Problem 34 - Precalculus : Building Concepts and Connections

Question

One zero of each polynomial is given. Use it to express the polynomial as a product of linear factors over the complex numbers. You may have already factored some of these polynomials into linear and irreducible quadratic factors in the previous group of exercises. $$2 x^{3}-9 x^{2}+7 x+6 ; \text { zero: } x=2$$

Step-by-Step Solution

Answer

The factorized form of the polynomial \(2x^3 - 9x^2 + 7x + 6\) over the complex numbers is \(2(x-2)(2x+1)(x - 3)\).

Step 1: Validate the given zero

Substitute \(x=2\) into the polynomial \(2x^3 - 9x^2 + 7x + 6\). If it equals to zero, then \(x=2\) is indeed a zero. Calculation: \(2*2^3 - 9*2^2 + 7*2 + 6 = 0\). So, \(x=2\) is a zero of the given polynomial.

Step 2: Use Polynomial Division

Divide the given polynomial by \(x-2\) (because x=2 is a zero) to find the other factor. This can be done through the process of polynomial division. Here, \(2x^3 - 9x^2 + 7x + 6\) is divided by \(x-2\).

Step 3: Determine the Quotient

After polynomial division, the quotient is found to be \(2x^2 - 5x - 3\). This implies that the given polynomial can now be expressed as \(2(x-2)\) multiplied by \(2x^2 - 5x - 3\).

Step 4: Factorizing the Quadratic

Now factorize the quadratic \(2x^2 - 5x - 3\). This can be written as \(2x^2 - 6x + x - 3\), which factors to \(2x(x - 3)+(x - 3)\), and then simplifying gives \((2x+1)(x - 3)\). Thus factorization of the given polynomial is \(2(x-2)(2x+1)(x - 3)\).

Key Concepts

Complex Numbers

Complex numbers are fundamental in understanding polynomial expressions, especially when dealing with linear factors and factorization. A complex number is a number that can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\). Complex numbers are crucial because they allow us to factor polynomials over a complete field, meaning every non-constant polynomial has a root.

• Complex numbers can help express roots that are not real.
• The fundamental theorem of algebra states that any polynomial of degree \(n\) has exactly \(n\) roots in the complex number system, counting multiplicities.
• The use of complex numbers ensures that all polynomials can be fully decomposed into linear factors in the complex plane.

Understanding complex numbers is essential for working with polynomials, as they can uncover unseen symmetries and alternative factorizations, expanding the tools available for finding roots of polynomials.

Linear Factors

Linear factors are the simplest building blocks of polynomials. A linear factor is a polynomial of degree 1, generally expressed in the form \(ax + b\). Every polynomial can be expressed as a product of linear factors, provided we allow the coefficients to be complex numbers.

• Linear factors are crucial because they represent each root of the polynomial.
• If \(c\) is a root of a polynomial, then \(x-c\) is a linear factor of the polynomial.
• Finding the linear factors involves identifying the roots of the polynomial, using methods such as synthetic division, factoring, or employing the quadratic formula for higher-degree polynomials.

In the context of the given exercise, once a polynomial is expressed in terms of its linear factors, it takes the form \((x-r_1)(x-r_2)...(x-r_n)\), where \(r_1, r_2, ..., r_n\) are the roots of the polynomial.

Quadratic Factorization

Quadratic factorization deals with breaking down a quadratic expression, a polynomial of degree 2, into a product of linear factors or irreducible quadratic factors. A quadratic polynomial is typically written as \(ax^2 + bx + c\). To factorize such an expression, one needs to find two numbers that multiply to \(ac\) and add up to \(b\).

• Quadratic factorization can simplify solving polynomial equations by reducing them to simpler, more manageable forms.
• Depending on the discriminant \(b^2 - 4ac\), a quadratic can have two real roots, one real root, or no real roots:
• If the discriminant is positive, the quadratic can be factorized into two distinct real linear factors.
• If the discriminant is zero, the quadratic factorizes into a repeated linear factor.
• If the discriminant is negative, the quadratic is irreducible over the real numbers and has complex roots.

In the exercise, factorizing the quadratic part \(2x^2 - 5x - 3\) involved rewriting it as \(2x(x-3) + (1)(x-3)\), revealing its factors as \((2x+1)(x-3)\), demonstrating a complete factorization of the given polynomial.