Problem 18 - Algebra and Trigonometry

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=4 x^{2}+9\)

Step-by-Step Solution

Answer

The polynomial doesn't factor into real numbers; its zeros are \(x = \frac{3i}{2}\) and \(x = -\frac{3i}{2}\), each of multiplicity 1.

Step 1: Recognize the polynomial type

The given polynomial is a quadratic polynomial: \(P(x) = 4x^2 + 9\). It is in the form \(ax^2 + bx + c\), where \(a = 4\), \(b = 0\), and \(c = 9\). Notably, there is no linear term.

Step 2: Check if the polynomial can be factored easily

Try to factor the quadratic. \(4x^2 + 9\) can be rewritten as a difference of squares, but since the terms are both positive, the usual difference of squares formula \(a^2 - b^2 = (a + b)(a - b)\) does not apply.

Step 3: Find the roots using complex numbers

Instead of factoring directly, calculate the zeros by solving the equation \(4x^2 + 9 = 0\) for \(x\). 1. Subtract 9 from both sides: \(4x^2 = -9\).2. Divide each term by 4: \(x^2 = -\frac{9}{4}\).3. Take the square root of both sides, remembering to include \(i\), the imaginary unit: \(x = \pm \frac{3i}{2}\).

Step 4: Identify the multiplicity of each zero

The zeros obtained are \(x = \frac{3i}{2}\) and \(x = -\frac{3i}{2}\). Each zero appears once in the solution, indicating that each has a multiplicity of 1.

Key Concepts

Quadratic Polynomial

A quadratic polynomial is a polynomial of degree 2. It has the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the case of our exercise, the polynomial is \( P(x) = 4x^2 + 9 \). Here, \( a = 4 \), \( b = 0 \), and \( c = 9 \).
Quadratic polynomials are unique because they form a parabola when graphed. Depending on the value of the coefficient \( a \), the parabola can open upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)). In this example, the absence of a linear term (\( b = 0 \)) indicates that the vertex of the parabola is on the y-axis.
When dealing with quadratic polynomials like \(4x^2 + 9\), it is important to check if they can be factored into simpler expressions. This requires looking for patterns, such as differences of squares, or employing methods like the quadratic formula for finding roots.

Imaginary Numbers

Imaginary numbers are a special class of numbers used to extend the real number system. This is done to provide solutions to equations that do not have real solutions. The imaginary unit is represented as \( i \), where \( i^2 = -1 \).
For the polynomial \( 4x^2 + 9 \), setting it to zero gives \( 4x^2 + 9 = 0 \), which simplifies to \( 4x^2 = -9 \). Dividing by 4, we get \( x^2 = -\frac{9}{4} \). Taking the square root of both sides introduces the imaginary number \( i \), resulting in \( x = \pm \frac{3i}{2} \).
Imaginary numbers help in addressing situations where we need square roots of negative numbers. In mathematics, this allows for a broader perspective on numbers, enabling calculations that are beyond the scope of just real numbers.

Complex Roots

Complex roots often occur in quadratic equations where the discriminant (the part of the quadratic formula under the square root, \( b^2 - 4ac \)) is negative. This indicates the need for imaginary numbers since you cannot take the square root of a negative number within the real number system.
In the case of the quadratic polynomial \( 4x^2 + 9 \), the discriminant is \( 0^2 - 4(4)(9) = -144 \). Because the discriminant is negative, the polynomial has complex roots. We found these roots to be \( x = \pm \frac{3i}{2} \), where \( i \) is the imaginary unit. Each complex root occurs only once, meaning each has a multiplicity of 1.
Completing tasks involving complex roots requires a good understanding of both real and imaginary components, as they often appear as conjugate pairs, such as \( \frac{3i}{2} \) and \( -\frac{3i}{2} \). This concept is pivotal not only in mathematics but also in engineering and physics, where complex numbers are used to solve real-world problems.