 + 6x^2 + 9 \] Here, it looks complicated at first glance, but with the right approach, it can be factored smoothly.

Start by identifying any common patterns or potential substitutions that can simplify the process. In the given polynomial, notice it's close to a quadratic trinomial if we let $ z = x^2 $.
Substitute $ z = x^2 $ to convert the polynomial into a simpler quadratic form: $ z^2 + 6z + 9 $.
Factor this quadratic easily as $ (z + 3)^2 $, which shows that the original polynomial is a perfect square.

Breaking down polynomials through factorization is useful not only for finding zeros but also for simplifying algebraic expressions.

Complex Numbers

Complex numbers expand the idea of one-dimensional number lines to a two-dimensional complex plane. They are represented in the form $ a + bi $, where $ a $ and $ b $ are real numbers, and $ i $ is the imaginary unit satisfying $ i^2 = -1 $. In the context of this exercise, complex numbers arise when solving the equation $ x^2 = -3 $. Since no real square roots exist for negative numbers, we need the concept of complex numbers to proceed.

The imaginary root of a negative number $ c $ is expressed as $ i\sqrt{|c|} $.
In the exercise, solving $ x^2 + 3 = 0 $ leads to $ x^2 = -3 $, giving complex solutions $ x = \pm i\sqrt{3} $.

Complex numbers are essential in algebra because they ensure polynomials can always be factored into linear factors (over the complex numbers), following the Fundamental Theorem of Algebra.

Quadratic Substitution

Quadratic substitution is a powerful technique used to solve higher-degree polynomials by simplifying them into quadratic-like forms. This is especially useful when a polynomial can be rearranged to mimic a quadratic equation. With the polynomial $ x^4 + 6x^2 + 9 $, a substitution makes this task easier:

Set $ z = x^2 $, turning the polynomial into the simpler form $ z^2 + 6z + 9 $. This substitution leverages the quadratic form $ az^2 + bz + c $ for easier factoring.
Once in this recognized form, factor it as $ (z + 3)^2 $. Then, substitute back $ z = x^2 $ to return to the original variable and maintain the equation's integrity.
Apply the resulting expression to solve for zeros or simplify further computations: from $ (x^2 + 3)^2 $, solve $ x^2 + 3 = 0 $ for complex zeros.

Quadratic substitution is thus incredibly useful in manipulating polynomials, providing a pathway to simpler and more recognizable forms, leading to effective solutions.

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91影视

\(1-12\) . A polynomial \(P\) is given. (a) Find all zeros of \(P\) , real and complex. (b) Factor \(P\) completely. $$ P(x)=x^{4}+6 x^{2}+9 $$

Short Answer

Step by step solution

Convert Polynomial to Quadratic Form

Factor Quadratic Expression

Replace Substitution Variable

Solve for Zeros of Components

Write Final Answers

Key Concepts

Factorization of Polynomials

Complex Numbers

Quadratic Substitution

One App. One Place for Learning.

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