Problem 29 - Algebra and Trigonometry

Question

13- 30 . Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$ P(x)=x^{5}+6 x^{3}+9 x $$

Step-by-Step Solution

Answer

The polynomial has one real zero \(0\) with multiplicity 1. Complex zeros \(x = i\sqrt{3}\) and \(x = -i\sqrt{3}\) with multiplicity 2 each.

Step 1: Factor out the Greatest Common Factor (GCF)

First, we look for the greatest common factor in the polynomial \(P(x) = x^5 + 6x^3 + 9x\). We can see that each term contains a factor of \(x\). Thus, we factor \(x\) out: \[ P(x) = x(x^4 + 6x^2 + 9) \].

Step 2: Factor the Quadratic Trinomial

Next, we focus on factoring the quadratic trinomial \(x^4 + 6x^2 + 9\). Let \(y = x^2\), then the expression becomes \(y^2 + 6y + 9\), which is recognizable as a perfect square trinomial:\[ y^2 + 6y + 9 = (y + 3)^2 \]Replace \(y\) back with \(x^2\):\[ x^4 + 6x^2 + 9 = (x^2 + 3)^2 \].

Step 3: Write Complete Factorized Form

Combine the factors from the previous steps together to get the completely factorized form of the polynomial:\[ P(x) = x(x^2 + 3)^2 \].

Step 4: Find the Zeros of the Polynomial

To find the zeros of \(P(x) = x(x^2 + 3)^2\), set each factor equal to zero:1. \(x = 0\): This gives a zero of \(0\).2. \((x^2 + 3)^2 = 0\): Solve for \(x^2 + 3 = 0\), which gives \(x^2 = -3\). Since \(x^2 = -3\) has no real solutions (as the square roots are imaginary), there are no additional real zeros from this factor.

Step 5: Determine Multiplicities

Next, determine the multiplicity of each zero. - The zero \(x = 0\) comes from the factor \(x\) which appears once, so it has a multiplicity of 1.- The zeros \(x = \sqrt{-3}\) and \(x = -\sqrt{-3}\), are complex with multiplicity 2 each because \((x^2 + 3)^2 = 0\) implies each solution appears twice.

Key Concepts

Greatest Common Factor

When you first encounter the task of factoring a polynomial, finding the greatest common factor (GCF) is a vital initial step. The GCF is the highest number or expression that evenly divides all the terms in the polynomial. This is like finding the largest common "ingredient" shared by each term. For \(P(x) = x^5 + 6x^3 + 9x\), each term contains the factor \(x\). Hence, \(x\) is the GCF.By factoring \(x\) out, you streamline the polynomial into: \[ P(x) = x(x^4 + 6x^2 + 9) \]This simplification is crucial to making further factorization feasible and to study the polynomial's roots. Once the GCF is out, you are left with a polynomial that is easier to handle in subsequent steps.

Zeros of Polynomial

Finding the zeros of a polynomial is all about discovering where the polynomial equals zero. These zeros are the solutions to the equation formed by setting the polynomial equal to zero, \(P(x) = 0\).With the polynomial expressed as \(P(x) = x(x^2 + 3)^2\), the zeros emerge from solving each distinct factor:

• The factor \(x\) gives a zero at \(x = 0\).
• The factor \((x^2 + 3)^2\) must solve for the equation \(x^2 + 3 = 0\).

However, \(x^2 = -3\) leads to imaginary solutions because negative numbers lack real square roots, hence are complex numbers. Therefore, the zeros \(\pm\sqrt{-3}\) are considered complex and won't appear on the real number line when graphed.

Multiplicity of Zeros

The multiplicity of a zero refers to how many times a particular zero appears in the factored form of the polynomial. It is like asking, "How many times do you visit the same zero?" Understanding multiplicity is crucial as it affects the graph's behavior at the zero, determining whether it just touches and turns around or passes through the x-axis.For \(P(x) = x(x^2 + 3)^2\), we broke it down into simpler factors:

• \(x = 0\) with multiplicity 1, as the factor \(x\) appears once.
• The zeros \(\pm\sqrt{-3}\), stemming from \((x^2 + 3)^2 = 0\), have multiplicity 2 because each zero is part of a squared term.

Thus, the polynomial will intersect the x-axis at \(x=0\) and have a different graphical behavior for its complex roots.

Quadratic Trinomial

A quadratic trinomial is a polynomial expression with the form \(ax^2 + bx + c\), which consists of three terms and a degree of 2. It plays a significant role in factorization problems because many complex polynomials can be reduced to factoring these simpler expressions.In the original polynomial \(x^4 + 6x^2 + 9\), substituting \(y = x^2\) transforms it into the trinomial \(y^2 + 6y + 9\). Recognizing that \(y^2 + 6y + 9\) forms a perfect square trinomial reveals that:\[y^2 + 6y + 9 = (y + 3)^2 \]Switching \(y\) back to \(x^2\) gives us:\[ x^4 + 6x^2 + 9 = (x^2 + 3)^2 \]This method simplifies more daunting expressions and connects them to recognizable forms that make factorization approachable. It's like breaking down a hard task into smaller, friendlier pieces.