Problem 32 - Precalculus : Building Concepts and Connections

Question

Determine together \(q(x)\) is a factor of \(p(x)\) Here, \(p(x)\) is the first polynomial and \(q(x)\) is the second polynomial. justify your answer. $$x^{3}-5 x^{2}+8 x-4 ; x-2$$

Step-by-Step Solution

Answer

After performing the substitution and calculations in the polynomial, if the result equals to 0, then \(x - 2\) is a factor of \(x^{3} - 5x^{2} + 8x - 4\). If the result is not 0, then \(x - 2\) is not a factor of the polynomial.

Step 1: Apply the Factor Theorem

To verify if \(x - 2\) is a factor of the polynomial \(x^{3} - 5x^{2} + 8x - 4\), we will use the Factor Theorem. We substitute \(x = 2\) into the polynomial \(p(x)\).

Step 2: Substitution and Calculation

Perform the substitution where \(x = 2\) in the polynomial \(x^{3} - 5x^{2} + 8x - 4\). So, \(p(2) = (2)^{3} -5 *(2)^{2} + 8 * 2 - 4\). Perform the calculations to determine the result.

Step 3: Analyse the Results

If the result of the calculations in Step 2 equals to 0, then according to the Factor Theorem, \(x - 2\) is a factor of the polynomial \(x^{3} - 5x^{2} + 8x - 4\). If the result is not equal to 0, then \(x - 2\) is not a factor of the polynomial.

Key Concepts

Polynomial Division

Polynomial division is essential to understanding how to simplify or break down complex algebraic expressions into more manageable parts. It works much like the division of numbers, where you divide a polynomial by another polynomial. This process can help in identifying factors of a polynomial, which means finding the simpler polynomials that multiply together to give the original polynomial.
If you imagine dividing the polynomial \(p(x)\) by \(q(x)\), the goal is to express \(p(x)\) as the product \(q(x) \cdot d(x) + r(x)\), where \(d(x)\) is the quotient, and \(r(x)\) is the remainder polynomial.

• Long Division: A method useful for dividing higher-degree polynomials by lower-degree equations, similar to numerical long division.
• Synthetic Division: A shortcut method that is quicker and easier, but only applicable when dividing by linear expressions, like \(x - c\).

Using polynomial division can simplify understanding whether a polynomial like \(x - 2\) is a factor of the given polynomial \(x^3 - 5x^2 + 8x - 4\). If the division results in a zero remainder, it suggests that \(x - 2\) is indeed a factor.

Remainder Theorem

The Remainder Theorem is a straightforward yet powerful principle that helps to determine the remainder of the division of a polynomial \(p(x)\) by a linear factor \(x - c\). According to this theorem, the remainder of this division is simply \(p(c)\).
For example, in the exercise with \(p(x) = x^3 - 5x^2 + 8x - 4\) and \(q(x) = x - 2\), we substitute \(x = 2\) into \(p(x)\). This is because \(x - 2\) implies \(c = 2\) in the context of the Remainder Theorem. Substituting, if \(p(2) = 0\), the theorem indicates there is no remainder, which also substantiates \(x - 2\) is a factor.

• If \(p(c) = 0\), the remainder is zero, confirming that \(x - c\) is a factor of \(p(x)\).
• Otherwise, if \(p(c)\) is not zero, \(x - c\) is not a factor, and \(p(c)\) is the remainder.

This theorem streamlines the process of testing multiple potential factors without performing complete polynomial division, especially when dealing with linear divisors.

Roots of Polynomials

The roots of a polynomial are the solutions to the equation \(p(x) = 0\). Finding these roots is crucial as they help to understand the behavior of polynomials graphically and numerically.
For a polynomial like \(x^3 - 5x^2 + 8x - 4\), a root corresponds to a value \(x = r\), such that substituting \(r\) into the polynomial makes it zero. These roots can be real numbers or complex numbers depending on the polynomial's degree and nature.

• Real Roots: These are values that satisfy the polynomial equation and can be plotted on a real number line.
• Complex Roots: Occur in conjugate pairs if the polynomial has real coefficients and the root itself is not real.

To test if \(x - 2\) contributes a root, Remainder Theorem comes in handy — check if substituting 2 yields zero. Therefore, when \(p(2) = 0\), it confirms \(x = 2\) is a root, and thus, \(x - 2\) is a factor.