Problem 245 - Precalculus

Question

For the following exercises, find the \(x\) - or \(t\) -intercepts of the polynomial functions. $$ f(x)=x^{5}-5 x^{3}+4 x $$

Step-by-Step Solution

Answer

The x-intercepts are \( x = 0, \pm 1, \pm 2 \).

Step 1: Understand the Problem

To find the x-intercepts of the polynomial function \( f(x) = x^5 - 5x^3 + 4x \), we need to find the values of \( x \) where \( f(x) = 0 \).

Step 2: Set the Equation to Zero

Set \( f(x) = 0 \) so that we have: \[ x^5 - 5x^3 + 4x = 0 \] This equation is what we'll solve to find the values for \( x \).

Step 3: Factor the Polynomial

Factor out the common term \( x \) from each term in the polynomial to simplify the equation: \[ x(x^4 - 5x^2 + 4) = 0 \]. This gives us a factored equation that is easier to solve.

Step 4: Set Each Factor to Zero

The product of two factors equals zero if at least one of the factors is zero. Thus, set each factor to zero: \( x = 0 \) or \( x^4 - 5x^2 + 4 = 0 \).

Step 5: Solve for x in the Quadratic Equation

Recognize that \( x^4 - 5x^2 + 4 = 0 \) is a quadratic in terms of \( x^2 \). Let \( y = x^2 \), leading to \( y^2 - 5y + 4 = 0 \).

Step 6: Solve the Quadratic Equation

Factor the quadratic equation: \( (y - 1)(y - 4) = 0 \). Thus, we have two solutions: \( y = 1 \) and \( y = 4 \).

Step 7: Solve for x from y

Substitute back to \( x^2 \):- For \( y = 1 \), \( x^2 = 1 \), so \( x = \pm 1 \).- For \( y = 4 \), \( x^2 = 4 \), so \( x = \pm 2 \).

Step 8: Compile the x-intercepts

Now, collecting all solutions from the factors, the x-intercepts are \( x = 0, \pm 1, \pm 2 \).

Key Concepts

x-intercepts

Understanding x-intercepts is crucial when dealing with polynomial functions. An x-intercept of a function is a point where the graph crosses the x-axis. This is where the output of the function, often represented by an equation like \( f(x) = 0 \), equals zero.

To find x-intercepts for any polynomial function, you set the equation to zero and solve for x. It’s similar to finding the roots of the polynomial.

For example, consider the function \( f(x) = x^5 - 5x^3 + 4x \). To find its x-intercepts, you solve \( x^5 - 5x^3 + 4x = 0 \). This involves determining the x values where the function’s output is zero.

factoring polynomials

Factoring polynomials is a fundamental algebraic method for simplifying expressions and solving equations. When you factor a polynomial, you rewrite it as a product of simpler polynomials.

In the context of finding x-intercepts, factoring helps simplify the problem significantly. Consider our polynomial \( f(x) = x^5 - 5x^3 + 4x \). By factoring out the common term \( x \), we get \( x(x^4 - 5x^2 + 4) = 0 \). This simplifies the equation, making it more manageable.

Once factored, each separate factor (like \( x \) or \( x^4 - 5x^2 + 4 \)) can be solved individually to find possible values of x for the intercepts. When one of these factors equals zero, the entire expression is zero, revealing an x-intercept.

quadratic equations

Quadratic equations play a crucial role when dealing with polynomials, especially when breaking down higher degree polynomials into simpler components.

Recognize that some polynomials can be transformed into a quadratic form with substitution. For the polynomial \( x^4 - 5x^2 + 4 \) in our factored expression, you can substitute \( y = x^2 \) to simplify it into a quadratic equation \( y^2 - 5y + 4 \).

Solving this using factoring gives \( (y - 1)(y - 4) = 0 \), resulting in \( y = 1 \) and \( y = 4 \). You then convert these solutions back to x by solving \( x^2 = y \).

You find \( x = \pm 1 \) and \( x = \pm 2 \) through this back-substitution, showing how quadratic solving techniques help find roots or x-intercepts of higher degree polynomials.