Problem 22 - A Graphical Approach to College Algebra

Question

Find a polynomial function \(P(x)\) having leading coefficient \(1,\) least possible degree, real coefficients, and the given zeros. $$1+\sqrt{2}, 1-\sqrt{2}, \text { and } 3$$

Step-by-Step Solution

Answer

The polynomial is \(P(x) = x^3 - 5x^2 + 7x - 3\).

Step 1: Understanding the Zeros

The polynomial has zeros at \(1+\sqrt{2}, 1-\sqrt{2},\) and \(3\). Since the coefficients of the polynomial must be real numbers, any irrational roots must occur in conjugate pairs. In this case, \(1+\sqrt{2}\) and \(1-\sqrt{2}\) are conjugates, so we can proceed to forming factors from these zeros.

Step 2: Forming the Factors

For each zero \(c\), the corresponding linear factor of the polynomial is \((x-c)\). Thus, we have the factors as \((x-(1+\sqrt{2})), (x-(1-\sqrt{2})),\) and \((x-3)\).

Step 3: Combining Conjugate Factors

Combine the conjugate factors: \[(x-(1+\sqrt{2}))(x-(1-\sqrt{2})) = ((x-1)-\sqrt{2})((x-1)+\sqrt{2})\]. This expression is a difference of squares: \((a-b)(a+b) = a^2 - b^2\).

Step 4: Simplifying the Difference of Squares

Calculate the difference of squares: \[((x-1)-\sqrt{2})((x-1)+\sqrt{2}) = (x-1)^2 - (\sqrt{2})^2 = x^2 - 2x + 1 - 2\]. Simplifying this yields \(x^2 - 2x - 1\).

Step 5: Form the Complete Polynomial

Now multiply the simplified factor with the remaining factor \((x - 3)\): \[P(x) = (x^2 - 2x - 1)(x - 3)\].

Step 6: Expanding the Polynomial

Expand \((x^2 - 2x - 1)(x - 3)\) using distribution:\[\begin{aligned} P(x) &= x^2(x - 3) - 2x(x - 3) - 1(x - 3) \&= x^3 - 3x^2 - 2x^2 + 6x - x + 3 \&= x^3 - 5x^2 + 7x - 3.\end{aligned}\]

Step 7: Conclusion

Thus, the polynomial with the given properties is \(P(x) = x^3 - 5x^2 + 7x - 3\). It satisfies all conditions: it has a leading coefficient of 1, least degree of 3 (equal to the number of distinct zeros), and real coefficients.

Key Concepts

Zeros of Polynomials

In the realm of polynomial functions, zeros hold a fundamental place. A zero of a polynomial is simply a value of the variable for which the polynomial evaluates to zero. In our task, we were given the zeros as \(1+\sqrt{2}, 1-\sqrt{2}\), and \(3\). These represent the values where the polynomial touches or crosses the x-axis.

• Each zero corresponds to a factor of the polynomial. If \(c\) is a zero, then \((x-c)\) is a factor.
• The degree of the polynomial is related to the number of zeros it has. Here, we have three zeros, so the minimum degree of the polynomial is three.
• In polynomials with real coefficients, if one root, especially an irrational one involving a square root, is known, its conjugate must also be a root. Hence, \(1+\sqrt{2}\) and \(1-\sqrt{2}\) are used in pairs.

This understanding ensures the polynomial remains with real coefficients and helps in building it step by step by factoring.

Real Coefficients

Polynomials with real coefficients are crucial in many practical applications. Real coefficients mean each term in the polynomial is a real number, encompassing both rational and irrational numbers. This property influences the nature and appearance of the polynomial's zeros.

• When coefficients are real, any non-real roots must appear in conjugate pairs if present. For real polynomials, complex numbers' conjugates ensure that the final polynomial expression does not include imaginary components.
• An example of this property in action is using \(1+\sqrt{2}\) and \(1-\sqrt{2}\). Although \(\sqrt{2}\) is irrational, its conjugate pairs with its negative counterpart to form a linear combination that results in a real coefficient polynomial.
• This system ensures simpler interpretation and application of the polynomial function in real-life contexts, including physics and engineering.

Therefore, always considering the implication of real coefficients helps when formulating polynomial expressions and solving real-world problems.

Difference of Squares

The difference of squares is a fundamental algebraic identity used to simplify expressions involving squared terms subtracted from each other. This identity is expressed as \((a-b)(a+b) = a^2 - b^2\). It serves as a handy tool when dealing with roots, especially quadratic roots like \(\sqrt{2}\) as seen in our solution.

• The difference of squares helps simplify conjugate factor expressions, such as \( (x-(1+\sqrt{2}))(x-(1-\sqrt{2})) \).
• By recognizing this pattern, we simplify to \((x-1)^2 - (\sqrt{2})^2 = x^2 - 2x + 1 - 2\), thus cleanly reducing down to a simpler quadratic form \( x^2 - 2x - 1 \).
• Utilizing this technique greatly reduces computational effort and helps maintain accuracy. It's a strategic approach in polynomial manipulations.

With this in mind, employing the difference of squares is essential when simplifying polynomial factors, especially when conjugates or quadratic roots are involved.