Problem 440 - Precalculus

Question

Construct a polynomial function of least degree possible using the given information. Real roots: \(-1,1,3\) and \((2, f(2))=(2,4)\)

Step-by-Step Solution

Answer

The polynomial is \( p(x) = -\frac{4}{3}(x+1)(x-1)^2(x-3) \).

Step 1: Identify the Roots

The problem states that the polynomial has real roots at 1, -1, 1, and 3. These roots suggest that the polynomial has factors of the form (x + 1), (x - 1), and (x - 3). Since 1 is repeated, it will appear as a square in the factorization.

Step 2: Construct the Polynomial Using Roots

Given the roots, we can construct the polynomial as\[ p(x) = a(x + 1)(x - 1)^2(x - 3) \].Here, 'a' is a leading coefficient that we need to determine.

Step 3: Use the Given Point to Find 'a'

The polynomial passes through the point (2, 4). Substituting this point into the polynomial to solve for 'a',\[ 4 = a(2 + 1)(2 - 1)^2(2 - 3) \ => 4 = a(3)(1)(-1) \ => 4 = -3a \ => a = -\frac{4}{3} \].So, the leading coefficient 'a' is -4/3.

Step 4: Write the Final Polynomial

Substitute back 'a' into the polynomial equation:\[ p(x) = -\frac{4}{3}(x + 1)(x - 1)^2(x - 3) \].This is the polynomial of least degree that satisfies the given conditions.

Key Concepts

Real Roots

Real roots are the values of a variable, typically denoted as \( x \), for which the polynomial equals zero. These roots correspond to the solutions of the equation \( p(x) = 0 \). In our exercise, the real roots provided are \(-1, 1,\) and \(3\). Each root represents a point at which the polynomial intersects the x-axis.

• Root \(-1\) gives us a factor of \((x + 1)\).
• Root \(1\) gives us a factor of \((x - 1)\), and since it's repeated, it appears squared: \((x - 1)^2\).
• Root \(3\) gives us a factor of \((x - 3)\).

These factors multiply together to form the polynomial. Repeated roots increase the multiplicity but not the root value, affecting the shape of the graph at those points.

Leading Coefficient

The leading coefficient is the constant multiplier in front of a polynomial's highest degree term. It plays a crucial role in determining the end behavior and general orientation of the polynomial graph. In our case, the leading coefficient \(a\) is found using additional given points.
Given that the polynomial passes through the point \((2, 4)\), we can substitute these values into the polynomial to determine \(a\). After substituting and simplifying, we find \(a = -\frac{4}{3}\). This coefficient ensures that the polynomial takes on the desired shape and crosses the specified point.

Point Substitution

Point substitution involves inserting known values of \( x \) and \( y \) into the polynomial to solve for unknowns. For this exercise, we used the point \((2, 4)\) to determine the leading coefficient \(a\).
Substitute \(x = 2\) and \(p(x) = 4\) into the polynomial equation. This gives the equation \(4 = a(2 + 1)(2 - 1)^2(2 - 3)\). After simplifying, we solve for \(a\), yielding \(a = -\frac{4}{3}\). This is how the point allows us to fine-tune the polynomial according to specific requirements.

Degree of Polynomial

The degree of a polynomial is the highest sum of the powers of \(x\) in any of its terms. It dictates the general shape and complexity of the polynomial curve. For our polynomial, the degree is affected by the number and multiplicity of the roots involved.
Using roots of \(-1, 1\), and \(3\), with \(1\) repeated, leads us to a polynomial of at least degree 4. Here's how:

• \((x + 1)\) contributes a degree of 1.
• \((x - 1)^2\) contributes a degree of 2.
• \((x - 3)\) contributes a degree of 1.

Adding these up, the polynomial has a degree of \(1 + 2 + 1 = 4\). This degree specifies that our polynomial will have up to four x-intercepts and a reflective wave shape.