Problem 16 - Precalculus : Building Concepts and Connections

Question

Show that the given value of \(x\) is a zero of the polynomial. Use the zero to completely factor the polynomial. $$p(x)=2 x^{3}-x^{2}+6 x-3 ; x=\frac{1}{2}$$

Step-by-Step Solution

Answer

Therefore, it is confirmed that \(\frac{1}{2}\) is indeed a root of the polynomial \(p(x)=2 x^{3}-x^{2}+6 x-3\). The completely factored form of the polynomial is \(p(x)=(2x - 1)\left(x^{2} +2x +3\right)\).

Step 1: Confirm the given value is a zero

Plug \(\frac{1}{2}\) into the polynomial \(p(x)=2 x^{3}-x^{2}+6 x-3\) and check if the result equals zero. So, \(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^{3}-\left(\frac{1}{2}\right)^{2}+6\left(\frac{1}{2}\right)-3 = 0 \). Thus, it confirms that the given value \(\frac{1}{2}\) is indeed a root of the polynomial since \(p\left(\frac{1}{2}\right) = 0\).

Step 2: Factor the Polynomial

To factor the polynomial, write the polynomial in a factored form. If \(\frac{1}{2}\) is a zero of the polynomial, then \((2x - 1)\) is a factor of the polynomial. Divide the given polynomial \(p(x)\) by the factor \((2x - 1)\) using polynomial division or synthetic division. It will result in \(p(x)=(2x - 1)\left( x^{2} +2x +3\right)\). This is a completely factored form of the given polynomial.

Key Concepts

Zeros of Polynomials

Understanding the concept of the zeros of polynomials is an important skill in algebra. Often referred to as "roots," these are the values of \(x\) for which the polynomial equals zero. To determine if a given value is a zero, substitute it into the polynomial and simplify. For example, with the polynomial \(p(x) = 2x^3 - x^2 + 6x - 3\) and the test value \(x = \frac{1}{2}\), substitute \(\frac{1}{2}\) in place of \(x\) to see if the polynomial equals zero.

• If \(p\left(\frac{1}{2}\right) = 0\), then \(\frac{1}{2}\) is a zero.
• In this example, substituting \(\frac{1}{2}\) confirms it as a zero.

Identifying zeros is essential because they reveal critical points where the graph of the polynomial touches or crosses the x-axis. It's the first step in factoring and solving polynomial equations.

Synthetic Division

Synthetic division is a streamlined method to divide polynomials, especially useful when dealing with linear divisors. It simplifies the traditional process of polynomial division, making it faster and easier. Here's how it works:

• Write the coefficients of the polynomial you want to divide.
• Use the zero of the divisor's linear factor (like \(x - r\)) as the number outside the division setup.
• The result offers coefficients of the quotient polynomial.

In our exercise, using \(x = \frac{1}{2}\) as a zero means \((2x - 1)\) is a factor. Use synthetic division to divide \(2x^3 - x^2 + 6x - 3\) by \(2x - 1\), leading to a quotient \(x^2 + 2x + 3\). This method highlights the elegance and efficiency of synthetic division over traditional polynomial division. It's a powerful tool for quickly finding factors of a polynomial.

Polynomial Division

Polynomial division is the method of dividing two polynomials, producing a quotient and possibly a remainder. Think of it like long division with numbers, but instead with polynomials. You do this by aligning terms by their degree and performing operations to eliminate terms in sequence.For the polynomial \(p(x) = 2x^3 - x^2 + 6x - 3\), using polynomial division to divide by \(2x - 1\) involves several steps:- Divide the highest degree term of the dividend by the highest degree term of the divisor.- Multiply the entire divisor by this result and subtract from the original polynomial.- Repeat with the new polynomial until no terms remain or a lower degree term than the divisor is left.The result of dividing \(2x^3 - x^2 + 6x - 3\) by \(2x - 1\) is \((x^2 + 2x + 3)\) with no remainder, confirming that \((2x - 1)\) is indeed a factor of the polynomial.