Problem 12 - A Graphical Approach to College Algebra

Question

Find a polynomial function \(P(x)\) of degree 3 with real coefficients that satisfies the given conditions. Do not use a calculator. Zeros of \(-7\) and \(2-i ; \quad P(1)=9\)

Step-by-Step Solution

Answer

The polynomial is \(P(x) = \frac{9}{16}(x + 7)(x^2 - 4x + 5)\).

Step 1: Identify Zeros and Their Conjugate

Given the zeros are \(-7\) and \(2-i\). Because the coefficients of the polynomial are real numbers, the conjugate of any non-real zero must also be a zero. Therefore, the conjugate of \(2-i\), which is \(2+i\), is also a zero.

Step 2: Construct the Polynomial with Zeros

The polynomial can be written using its zeros as factors: \[ P(x) = a(x + 7)(x - (2-i))(x - (2+i)) \].

Step 3: Simplify the Complex Conjugate Pair

Simplify the product \((x - (2-i))(x - (2+i))\):- This is a difference of squares: \( (x-2)^2 - (i)^2 = (x-2)^2 + 1 \) because \(i^2 = -1\).- Expanding \( (x-2)^2 + 1 \) gives: \( x^2 - 4x + 4 + 1 = x^2 - 4x + 5 \).

Step 4: Expand the Polynomial

Now insert \( (x^2 - 4x + 5) \) into the polynomial: \[ P(x) = a(x + 7)(x^2 - 4x + 5) \].

Step 5: Use Given Condition to Determine 'a'

We know \(P(1)=9\). Substitute \(x=1\) into the polynomial equation:\[ 9 = a((1) + 7)((1)^2 - 4(1) + 5) \].Simplify inside the parentheses:- \(1+7 = 8\),- \((1)^2 - 4(1) + 5 = 1 - 4 + 5 = 2\).Thus,\[ 9 = a \cdot 8 \cdot 2 \].Solve for 'a':\[ 9 = 16a \implies a = \frac{9}{16} \].

Step 6: Write the Final Polynomial

Substitute the value of 'a' back into \(P(x)\): \[ P(x) = \frac{9}{16}(x + 7)(x^2 - 4x + 5) \].

Key Concepts

Complex Numbers

Complex numbers are numbers that can express both a real and an imaginary part. They are usually written in the form \(a + bi\), where \(a\) represents the real part and \(bi\) represents the imaginary part with \(i\) being the square root of \(-1\). This means \(i^2 = -1\).
For example, the complex number \(2 - i\) consists of a real part \(2\) and an imaginary part \(-i\). Complex numbers are crucial in solving polynomial equations, particularly when dealing with roots that are not real numbers.

• Real part: Found on the traditional number line.
• Imaginary part: Corresponds to multiples of \(i\).

Understanding how these parts work together allows us to handle equations that include square roots of negative numbers, which would otherwise be unsolvable with only real numbers.

Conjugate Zeros Theorem

The Conjugate Zeros Theorem states that if a polynomial has real coefficients and a complex root, like \(a + bi\), the complex conjugate \(a - bi\) must also be a root. This theorem ensures that complex roots occur in conjugate pairs to maintain the polynomial's property of having only real coefficients.
For example, in the exercise, the polynomial has zeros at \(2-i\) and \(2+i\). These zeros are conjugates, ensuring all parts of the polynomial are balanced with real coefficients.

• Conjugate pairs: Always appear together when coefficients are real.
• Maintains the polynomial's integrity of real numbers.

This theorem is an important tool for constructing polynomials from given zeros, particularly those dealing with complex numbers.

Degree of a Polynomial

The degree of a polynomial is the highest power of the variable in the polynomial expression, representing the complexity and the number of solutions a polynomial can have. For instance, a third-degree polynomial like the one in our exercise, can be expressed in the general form \(ax^3 + bx^2 + cx + d\).
The degree tells us three important things:

• The number of roots or zeros: A degree 3 polynomial can have up to 3 roots.
• The behavior of the polynomial graph at high powers: Higher degree polynomials have more twists and turns.
• The end behavior: Determines how the polynomial graph behaves as \(x\) approaches infinity or negative infinity.

Understanding the degree offers insight into the fundamental structure and possible solutions of the polynomial.

Real Coefficients

Polynomials with real coefficients mean that all the numbers multiplying or subtracting with the variable \(x\) are real numbers. This aspect is essential because it influences how we solve and understand the polynomial's roots.
With real coefficients, if a polynomial has a complex root, its conjugate must also be a root, ensuring all terms result in real numbers. This condition is vital when constructing a polynomial with given roots, such as in the exercise.
Why are real coefficients important?

• Ensure the polynomial remains solvable within the real number system.
• Indirectly dictate that complex roots appear in conjugate pairs, maintaining the real nature of the polynomial.

By working with real coefficients, we ensure the polynomial is aligned with real-world applications and can be graphed on a real coordinate plane.