Problem 20 - Precalculus with Calculus Previews

Question

Find all rational zeros of the given polynomial function \(f\). $$ f(x)=x^{4}+\frac{5}{2} x^{3}+\frac{3}{2} x^{2}-\frac{1}{2} x-\frac{1}{2} $$

Step-by-Step Solution

Answer

The only rational zero is \(x = -1\).

Step 1: Understand the problem

We need to find all rational zeros of the polynomial function. This means finding all rational numbers \(x\) such that \(f(x) = 0\). Use the Rational Root Theorem to guide us.

Step 2: Use the Rational Root Theorem

According to the Rational Root Theorem, any rational root, \( \frac{p}{q} \), of the polynomial must satisfy \(p\) as a factor of the constant term (-1/2) and \(q\) as a factor of the leading coefficient (1). Thus, possible rational roots are \( \pm 1, \pm \frac{1}{2} \).

Step 3: Test possible rational roots

Substitute the potential rational roots from Step 2 into the polynomial to see if any of them yields zero: For \( x = 1 \):$$f(1) = 1^4 + \frac{5}{2}(1^3) + \frac{3}{2}(1^2) - \frac{1}{2}(1) - \frac{1}{2} = 1 + \frac{5}{2} + \frac{3}{2} - \frac{1}{2} - \frac{1}{2} eq 0$$ For \( x = -1 \):$$f(-1) = (-1)^4 + \frac{5}{2}(-1)^3 + \frac{3}{2}(-1)^2 - \frac{1}{2}(-1) - \frac{1}{2} = 1 - \frac{5}{2} + \frac{3}{2} + \frac{1}{2} - \frac{1}{2} = 0$$\(x = -1\) is a rational root.

Step 4: Use synthetic division with found root

Perform synthetic division using \(x = -1\) to divide the polynomial, which helps reduce the polynomial degree. The coefficients are [1, 5/2, 3/2, -1/2, -1/2]. Performing the synthetic division gives us a quotient of a lower degree polynomial.

Step 5: Factor reduced polynomial

After synthetic division, the quotient polynomial obtained is \(x^3 + 2x^2 + x - \frac{1}{2}\). Repeat Steps 2-3 to find any rational roots of this new polynomial.

Step 6: Finding rational roots of reduced polynomial

Now, follow similar steps to find the rational roots of \(x^3 + 2x^2 + x - \frac{1}{2}\):For \(x = \frac{1}{2}\):$$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^2 + \frac{1}{2} \left(\frac{1}{2}\right) - \frac{1}{2} = \frac{1}{8} + \frac{1}{2} + \frac{1}{4} - \frac{1}{2} eq 0$$For \(x = -\frac{1}{2}\):$$f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^3 + 2\left(-\frac{1}{2}\right)^2 + \frac{1}{2} \left(-\frac{1}{2}\right) - \frac{1}{2} = -\frac{1}{8} + \frac{1}{2} - \frac{1}{4} - \frac{1}{2} eq 0$$None of these are roots, so check other terms or factor further.

Step 7: Finalize rational roots

After trying the possible rational roots, we saw that only \(x = -1\) is a rational root confirmed by actual division, as others did not result in zero. The reduced polynomial doesn't have any additional rational roots that satisfy.

Key Concepts

Synthetic Division

Synthetic division is a highly efficient and streamlined method to divide polynomials, particularly when one already knows a root of the polynomial. Imagine you have already found that for the polynomial \( f(x) = x^4 + \frac{5}{2}x^3 + \frac{3}{2}x^2 - \frac{1}{2}x - \frac{1}{2} \), \( x = -1 \) is a root. This means you can quickly divide the polynomial by \( x + 1 \) (since \( x + 1 \) equates to a root of \( x = -1 \)) using synthetic division.

Here's how it works:

• Write down the coefficients of the polynomial: [1, \(\frac{5}{2}\), \(\frac{3}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\)].
• Use the root \( x = -1 \) in the synthetic division setup. Bring down the first coefficient.
• Multiply this first coefficient by the root (\(-1\)) and add this product to the next coefficient. Repeat for the entire row of coefficients.
• The final result gives you a new polynomial of a lower degree, and the last number will indicate whether the remainder is zero, confirming the root.

Breaking down complex polynomial equations becomes much simpler with synthetic division, particularly after establishing one or more rational roots.

Polynomial Roots

Fascination with polynomial roots lies in finding values of \( x \) where the polynomial equals zero. For our example polynomial \( f(x) = x^4 + \frac{5}{2}x^3 + \frac{3}{2}x^2 - \frac{1}{2}x - \frac{1}{2} \), this involves finding all possible rational roots. We apply the Rational Root Theorem, which proposes a set of potential rational roots as fractions \( \frac{p}{q} \) where:

• \( p \) is a factor of the constant term (here \(-\frac{1}{2}\)).
• \( q \) is a factor of the leading coefficient (here 1).

Therefore, our possible rational roots include \( \pm 1 \) and \( \pm \frac{1}{2} \).

Testing these values in the polynomial, we determine if any of them satisfies \( f(x) = 0 \). As seen, \( x = -1 \) proves to be one of these roots, driving us to use synthetic division to further simplify our polynomial. Rational roots provide the gateway to factoring and solving polynomials, confirming solutions and simplifying expressions.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial of higher degree into simpler, easy-to-solve parts. With tools like the Rational Root Theorem, once a root such as \( x = -1 \) is determined, synthetic division allows for the polynomial to reduce in degree.

In our example, strategic simplification led to a new polynomial: \( x^3 + 2x^2 + x - \frac{1}{2} \). This reduced polynomial can then be analyzed further to find additional roots or factored into smaller components. Although further rational roots were not found in this case, the process of factoring continues until expressions cannot be simplified anymore.

Factoring not only helps find roots but also reveals insights into the behavior of the equation. In many mathematical problems, factoring is essential for simplifying polynomial expressions into manageable pieces, aiding in the understanding and solving of complex equations.