Problem 56 - Algebra and Trigonometry

Question

\(41-58=\) Find all zeros of the polynomial. $$ P(x)=4 x^{4}+2 x^{3}-2 x^{2}-3 x-1 $$

Step-by-Step Solution

Answer

The zeros of \( P(x) \) are \( x = 1\) and \( x = -\frac{1}{2} \) (double root).

Step 1: Identify the Polynomial

The given polynomial is \( P(x) = 4x^4 + 2x^3 - 2x^2 - 3x - 1 \). Our task is to find the zeros, or roots, of this polynomial function.

Step 2: Apply Rational Root Theorem

According to the Rational Root Theorem, any potential rational root of the polynomial \( P(x) \) is a factor of the constant term \(-1\) divided by a factor of the leading coefficient \(4\). This means we should test the potential roots: \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}\).

Step 3: Test Potential Roots Using Synthetic Division

Begin testing these candidates by performing synthetic division for each to check whether the remainder is zero. This indicates the number is a root. After testing, we find that \( x = 1 \) is a root, meaning the polynomial can be divided by \( x - 1 \).

Step 4: Divide Polynomial by \( x - 1 \)

Use synthetic division to divide the polynomial \( P(x) \) by \( x - 1 \). The result is a quotient polynomial with a lower degree: \( 4x^3 + 6x^2 + 4x + 1 \).

Step 5: Factor the Quotient Polynomial

To further find the zeros, we need to factor \(4x^3 + 6x^2 + 4x + 1\). Notice that the potential roots from Step 2 still apply, so using synthetic division, test \( \pm \frac{1}{2}, \pm \frac{1}{4}\). \( x = -\frac{1}{2} \) is a root, which means the quotient can be divided by \( x + \frac{1}{2} \).

Step 6: Divide by \( x + \frac{1}{2} \)

Perform synthetic division on \( 4x^3 + 6x^2 + 4x + 1 \) using \( x + \frac{1}{2} \). After division, the quotient polynomial is \( 4x^2 + 4x + 2\).

Step 7: Solve the Quadratic Equation

Now solve the quadratic equation \( 4x^2 + 4x + 2 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting gives \( x = \frac{-4 \pm \sqrt{0}}{8} \), resulting in the double root \( x = -\frac{1}{2} \).

Step 8: Collect All Zeros

Compile all found zeros from previous steps: \( x = 1 \), \( x = -\frac{1}{2} \), and the double root \( x = -\frac{1}{2} \) repeating once more.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool for identifying possible rational roots of a polynomial. It's based on the relationship between the constant term and the leading coefficient of the polynomial. Here's a simple breakdown:

• The theorem states that any rational root of a polynomial, when expressed in its lowest terms as \( \frac{p}{q} \), will have \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient.
• In our polynomial \( P(x) = 4x^4 + 2x^3 - 2x^2 - 3x - 1 \), the constant term is \(-1\), and the leading coefficient is \(4\).
• This gives us potential rational roots: \( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4} \).

Testing these roots is essential because it helps narrow down which root(s) to further investigate using methods like synthetic division. This process makes identifying zeros more manageable and efficient.

synthetic division

Synthetic division is a simplified form of polynomial division, specifically useful when dividing by linear factors, such as \(x - c\). It's a handy tool for testing potential roots identified using the Rational Root Theorem.Here's an overview of how synthetic division works:

• Arrange the coefficients of the polynomial in decreasing order of terms.
• Use the potential root (like \(1\) in the exercise) for performing the operation.
• The result of the division can provide a quotient and a remainder. A remainder of zero confirms the divisor is indeed a root.

In the exercise, a few rounds of synthetic division were performed to discover that \(x = 1\) and \(x = -\frac{1}{2}\) were roots. Each successful division results in a new, lower-degree polynomial, making it easier to find further roots.

quadratic formula

When a polynomial is reduced to a quadratic form, the quadratic formula becomes a useful method for finding its roots. This formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]It is applicable to any quadratic equation of the form \( ax^2 + bx + c = 0 \). Here's how it helps:

• The formula enables us to find both real and complex roots by calculating the square root of the discriminant \( b^2 - 4ac \).
• If the discriminant is zero, the roots are real and equal, as seen in the exercise where \( x = -\frac{1}{2} \) was a double root.

Ultimately, solving a quadratic equation using this formula allows for precision and clarity, even when polynomial equations become complex. For students, understanding and applying this formula is crucial for solving polynomial problems effectively.