Problem 13 - Algebra 2 and Trigonometry

Question

In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ f(x)=x^{4}-10 x^{2}+9 $$

Step-by-Step Solution

Answer

The roots are \( x = 1, -1, 3, -3 \).

Step 1: Recognize the Quadratic Form

The function given is \( f(x) = x^{4} - 10x^{2} + 9 \). Recognize that this is a quadratic in form if we let \( y = x^2 \). Thus, the function becomes \( y^2 - 10y + 9 \).

Step 2: Factor the Quadratic

Attempt to factor the quadratic equation \( y^2 - 10y + 9 \). Look for two numbers that multiply to 9 and add to -10. These numbers are -1 and -9. So, \( y^2 - 10y + 9 = (y - 1)(y - 9) \).

Step 3: Substitute Back

Since \( y = x^2 \), substitute back to get \( (x^2 - 1)(x^2 - 9) \).

Step 4: Factor Completely

Factor each quadratic expression: \( x^2 - 1 = (x - 1)(x + 1) \) and \( x^2 - 9 = (x - 3)(x + 3) \). Thus, the complete factorization is \( (x - 1)(x + 1)(x - 3)(x + 3) \).

Step 5: Solve for Roots

Set each factor equal to zero to find the roots: \( x - 1 = 0 \), \( x + 1 = 0 \), \( x - 3 = 0 \), \( x + 3 = 0 \). Solving these equations, we get \( x = 1, -1, 3, -3 \).

Key Concepts

Factoring Polynomials

Factoring polynomials is a powerful method for simplifying equations, enabling us to find solutions more easily. When you factor a polynomial, you are essentially breaking it down into simpler "building block" expressions. This is much like taking a big, complicated Lego structure apart to find the smaller pieces that make it up.

In this exercise, we started with the polynomial \(x^4 - 10x^2 + 9\). Recognizing it as a quadratic in disguise, we dimmed it to \(y^2 - 10y + 9\) by setting \(y = x^2\). This makes it much simpler to solve. Factoring this results in \((y - 1)(y - 9)\) because \(-1\) and \(-9\) multiply to \(9\) and add to \(-10\).

Once factored, we substitute back \(y = x^2\) and factor further into \((x^2 - 1)(x^2 - 9)\), splitting these into the even simpler parts \((x - 1)(x + 1)\) and \((x - 3)(x + 3)\). Each of these polynomials represents a root or solution of our original equation. Factoring is often the easiest way to handle complex polynomials, and practice can make recognizing these patterns much easier.

Quadratic Formula

The quadratic formula is a universal tool used to find the roots of any quadratic equation. A quadratic equation is any polynomial that can be written in the form \(ax^2 + bx + c = 0\). The roots of this equation are given by the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula works because it is derived from completing the square of the general quadratic equation, which is a method of re-arranging the equation until it becomes a perfect square trinomial. The part under the square root, \(b^2 - 4ac\), is called the "discriminant." It indicates how many real roots the equation has:

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is one real double root.
• If \(b^2 - 4ac < 0\), there are no real roots, but two complex roots.

In the original exercise, we didn't need to use the quadratic formula since factoring worked perfectly, but understanding this formula is crucial. It's particularly useful when factoring is difficult or impossible.

Finding Roots of Equations

Finding roots—the solutions of polynomial equations where the polynomial equals zero—is the ultimate goal when solving these equations. Roots are the values of \(x\) that satisfy \(f(x) = 0\). In simple terms, they are where the graph of the polynomial crosses the x-axis.

Using the step-by-step solution provided, we found the roots of the polynomial \(x^4 - 10x^2 + 9\) through factoring. By breaking it down in parts, we found the complete factorization \((x - 1)(x + 1)(x - 3)(x + 3)\).

Setting each factor equal to zero, we solve:

• \(x - 1 = 0\), giving us \(x = 1\)
• \(x + 1 = 0\), giving us \(x = -1\)
• \(x - 3 = 0\), giving us \(x = 3\)
• \(x + 3 = 0\), giving us \(x = -3\)

These solutions—1, -1, 3, and -3—are where the polynomial equation \(x^4 - 10x^2 + 9 = 0\) has roots. Understanding how to find these roots helps us unlock the behavior of polynomial functions.