Problem 63 - Precalculus Mathematics for Calculus

Question

Find a polynomial of degree 3 that has zeros \(1,-2,\) and 3 and in which the coefficient of \(x^{2}\) is \(3 .\)

Step-by-Step Solution

Answer

The polynomial is \(-\frac{3}{2}x^3 + 3x^2 + \frac{15}{2}x - 9\).

Step 1: Understanding Polynomial Zeros

To find a polynomial with specific zeros, use the fact that if a polynomial has zeros at \(x = a\), \(x = b\), and \(x = c\), then it can be expressed as \((x-a)(x-b)(x-c)\). For our polynomial, the zeros are \(1\), \(-2\), and \(3\), so the polynomial can initially be written as \((x-1)(x+2)(x-3)\).

Step 2: Expand the Polynomial

Next, we need to expand \((x-1)(x+2)(x-3)\). First multiply the first two terms: \((x-1)(x+2) = x^2 + x - 2\). Now, multiply this with the third term \((x-3)\):\[ (x^2 + x - 2)(x-3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 \]Simplify this to get:\[ x^3 - 2x^2 - 5x + 6 \].

Step 3: Adjust Coefficient of \(x^2\)

The problem requires that the coefficient of \(x^2\) is \(3\). In the polynomial \(x^3 - 2x^2 - 5x + 6\), the coefficient of \(x^2\) is \(-2\). To adjust this to \(3\), multiply the entire polynomial by a constant factor \(k\) such that \(-2k = 3\). Thus, \(k = -\frac{3}{2}\).

Step 4: Adjust the Polynomial

Multiply the polynomial \(x^3 - 2x^2 - 5x + 6\) by \(k = -\frac{3}{2}\):\[ -\frac{3}{2}(x^3 - 2x^2 - 5x + 6) = -\frac{3}{2}x^3 + 3x^2 + \frac{15}{2}x - 9 \]This gives the final polynomial: \[ -\frac{3}{2}x^3 + 3x^2 + \frac{15}{2}x - 9 \].

Key Concepts

Understanding Polynomial Zeros

Polynomial zeros are the values of the variable for which the polynomial equals zero. In other words, they are the solutions or roots of the polynomial equation. For a polynomial of degree 3, you can expect up to three zeros if you consider all multiplicities. For the provided exercise, the zeros are given as 1, -2, and 3.

This means that for the polynomial to equal zero, substituting x with any of these values should give zero. Mathematically, if the zeros are \(x = a\), \(x = b\), and \(x = c\), the polynomial can be expressed using linear factors like this:

• \((x - a)(x - b)(x - c)\)

For our problem, the polynomial with zeros 1, -2, and 3 can be expressed as:

• \((x-1)(x+2)(x-3)\)

This initial form is critical, as multiplying these factors together will give us a polynomial from which we can work further.

Polynomial Expansion

Polynomial expansion is the process of multiplying out the factors of the polynomial to create a standard form expression. This involves multiplying the factors sequentially until you arrive at a single polynomial expression.

Let's start with the given factors: \((x-1)(x+2)(x-3)\). We first multiply the first two factors:

• \((x-1)(x+2) = x^2 + x - 2\)

Next, we take this result, \(x^2 + x - 2\), and multiply it by the third factor \((x-3)\):

• \((x^2 + x - 2)(x-3)\) expands to:
  \[ (x^2 + x - 2)(x - 3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 \]

Simplifying gives the expanded polynomial:

• \(x^3 - 2x^2 - 5x + 6\)

The expansion process ensures each term of the polynomial and its respective coefficients are correctly calculated based on the original factors.

Coefficient Adjustment

Coefficient adjustment is necessary when specific conditions are given for the polynomial's coefficients. In this exercise, the condition is that the coefficient of \(x^2\) must be 3.

Currently, in the polynomial \(x^3 - 2x^2 - 5x + 6\), the coefficient of \(x^2\) is -2. To make this 3, we must find a multiplication factor \(k\) that adjusts all coefficients accordingly.

We set up the equation \(-2k = 3\) to find \(k\). Solving gives:

• \(k = -\frac{3}{2}\)

By multiplying the entire polynomial by \(-\frac{3}{2}\), each coefficient gets adjusted while maintaining the polynomial's zeros:

• \(-\frac{3}{2}(x^3 - 2x^2 - 5x + 6) = -\frac{3}{2}x^3 + 3x^2 + \frac{15}{2}x - 9 \)

This adjustment gets the coefficient to the desired value while preserving the roots, giving us the final polynomial expression that meets all requirements.