Question

Find a polynomial function with the given real zeros whose graph contains the given point. Zeros: -5,-1,2,6 Degree 4 Point: \(\left(\frac{5}{2}, 15\right)\)

Short answer

The polynomial is \( f(x) = -\frac{16}{49}(x + 5)(x + 1)(x - 2)(x - 6) \).

Step-by-step solution

Write the polynomial with the given zeros

A polynomial with zeros (-5), (-1), 2, and 6 can be written as \( f(x) = k(x + 5)(x + 1)(x - 2)(x - 6) \), where \( k \) is a constant to be determined.

Use the given point to determine the constant k

Substitute the coordinates of the point \( \left( \frac{5}{2}, 15 \right) \) into the polynomial function and solve for \( k \) : \( 15 = k \left( \frac{5}{2} + 5 \right) \left( \frac{5}{2} + 1 \right) \left( \frac{5}{2} - 2 \right) \left( \frac{5}{2} - 6 \right) \).

Simplify the expression inside the brackets

Evaluate each term: \( \frac{5}{2} + 5 = \frac{5}{2} + \frac{10}{2} = \frac{15}{2} \), \( \frac{5}{2} + 1 = \frac{5}{2} + \frac{2}{2} = \frac{7}{2} \), \( \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2} \), and \( \frac{5}{2} - 6 = \frac{5}{2} - \frac{12}{2} = -\frac{7}{2} \).

Substitute the simplified values back into the equation

The expression is now: \( 15 = k \left( \frac{15}{2} \right) \left( \frac{7}{2} \right) \left( \frac{1}{2} \right) \left( -\frac{7}{2} \right) \).

Solve for k by simplifying further

First calculate the product inside: \( \left( \frac{15 \times 7 \times 1 \times -7}{2 \times 2 \times 2 \times 2} \right) = \frac{-735}{16} \), Then equates: \( 15 = k \cdot \frac{-735}{16} \)Thus, \( k = \frac{15 \times 16}{-735} = \frac{240}{-735} = -\frac{16}{49} \).

Write the final polynomial equation

Substitute \( k = -\frac{16}{49} \) back into the polynomial equation to get: \( f(x) = -\frac{16}{49} (x + 5)(x + 1)(x - 2)(x - 6) \).

Key concepts

polynomial equation

A polynomial equation is an expression involving a sum of powers in one or more variables multiplied by coefficients. The general form of a polynomial equation in one variable is given as: \(a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0 = 0\) Here, \(a_n, a_{n-1}, ..., a_0\) are coefficients, and \(n\) represents the degree of the polynomial, which is the highest power of the variable \(x\) in the equation.
In our exercise, because we are given the real zeros, the polynomial can be constructed by using these zeros. You can construct such a polynomial by setting each zero as a root in the form of factorized terms. Each root (zero) shifts the polynomial by a certain value. For instance, a zero at \(x = -5\) would translate to \( (x + 5) \) and similarly for each zero.
Once you have each factor, you then multiply them together to form the polynomial. The result for our exercise, without the constant, is: \((x + 5)(x + 1)(x - 2)(x - 6)\). This ensures that the polynomial touches or crosses the x-axis at the provided zeros.

real zeros

Real zeros of a polynomial are the values of \(x\) for which the polynomial function \(f(x)\) equals zero. They are also known as the roots of the polynomial.
To determine where a polynomial will cross the x-axis, you need the real zeros. These zeros split the polynomial into binomial factors. Here, given zeros are \(-5, -1, 2, 6\). Each zero can be written as a factor:

• \(x = -5 \rightarrow (x + 5)\)
• \(x = -1 \rightarrow (x + 1)\)
• \(x = 2 \rightarrow (x - 2)\)
• \(x = 6 \rightarrow (x - 6)\)

These factors can then be multiplied to form the polynomial equation. Thus, for our exercise, the polynomial without the constant is \((x + 5)(x + 1)(x - 2)(x - 6)\).
Notice that each factor corresponds to making the polynomial equation equal zero, fulfilling our requirement for real zeros. This method simplifies the polynomial construction by clearly associating the zeros with the equation factors.

determine constant

The constant \(k\) in a polynomial equation \(f(x) = k(x - \text{zero}_1)(x - \text{zero}_2)...\) determines the vertical stretch or compression of the graph. To find this constant, we use a given point that the polynomial passes through. This method ensures the polynomial not only has the right zeros but also fits the specific point.
In our exercise, the point \( \left( \frac{5}{2}, 15 \right) \) is used. We substitute \(x = \frac{5}{2}\) and \(f(x) = 15\) into our polynomial, then solve for \(k\).
The steps are as follows:

• Start with the polynomial equation: \(f\left( \frac{5}{2} \right) = k \times \frac{735}{16}\)
• Since \( f\left( \frac{5}{2} \right) = 15 \)
• Set up the equation: \(15 = k \times \frac{-735}{16}\)
• Solve: \(k = \frac{15 \times 16}{-735} = \frac{-240}{735} = -\frac{16}{49}\)

By finding the right \(k\), we ensure our polynomial is the correct shape, as described by passing through the given point. For our final polynomial, we thus get: \(f(x) = -\frac{16}{49} (x + 5)(x + 1)(x - 2)(x - 6)\). This process guarantees that the function fits all given conditions.