Question

1. find all the zeros of the function f(x) = 2x⁴ + x³ + x² + x − 1

Explanation:

Step1: Try Rational Root Theorem

The Rational Root Theorem states that possible rational roots are factors of the constant term over factors of the leading coefficient. For \(f(x) = 2x^4 + x^3 + x^2 + x - 1\), possible rational roots are \(\pm1, \pm\frac{1}{2}\).

Test \(x = \frac{1}{2}\): \(f(\frac{1}{2}) = 2(\frac{1}{2})^4 + (\frac{1}{2})^3 + (\frac{1}{2})^2 + \frac{1}{2} - 1 = 2(\frac{1}{16}) + \frac{1}{8} + \frac{1}{4} + \frac{1}{2} - 1 = \frac{1}{8} + \frac{1}{8} + \frac{2}{8} + \frac{4}{8} - \frac{8}{8} = 0\). So \(x = \frac{1}{2}\) is a root, and \((2x - 1)\) is a factor.

Step2: Perform Polynomial Division or Factoring

Divide \(2x^4 + x^3 + x^2 + x - 1\) by \((2x - 1)\) or factor by grouping. Let's try factoring:

\(2x^4 + x^3 + x^2 + x - 1 = 2x^4 - x^3 + 2x^3 - x^2 + 2x^2 - x + 2x - 1\) (rewriting terms to factor)

Group as \((2x^4 - x^3) + (2x^3 - x^2) + (2x^2 - x) + (2x - 1)\)

Factor each group: \(x^3(2x - 1) + x^2(2x - 1) + x(2x - 1) + 1(2x - 1) = (2x - 1)(x^3 + x^2 + x + 1)\)

Step3: Factor the Cubic

Factor \(x^3 + x^2 + x + 1\) by grouping: \((x^3 + x^2) + (x + 1) = x^2(x + 1) + 1(x + 1) = (x + 1)(x^2 + 1)\)

Step4: Find Zeros

Set each factor to zero:

• \(2x - 1 = 0 \implies x = \frac{1}{2}\)
• \(x + 1 = 0 \implies x = -1\)
• \(x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm i\)

Answer:

The zeros of the function \(f(x) = 2x^4 + x^3 + x^2 + x - 1\) are \(x = \frac{1}{2}\), \(x = -1\), \(x = i\), and \(x = -i\).