Question

Write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros. \(-6,0,3-\sqrt{5}\)

Short answer

The polynomial function of least degree with a leading coefficient of 1 and given the roots -6, 0, and \(3-\sqrt{5}\) is \(f(x) = x^4+24x\).

Step-by-step solution

Understanding the Root Pair

We first note that there is an irrational root \(3 - \sqrt{5}\) . For a polynomial with real-coefficients, the conjugate of this root, \(3 + \sqrt{5}\) must also be a root.

Construct Polynomial from Roots

We next construct the polynomial from the roots. This is done by setting each factor of the polynomial equal to 0 and solving for x, giving us: \(f(x) = (x+6)(x)(x-3+\sqrt{5})(x-3-\sqrt{5})\).

Simplify Polynomial

The next step is to simplify the polynomial by multiplying the terms out. For this we will first multiplies the last two factors to remove the square root term: \(f(x) = (x+6)(x)((x-3)^2 - \sqrt{5}^2)\). Which simplifies to: \(f(x) = (x+6)x(x^2 - 6x + 9 - 5)\) and finally: \(f(x) = (x+6)x(x^2 - 6x + 4)\).

Expanding Polynomial

Finally, we expand the polynomial by multiplying all the terms out: \(f(x) = x(x+6)(x^2-6x+4) = x(x^3-6x^2+4x+6x^2-36x+24) = x(x^3+24) = x^4+24x\).

Key concepts

Root Pairs

When dealing with polynomial functions, especially if they have real coefficients, it's essential to understand the concept of root pairs. If the polynomial has an irrational root (a root that involves a square root or any non-quadratic number), then its conjugate must also exist as a root. In simple terms, if you have a root in the form of \(a + \,\sqrt{b}\), you will also have \(a - \,\sqrt{b}\) as a root. This happens because real polynomial functions can't have roots that produce complex parts without their pairs balancing the equation when expanded.

• For the given root \(3 - \sqrt{5}\), its pair will automatically be \(3 + \sqrt{5}\).
• This guarantees that no complex or imaginary part remains in the final polynomial expression.

This ensures the polynomial remains real when expanded.

Leading Coefficient

The leading coefficient is the coefficient of the term with the highest degree in a polynomial function. In the polynomial created from given roots, it defines the overall orientation and scale of the graph. If you are asked to ensure that the leading coefficient is 1, it means the equation should not have any scalar multiplier affecting the highest degree term. This is often a requirement for simplicity and standardization. For the equation created using the roots given in the problem, we start with \(f(x) = (x+6)(x)(x-3+\sqrt{5})(x-3-\sqrt{5})\).

• Since each factor is multiplied together, and none of them has a leading coefficient other than 1, the overall leading coefficient of the polynomial naturally is 1.
• This means, in the expanded polynomial \(x^4+24x\), the term \(x^4\) correctly has a leading coefficient of 1.

This provides a standard form most polynomial applications require.

Real-Coefficients

Real coefficients are numbers that are not imaginary, meaning they have no imaginary component (involving \(i\), the square root of -1). For polynomials with real coefficients, if one root is irrational or complex, its conjugate must also be a root. This requirement ensures that when multiplied and expanded, any imaginary components cancel out, leaving an expression comprised entirely of real numbers.

• In this exercise, all coefficients are real, representing the fact that irrational roots and their conjugates will simplify to real coefficients when expanded.
• The simplified polynomial \(x^4+24x\) contains only real numbers.

This aligns with the expectation that polynomial functions used in most forms should be expressed in terms of real numbers.

Irrational Roots

Irrational roots are roots that cannot be expressed as simple fractions because they include irrational numbers. In our context, they involve square roots of non-perfect squares, resulting in non-repeating, non-terminating decimals. Given the root \(3 - \sqrt{5}\), this is an example of an irrational root as \(\sqrt{5}\) cannot be simplified to a rational number. Working with polynomials having real coefficients, if one irrational root is present, its conjugate, \(3 + \sqrt{5}\), also needs to be included.

• This ensures that when the polynomial is expanded, the irrational parts cancel each other.
• When roots \((x - 3 + \sqrt{5})(x - 3 - \sqrt{5})\) are expanded, they simplify to a form containing only real numbers.

Understanding how to manage irrational roots is essential for forming correct polynomial functions that meet given criteria without leaving irrational numbers in the coefficients.