Question

Use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor \(f\) over the real numbers. $$ f(x)=2 x^{3}-x^{2}+2 x-1 $$

Short answer

The real zero is \(x = \frac{1}{2}\), and the factorization is \(2 (x - \frac{1}{2}) (x^2 + 1)\).

Step-by-step solution

- Identify the Coefficients

First, identify the coefficients of the polynomial. For the polynomial \(f(x) = 2x^3 - x^2 + 2x - 1\), the coefficients are: \(a_3 = 2\), \(a_2 = -1\), \(a_1 = 2\), and \(a_0 = -1\).

- Find the Factors of the Constant Term and Leading Coefficient

From the Rational Zeros Theorem, we need to find the factors of both the constant term \(a_0 = -1\) and the leading coefficient \(a_3 = 2\). Factors of \(a_0\) are \(\pm 1\). Factors of \(a_3\) are \(\pm 1, \pm 2\).

- List the Possible Rational Zeros

The possible rational zeros are the ratios of the factors of the constant term to the factors of the leading coefficient. Thus, the possible rational zeros are: \( \pm 1, \pm \frac{1}{2} \).

- Test the Possible Zeros

Substitute each possible zero into the polynomial \(f(x)\) to see if it yields zero. Start with \(x = 1\): \ \[f(1) = 2(1)^3 - 1(1)^2 + 2(1) - 1 = 2 - 1 + 2 - 1 = 2 \ \]Since \(f(1)\) is not zero, \(x = 1\) is not a zero.

- Continue Testing Possible Zeros

Next, test \(x = -1\): \ \[f(-1) = 2(-1)^3 - (-1)^2 + 2(-1) - 1 = -2 - 1 - 2 - 1 = -6 \ \]Since \(f(-1)\) is not zero, \(x = -1\) is not a zero. Continue with \(x = \frac{1}{2}\): \ \[f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 1 = 2\left(\frac{1}{8}\right) - \left(\frac{1}{4}\right) + 1 - 1 = \frac{1}{4} - \frac{1}{4} = 0 \ \]Since \(f\left(\frac{1}{2}\right) = 0\), \(x = \frac{1}{2}\) is a zero.

- Factor the Polynomial Using the Found Zero

Since \(\frac{1}{2}\) is a zero, divide \(f(x)\) by \(x - \frac{1}{2}\). Perform synthetic division: \ \[ \begin{array}{r|rrr} \frac{1}{2} & 2 & -1 & 2 & -1 \ & & 1 & 0.5 & 1.25 \ \hline \ & 2 & 0 & 2 & 0 \end{array}\ \]After division, the quotient is \(2x^2 + 2 = 2(x^2 + 1)\). The factorization is: \[2 \left(x - \frac{1}{2}\right) \left(x^2 + 1\right) \].

- Final Factorization

The final factorization of the polynomial \(f(x) = 2x^3 - x^2 + 2x - 1\) over the real numbers is: \[ \text{Factors: } 2 \left(x - \frac{1}{2}\right) \left(x^2 + 1\right). \]

Key concepts

Polynomial Zeros

Understanding polynomial zeros is crucial for solving polynomial equations effectively. A zero of a polynomial is a value of \(x\) that makes the polynomial equal to zero. In other words, if \(f(x)\) is a polynomial, then \(x = r\) is a zero if \(f(r) = 0\). Here, we utilized the Rational Zeros Theorem to find possible real zeros of the polynomial \(f(x) = 2x^3 - x^2 + 2x - 1\). The theorem states that any rational zero of the polynomial is a fraction \(p/q\), where \(p\) is a divisor of the constant term and \(q\) is a divisor of the leading coefficient.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear binomial of the form \((x − c)\). It is especially useful for checking whether a candidate obtained from the Rational Zeros Theorem is actually a zero, and for polynomial factorization. After identifying \(x = \frac{1}{2}\) as a zero, we performed synthetic division to divide the polynomial \(f(x)\) by \((x - \frac{1}{2})\). The division process verified that \(\frac{1}{2}\) is indeed a zero and helped in reducing the polynomial to a simpler form: \(2x^2 + 2\).

Factorization

Factorization involves expressing a polynomial as the product of its factors. After confirming \(\frac{1}{2}\) as a zero and performing synthetic division, we factorized \(f(x) = 2(x - \frac{1}{2})(x^2 + 1)\). This final factorization expresses the original cubic polynomial as the product of a linear factor and a quadratic factor, making it easier to work with. The ability to factorize polynomials simplifies solving equations, integrating functions, and understanding polynomial graphs. By breaking down into simpler factors, we gain valuable insights into the polynomial's behavior and roots.