Question

Write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros. \(-2,1+\sqrt{7}\)

Short answer

The polynomial function of least degree with a leading coefficient of 1 and the given zeros \(-2\), \(1 + \sqrt{7}\), and its conjugate root is \(f(x) = x^{3} - 4x + 6\).

Step-by-step solution

Recognize the conjugate

Firstly, identify that when a polynomial has real coefficients and one of its roots is a complex number, the conjugate of this root is also a root. Therefore, given one root as \(1 + \sqrt{7}\), its conjugate \(1 - \sqrt{7}\) is also a root of the polynomial.

Formulate the polynomial

After identifying all roots and confirming that they are all real numbers, construct a polynomial that has these roots. This can be done by forming binomials, each of which equals zero when x equals each root. So the polynomial will be \(f(x) = (x - (-2))(x - (1 + \sqrt{7}))(x - (1 - \sqrt{7}))\).

Simplify the polynomial

Now, simplify the polynomial by multiplying out the factors. \(f(x) = (x + 2)(x - 1 - \sqrt{7})(x - 1 + \sqrt{7})\). Simplifying gives \(f(x) = (x + 2)((x - 1)^2 - (\sqrt{7})^2)\), which simplifies to \(f(x) = (x + 2)(x^{2} - 2x + 1 - 7)\), and finally to \(f(x) = x^{3} - 4x + 6\).

Key concepts

Conjugate Roots in Polynomial Functions

Conjugate roots play a critical role when dealing with polynomial functions, especially those with complex roots. When a polynomial has real coefficients, and one of its roots is complex (involving a square root, for example), its conjugate must also be a root. This is because the product of these roots is real. For example, if your polynomial has a root of \(1 + \sqrt{7}\), the conjugate \(1 - \sqrt{7}\) also becomes a root. This ensures the polynomial remains with real coefficients.

Conjugate roots appear in pairs and help in maintaining the symmetry needed in polynomial equations. By including the conjugate, the polynomial's integrity is maintained, ensuring no imaginary parts remain once the polynomial is expanded.

Real Coefficients and Their Importance

Real coefficients in a polynomial function mean that all coefficients (the numbers in front of the variables) are real numbers, with no imaginary parts. This is important in polynomial functions because for any non-real root, its conjugate needs to be present, ensuring all resulting coefficients remain real.

Real coefficients often come into play in physics and engineering problems, where complex numbers represent quantities like voltages or currents, but final answers must be real numbers. For a polynomial in its simplest form with real coefficients, if one root is introduced as a complex number like \(1 + \sqrt{7}\), its conjugate \(1 - \sqrt{7}\) balances the equation. Thus, it ensures that the resulting polynomial expression has purely real-number terms after all roots are accounted for.

Simplifying Polynomials

Simplifying polynomials involves combining like terms and removing any unnecessary complexity from the polynomial expression. This is often the final step in formulating a polynomial function from its roots. Once you have identified and multiplied the factors corresponding to each root, you need to simplify the expression.

For instance, consider the polynomial \((x+2)((x-1)-\sqrt{7})((x-1)+\sqrt{7})\). The goal is to multiply these factors and simplify. Start by pairing the conjugate roots: \((x-1)^2 - (\sqrt{7})^2\), which simplifies using the difference of squares to \(x^2 - 2x - 6\). Now, multiply this simplified result with the remaining factor, \((x+2)\), to achieve the final simplified polynomial expression: \(x^3 - 4x + 6\).

Through simplification, the polynomial is more accessible and exhibits its degree clearly, making further mathematical operations easier to perform.