Problem 12 - A Graphical Approach to Precalculus with Limits

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 function is \( P(x) = \frac{9}{16}(x^3 + 3x^2 - 23x + 35) \).

Step 1: Identify Necessary Information

We are given zeros of the polynomial and a value for the polynomial at a specific point. Specifically, the zeros are \(-7\) and \(2-i\). Since polynomials with real coefficients have complex roots in conjugate pairs, \(2+i\) must also be a zero.

Step 2: Write the Polynomial in Factored Form Using Zeros

The polynomial can be expressed as \(P(x) = a(x + 7)(x - (2 - i))(x - (2 + i))\), where \(a\) is a constant coefficient we need to determine. The factors \((x - (2-i))(x - (2+i))\) form a quadratic when expanded.

Step 3: Expand the Quadratic Factor

Expand the expression for the quadratic factor. \[(x - (2-i))(x - (2+i)) = (x - 2 + i)(x - 2 - i) = (x - 2)^2 - (i^2) = (x - 2)^2 + 1\]Calculate the expanded form: \[(x - 2)^2 = x^2 - 4x + 4\]So we have:\[(x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5\]

Step 4: Expand the Entire Polynomial

Substitute the expanded quadratic back into the polynomial:\[P(x) = a(x + 7)(x^2 - 4x + 5)\]Now, expand \((x + 7)(x^2 - 4x + 5)\) to get:\[(x + 7)(x^2 - 4x + 5) = x(x^2 - 4x + 5) + 7(x^2 - 4x + 5)\]\[= x^3 - 4x^2 + 5x + 7x^2 - 28x + 35 \]Combine like terms:\[= x^3 + 3x^2 - 23x + 35\]

Step 5: Use Given Condition to Find 'a'

We know \(P(1) = 9\). Substitute \(x = 1\) into \(P(x) = a(x^3 + 3x^2 - 23x + 35)\): \[P(1) = a(1^3 + 3(1)^2 - 23(1) + 35) = a(1 + 3 - 23 + 35) = a \cdot 16 = 9\]So, \(a = \frac{9}{16}\).

Step 6: State the Complete Polynomial Function

Place \(a = \frac{9}{16}\) back into the polynomial expression: \[P(x) = \frac{9}{16}(x^3 + 3x^2 - 23x + 35)\]

Key Concepts

Complex Conjugate Zeros

When dealing with polynomial functions that have real coefficients, it's important to remember that they often involve complex zeros. These zeros occur in conjugate pairs. A complex conjugate pair consists of two numbers of the form

• A + Bi
• A - Bi

Here, 'A' is the real part and 'Bi' is the imaginary part.
In the context of our exercise, we have a complex zero at \(2 - i\).
Since the polynomial has real coefficients, its complex conjugate \(2 + i\) must also be a zero.
This characteristic ensures that any imaginary parts will cancel out during multiplication, maintaining the polynomial's real nature.
This is a crucial step for solving these types of problems.

Factored Form

The factored form of a polynomial is a way of expressing the polynomial by its zeros.
In simpler terms, if you know the zeros of a polynomial, you can write it as a series of factors.
Each zero leads to a factor of the form \((x - c)\), where \(c\) is a given zero.

• For zero \(-7\), the factor is \((x + 7)\)
• For zero \(2 - i\), the factor is \((x - (2-i))\)
• For zero \(2 + i\), the factor is \((x - (2+i))\)

Hence, the polynomial can be expressed as \(P(x) = a(x + 7)(x - (2 - i))(x - (2 + i))\), where \(a\) is a constant that needs to be determined.
This form helps us to easily expand and solve the polynomial by focusing on the zeros.

Coefficient Determination

Determining the coefficient of a polynomial is essential for defining its specific behavior.
Here, the coefficient \(a\) adjusts the polynomial so that it meets a given condition.
We are given that \(P(1) = 9\). This means when you substitute \(x = 1\) into the polynomial, the result should be 9.

• By substituting \(x = 1\) into our expanded polynomial \(P(x) = a(x^3 + 3x^2 - 23x + 35)\), we compute a value for \(P(1)\)
• This simplifies to \(a \cdot 16 = 9\)
• Solving for \(a\), you get \(a = \frac{9}{16}\)

With this coefficient, the polynomial is completely defined as \(P(x) = \frac{9}{16}(x^3 + 3x^2 - 23x + 35)\).
Such determination ensures that the polynomial behaves as needed at specific points, adhering to given conditions.