Problem 49 - Algebra and Trigonometry

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=x^{4}-6 x^{3}+4 x^{2}+15 x+4 $$

Step-by-Step Solution

Answer

The real zeros are \( x = 4, x = -1, x = \frac{3 + \sqrt{13}}{2}, x = \frac{3 - \sqrt{13}}{2}.\)

Step 1: Set the polynomial to zero

Begin by setting the polynomial equal to zero to find its zeros: \[ P(x) = x^4 - 6x^3 + 4x^2 + 15x + 4 = 0 \]

Step 2: Perform synthetic division

We first try to find possible rational roots using the Rational Root Theorem. The potential roots are the factors of the constant term, which are \( \pm 1, \pm 2, \pm 4 \). We use synthetic division to test these roots and find that \( x = 4 \) is a root.

Step 3: Factor the polynomial using the root

Since \( x = 4 \) is a root, divide the polynomial by \( (x - 4) \) using synthetic division.After division, we get \[ (x - 4)(x^3 - 2x^2 - 4x - 1) \].

Step 4: Use the Rational Root Theorem again

Apply the Rational Root Theorem to \( x^3 - 2x^2 - 4x - 1 \). The possible rational roots are \( \pm 1 \). Again, use synthetic division to test these roots. Find that \( x = -1 \) is another root.

Step 5: Factor again using the new root

Divide \( x^3 - 2x^2 - 4x - 1 \) by \( (x + 1) \) to obtain \[ (x - 4)(x + 1)(x^2 - 3x - 1) \].

Step 6: Solve the remaining quadratic equation

To solve \( x^2 - 3x - 1 = 0 \), use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 1, b = -3, \) and \( c = -1 \): \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 1 \times (-1)}}{2 \times 1} \] \[ x = \frac{3 \pm \sqrt{9 + 4}}{2} \] \[ x = \frac{3 \pm \sqrt{13}}{2} \]

Step 7: Write all roots

The roots of the polynomial are \( x = 4, x = -1, x = \frac{3 + \sqrt{13}}{2}, x = \frac{3 - \sqrt{13}}{2} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool in analyzing polynomials, especially when searching for rational zeros. This theorem tells us that any rational solution, or root, of a polynomial of the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\) must be in the form of \(\frac{p}{q}\). Here, \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).
In the given polynomial \(P(x) = x^4 - 6x^3 + 4x^2 + 15x + 4\), the leading coefficient is 1, and the constant term is 4. Thus, the possible rational roots are the factors of 4, which are \(\pm 1, \pm 2, \pm 4\).
Applying the theorem greatly reduces potential candidates for roots, making it easier to identify genuine rational solutions with further methods like Synthetic Division.

Synthetic Division

Synthetic Division is a streamlined method to divide polynomials, particularly when testing potential roots provided by the Rational Root Theorem. This technique is generally quicker and simpler compared to the traditional long division.
Here's how you use synthetic division:

• Write down the coefficients of the polynomial.
• Place the potential root outside the division box.
• Perform the synthetic division process which involves multiplying, adding, and writing down the results in a new row.

For example, in our polynomial \(x^4 - 6x^3 + 4x^2 + 15x + 4\), we use synthetic division to test first the root \(x = 4\). This root is verified when the division yields a remainder of zero, confirming that \(x - 4\) is a factor of the polynomial. This technique drastically simplifies the factoring process.

Quadratic Formula

The Quadratic Formula is an essential tool for solving second-degree polynomial equations, specifically quadratic equations of the form \(ax^2 + bx + c = 0\). It provides us with a direct way to find roots and is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula works universally for any quadratic polynomial and is especially useful when the polynomial does not factor easily.
In our exercise, after breaking down the polynomial into factors \((x - 4)(x + 1)(x^2 - 3x - 1)\), we apply the quadratic formula to the remaining quadratic component \(x^2 - 3x - 1\). By substituting \(a = 1\), \(b = -3\), and \(c = -1\), we find the roots \(x = \frac{3 \pm \sqrt{13}}{2}\), completing the solution.

Factoring Polynomials

Factoring Polynomials is a crucial skill for simplifying and solving polynomial equations. It involves expressing a polynomial as a product of its factors. Recognizing patterns and using root-findings like the Rational Root Theorem help in physically breaking down complex polynomials into simple, manageable factors.
In our example polynomial \(P(x) = x^4 - 6x^3 + 4x^2 + 15x + 4\), we first identified that \(x - 4\) and \(x + 1\) are factors through the roots \(x = 4\) and \(x = -1\) respectively. This process gives a strong start to simplifying the higher degree polynomial to a product of lower-degree parts, specifically leading us to \((x - 4)(x + 1)(x^2 - 3x - 1)\).
Factoring remains a foundational technique in algebra, as it lays the groundwork for finding polynomial roots effectively.