Question

Find a polynomial function with the given real zeros whose graph contains the given point. Zeros: -2,3,5 Degree 3 Point: (2,36)

Short answer

The polynomial function is \( f(x) = 3(x + 2)(x - 3)(x - 5) \).

Step-by-step solution

Write the general polynomial form based on zeros

Since the polynomial has zeros at -2, 3, and 5 and is of degree 3, the polynomial can be written in its factored form as: \[ f(x) = a(x + 2)(x - 3)(x - 5) \]. Here, \(a\) is a constant that we need to determine.

Substitute the given point into the equation

The polynomial needs to pass through the point (2, 36). Substitute \(x = 2\) and \(f(x) = 36\) into the equation to find \(a\): \[ 36 = a(2 + 2)(2 - 3)(2 - 5) \].

Solve for the constant \(a\)

Evaluate the expression inside the parentheses: \[ 36 = a(4)(-1)(-3) \]. This simplifies to: \[ 36 = 12a \]. Solve for \(a\): \[ a = \frac{36}{12} = 3 \].

Write the final polynomial function

Substitute the value of \(a = 3\) back into the polynomial function: \[ f(x) = 3(x + 2)(x - 3)(x - 5) \]. This is the polynomial function that has the given zeros and passes through the point (2, 36).

Key concepts

zeros

In mathematics, the 'zeros' of a polynomial refer to the values of x that make the polynomial equal to zero. In simpler terms, they are the points where the graph of the polynomial crosses the x-axis. For example, if a polynomial has zeros at -2, 3, and 5, it means that these are the x-values where the polynomial function equals zero.
Finding the zeros is crucial in deriving the factored form of the polynomial.

degree of polynomial

The 'degree of a polynomial' is the highest power of the variable x that appears in the polynomial. For example, in the polynomial \(x^3 - 4x^2 + 6x - 5\), the degree is 3 because the highest power of x is 3. Determining the degree helps in understanding the polynomial's general shape and behavior.
In our given exercise, the polynomial is of degree 3, indicating that the expression in its expanded form will have an x to the power of 3 term.

factored form

The 'factored form' of a polynomial represents it as a product of its linear factors. It is an essential aspect because it simplifies solving the polynomial equations and understanding its zeros. For example, if a polynomial has zeros -2, 3, and 5, its factored form can be written as \( f(x) = a(x + 2)(x - 3)(x - 5) \).
In this format, each factor corresponds to a zero and the constant ‘a’ scales the polynomial.

substituting points

Substituting points involves replacing the x-value and the corresponding y-value (f(x)) in the polynomial equation to find unknown constants. This is particularly useful when you have a specific point that the polynomial must pass through.
For instance, given the point (2,36) and our polynomial \( f(x) = a(x + 2)(x - 3)(x - 5)\), you can plug in 2 for x and 36 for f(x) to solve for ‘a’. This gives the equation: \( 36 = a(2 + 2)(2 - 3)(2 - 5)\).

solving for constants

Finally, 'solving for constants' involves calculating the specific values of unknown variables (constants) in your polynomial equation. This step ensures the polynomial passes through a particular point. Using our example, after substituting the point (2,36), we get the equation:\( 36 = a(4)(-1)(-3)\).
This simplifies to \( 36 = 12a\), and further solving gives \( a = 3\). Now, substituting this value of 'a' back into the original factored form provides the complete polynomial equation: \( f(x) = 3(x + 2)(x - 3)(x - 5)\).