Problem 19 - College Algebra with Corequisite Support

Question

For the following exercises, find the \(x\) - or t-intercepts of the polynomial functions. $$ f(x)=2 x^{4}+6 x^{2}-8 $$

Step-by-Step Solution

Answer

The x-intercepts are at \(x = 1\) and \(x = -1\).

Step 1: Set the Function to Zero

To find the x-intercepts, we need to set the polynomial function equal to zero. This means solving the equation: \[2x^4 + 6x^2 - 8 = 0\]

Step 2: Factor by Grouping

Notice that each term in the polynomial is even, allowing us to factor out a 2:\[2(x^4 + 3x^2 - 4) = 0\]Next, we need to factor the quadratic trinomial \(x^4 + 3x^2 - 4\). Let \(y = x^2\), then substitute:\[y^2 + 3y - 4 = 0\]

Step 3: Solve the Quadratic Equation

Now solve the quadratic equation \(y^2 + 3y - 4 = 0\). We factor it:\[(y+4)(y-1) = 0\]This gives the solutions \(y = -4\) and \(y = 1\). Since \(y = x^2\), we have:\[x^2 = -4\] and \[x^2 = 1\].

Step 4: Solve for x

The equation \(x^2 = -4\) has no real solutions, so we ignore it for real x-intercepts. The equation \(x^2 = 1\) gives:\[x = \pm 1\]. Thus, the real x-intercepts are \(x = 1\) and \(x = -1\).

Key Concepts

x-intercepts

When dealing with polynomial functions, finding the x-intercepts is a crucial step. The x-intercepts are the points where the graph of the function crosses the x-axis. In other words, these are the values of \( x \) where the function \( f(x) \) equals zero. To find them, you need to solve the equation \( f(x) = 0 \).

For example, in the case where \( f(x) = 2x^4 + 6x^2 - 8 \), setting it to zero is the first step. Solving this equation will reveal the x-values that make the polynomial zero. This equation will sometimes need methods like factoring or using the quadratic formula, especially when it’s not simple to observe the solutions directly. Finding x-intercepts is essential in understanding the behavior and shape of the polynomial graph.

solving equations

Solving equations involves finding the value of the variable that makes the equation true. In the context of a polynomial equation like \( 2x^4 + 6x^2 - 8 = 0 \), the goal is to isolate \( x \) and find its values. This often requires breaking down the equation into simpler parts.

• First, simplify the equation if possible by factoring out common elements. This makes the equation easier to solve.
• Next, attempt to use algebraic techniques such as substitution, especially if the degree is higher than two.
• Lastly, make sure to check the solutions in the context of the problem, as some solutions might be extraneous, particularly if you deal with square roots.

Solving polynomial equations is a skill that enhances problem-solving abilities and leads to a deeper understanding of mathematical functions.

factoring polynomials

Factoring polynomials is a method of expressing a polynomial as a product of its simplest parts. This technique is particularly useful when solving polynomial equations, as it breaks them down into more manageable pieces.

For example, in \( 2x^4 + 6x^2 - 8 \), you first factor out 2 to simplify: \( 2(x^4 + 3x^2 - 4) = 0 \). The next step involves factoring the quadratic trinomial inside the parentheses. Using substitution by letting \( y = x^2 \), it becomes \( y^2 + 3y - 4 = 0 \).

• This leads to finding factors like \((y+4)(y-1)\). Once factored, each expression set to zero can be solved to find \( y \), and subsequently \( x \) from \( y = x^2 \).

Mastering polynomial factoring is not only necessary for solving equations but also useful for analyzing polynomials' roots and their graphs.

quadratic equations

Quadratic equations often appear within larger polynomial functions and follow the general form \( ax^2 + bx + c = 0 \). Solving quadratics is essential when working with higher-powered polynomials.

In our example, after simplifying \( 2x^4 + 6x^2 - 8 = 0 \), we deal with the quadratic \( y^2 + 3y - 4 = 0 \). The solutions to this quadratic will lead us to solutions for \( x \).

• Using methods like factoring, completing the square, or the quadratic formula, we can solve for \( y \) and then revert to \( x \) by substituting back.
• The solutions of a quadratic, such as \( y = 1 \) and ignoring \( y = -4 \) for no real solutions when \( y = x^2 \), lead directly back to finding solutions for the original polynomial equation.

Understanding quadratic equations is crucial as they provide foundational skills necessary to tackle more complex algebraic problems.