Problem 50 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The real zeros are: 1, \(-2\), \(\sqrt{2}\), \(-\sqrt{2}\).

Step 1: Identify the Polynomial

We have the polynomial function \(P(x) = x^4 + 2x^3 - 2x^2 - 3x + 2\). Our task is to find all real zeros of this function.

Step 2: Check for Easy Rational Zeros

To find potential rational zeros, we use the Rational Root Theorem. The possible rational zeros are the factors of the constant term 2, which are \(\pm 1, \pm 2\). Evaluate \(P(x)\) at these points to see if any are zeros.

Step 3: Evaluate Potential Zeros

Substitute \(x = 1\) into \(P(x)\) to check: \[ P(1) = (1)^4 + 2(1)^3 - 2(1)^2 - 3(1) + 2 = 1 + 2 - 2 - 3 + 2 = 0 \] Since \(P(1) = 0\), \(x = 1\) is a zero. Check other candidates similarly.

Step 4: Polynomial Division

Use synthetic division to divide \(P(x)\) by \(x - 1\) to reduce the polynomial: Coefficients are \(1, 2, -2, -3, 2\): After synthetic division by \(x - 1\), we are left with \(x^3 + 3x^2 + x - 2\).

Step 5: Check Remaining Polynomial for More Zeros

Apply the Rational Root Theorem again on \(x^3 + 3x^2 + x - 2\). Possible rational zeros are \(\pm 1, \pm 2\). Evaluate these values.

Step 6: Evaluating Remaining Candidates

Substitute \(x = -1\) into the reduced polynomial: \[ P(-1) = (-1)^3 + 3(-1)^2 + (-1) - 2 = -1 + 3 - 1 - 2 = -1 \] This is not zero, so continue with other candidates.

Step 7: Verification of Another Zero

Use synthetic division with \(x - 2\) for the remaining polynomial. For \(x = 2\), the synthetic division doesn't give zero. Use \(x = -2\) and it yields zero, leaving \(x^2 - 2\) as the quotient.

Step 8: Solve Quadratic Equation

The remaining polynomial is \(x^2 - 2\). It has two real zeros: \[ x^2 - 2 = 0 \] Adding 2 gives \(x^2 = 2\). Taking the square root yields \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). These are the final two real zeros.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool for finding potential rational zeros of a polynomial. This theorem states that for a polynomial function with integer coefficients, any possible rational root, expressed in the form \( \frac{p}{q} \), contains factors of the constant term as the numerator \( p \) and factors of the leading coefficient as the denominator \( q \).
This means if you have a polynomial like \( P(x) = x^4 + 2x^3 - 2x^2 - 3x + 2 \), you first identify possible rational zeros by taking the factors of the constant term, which is 2 in this example. The potential candidates for zeros are \( \pm 1, \pm 2 \). You check these by substituting them into the polynomial to see if they produce a zero value.
While the Rational Root Theorem helps narrow down options, it doesn't guarantee actual roots, only possibilities. We confirm these potential roots by further substitutions and calculations.

Synthetic Division

Synthetic division is a streamlined method of dividing polynomials, which is particularly useful when you need to divide by a linear factor like \( x - k \). It allows you to quickly determine whether a suspected zero of a polynomial actually divides evenly, confirming it as a root.
In the given polynomial, we used synthetic division to divide by \( x - 1 \) because we already verified that 1 is a zero of the polynomial \( P(x) \). The process involves simplifying the polynomial division by focusing on coefficients, making it faster than long division. Place the coefficients of the polynomial \( 1, 2, -2, -3, 2 \) in a row, and perform calculations to transform them into the reduced polynomial's coefficients.
Through synthetic division, we confirmed that \( x - 1 \) divides evenly, leaving us with a new polynomial of lower degree: \( x^3 + 3x^2 + x - 2 \). This technique is not only more efficient but also essential in poly pre-calculus for quickly solving polynomial equations.

Quadratic Formula

The Quadratic Formula is an algebraic solution for finding roots of a quadratic equation \( ax^2 + bx + c = 0 \). When you cannot easily factor a quadratic, this formula offers a reliable solution. The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] It considers all possible scenarios for the roots and works regardless of the discriminant's sign.
In our polynomial journey, after employing synthetic division, we were left with a simpler equation \( x^2 - 2 \). Although factorizable for simple square roots, thinking about more complex quadratics needing the Quadratic Formula is instructive.
For \( x^2 - 2 = 0 \), solving gives \( x = \sqrt{2} \) and \( x = -\sqrt{2} \), demonstrating the connection between theoretical roots calculated through different methods and their verification using the formula.

Polynomial Division

Polynomial Division is another essential tool mathematicians use to break down and simplify complex polynomial expressions, often used when synthetic division is not applicable, or a clearer solution is needed.
The process resembles long division, but instead of numbers, you work with polynomial terms. For a polynomial like \( P(x) \), you divide according to polynomial power and coefficients from highest to lowest.
While we mainly used synthetic division in the exercise to swiftly narrow down zeros, understanding polynomial division helps build a robust mathematical foundation. It reinforces how operations we learn in arithmetic extend into handling variables and algebraic expressions, a key aspect of polynomial analysis. This skill also directly supports future learning in calculus and higher-level math.