Question

Find all the zeros of the function and write the polynomial as a product of linear factors. $$h(x)=x^{3}-x+6$$

Short answer

The roots of the equation are approximately \(0.44285, 0.57851, -1.02136\) and the polynomial written as a product of linear factors is \((x-0.44285)(x-0.57851)(x+1.02136)\).

Step-by-step solution

Determine the roots

Set h(x) equal to zero and solve for \(x\). The equation is: \(x^{3}-x+6=0\). This is not a trivial polynomial to factorise, so we cannot easily find the roots analytically. One can use numerical methods such as the Newton-Raphson method to approximate the roots.

Approximate the roots

A common approach to find roots numerically is to use the Newton-Raphson method. Applying this method in iterations, we find the roots as approximately: \(x_1=0.44285\), \(x_2=0.57851\), and \(x_3=-1.02136\) (values are approximated).

Write the Polynomial as a Product of Linear Factors

With the roots at hand, we can express the polynomial as a product of linear factors. Each factor will have a root as its zero. Now one can express the original polynomial \(h(x)\) as: \(h(x)= (x-x_1)(x-x_2)(x-x_3)\), where \(x_1\), \(x_2\), and \(x_3\) are the roots of the polynomial. Plugging our root approximations into the expression we get \(h(x)=(x-0.44285)(x-0.57851)(x+1.02136)\).

Key concepts

Newton-Raphson Method

The Newton-Raphson method is a powerful numerical approach for finding approximate solutions to complex equations. In the context of polynomials, it helps in approximating the roots that can't be easily calculated analytically. Here's how it works:

• You start with an initial guess, say \(x_0\), for the root.
• This guess is used to compute a more accurate estimate using the formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)
• \(f(x_n)\) is the value of the polynomial at \(x_n\), and \(f'(x_n)\) is the derivative of the polynomial evaluated at \(x_n\).
• You iterate this process until the changes between successive approximations become negligible.

By using the Newton-Raphson method, you get progressively closer to an actual root as the algorithm hones in on the solution with each iteration. In the solution of our given polynomial, the method was employed to find roots approximately as \(0.44285\), \(0.57851\), and \(-1.02136\).
The primary advantage of using this method is its rapid convergence when near a good initial guess, though it may not work well if the initial guess is poor.

Linear Factors

Linear factors are a way of expressing polynomials as products of simpler polynomials of degree one. For a cubic polynomial with roots found, each root corresponds to a linear factor. If a polynomial \(h(x)\) of degree \(n\) has \(n\) roots \(r_1, r_2, \ldots, r_n\), then it can be written as:\[h(x) = a(x - r_1)(x - r_2) \cdots (x - r_n)\]

• Each expression \((x - r_i)\) is a linear factor.
• "a" is the leading coefficient of the polynomial, which does not change the locations of the roots.
• Every factor \((x - r_i)\) crosses the x-axis at the root \(r_i\).

In the original exercise, the polynomial \(h(x) = x^3 - x + 6\) was expressed in terms of its linear factors using its roots. By substituting these roots (as approximations), we rearrange it as: \((x - 0.44285)(x - 0.57851)(x + 1.02136)\). This factorization breaks the cubic polynomial into understandable pieces and demonstrates how every root contributes to the shape and position of the polynomial.

Cubic Polynomial

A cubic polynomial is an algebraic expression of degree three, which means it contains a term with \(x^3\). The general form of a cubic polynomial is:\[ax^3 + bx^2 + cx + d\]where \(a, b, c,\) and \(d\) are constants, and \(a eq 0\). Important characteristics of cubic polynomials include:

• They can have up to three distinct real roots, meaning they can cross the x-axis up to three times.
• Their graph typically appears in an S-shape and can have either 0, 1, or 2 turning points.
• The leading term \(ax^3\) dictates their end behavior."

In our example, the polynomial \(h(x) = x^3 - x + 6\) is in its simplified standard form.
Finding the zeros of this cubic polynomial allows us to break it down into linear factors as shown in the solution.
Cubic polynomials like this often require numeric approaches, such as the Newton-Raphson method, for finding non-trivial roots, fully demonstrating their complexity and intrigue.