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Chapter 4: Problem 16

For the function \(g(x)=x^{5}-9 x^{3},\) solve each of the following. $$g(x)=0$$

Short Answer

Expert verified
The solutions are \(x = 0, x = 3, x = -3 \).

Step by step solution

01

- Set the function equal to zero

First, we need to set the given function equal to zero: \(g(x) = x^5 - 9x^3 = 0 \)
02

- Factor the equation

Next, factor out the common term in the equation: \(x^3 (x^2 - 9) = 0 \)
03

- Solve each factor separately

Solve for each factor separately. Start with the first factor: \( x^3 = 0 \) The solution is \(x = 0 \). Now, solve the second factor: \(x^2 - 9 = 0 \)
04

- Solve the quadratic equation

Solve \(x^2 - 9 = 0 \) by adding 9 to both sides: \(x^2 = 9 \) Take the square root of both sides: \( x = \pm\sqrt{9} \)
05

- Find the final solutions

The solutions to \( x^2 = 9 \) are \( x = 3 \) and \( x = -3 \). Therefore, all solutions to the equation \( g(x) = 0 \) are: \( x = 0, x = 3, x = -3 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring polynomials
Factoring polynomials involves breaking down a polynomial into simpler components called factors. This makes equations easier to solve. In our example, we started with the function \(g(x) = x^5 - 9x^3\). To factor this equation, the first step is to find common terms. We noticed that both terms, \(x^5\) and \(9x^3\), have a common factor of \(x^3\).

By factoring \(x^3\) out, the equation turns into \(x^3(x^2 - 9) = 0\). Now, solving the quadratic part, \(x^2 - 9\), becomes much simpler. Remember, always look for common factors and factor them out to simplify your work. This method is essential for solving polynomial equations efficiently.
quadratic equations
Quadratic equations are in the form \(ax^2 + bx + c = 0\). They can be solved using various methods such as factoring, completing the square, or using the quadratic formula.

In this exercise, we encountered a quadratic equation \(x^2 - 9 = 0\). This equation can be factored since it's a difference of squares. The difference of squares means something like \(a^2 - b^2 = (a + b)(a - b)\).

For our equation, \(x^2 - 9\) can be rewritten as \((x + 3)(x - 3) = 0\). Factoring makes it simple to find the solutions: \(x = 3\) and \(x = -3\). This method is sometimes quicker than using the quadratic formula and helps in recognizing special patterns in quadratic equations.
roots of equations
Roots of equations, also known as solutions, are the values of the variable that satisfy the equation. When we solve an equation like \(g(x) = 0\), we're finding its roots.

In our example, through factoring and solving the quadratic equation, we found the roots. First, we solved \(x^3 = 0\), giving \(x = 0\). Then, for the quadratic \(x^2 - 9 = 0\), we solved it to find \(x = 3\) and \(x = -3\).

So, the roots of \(x^5 - 9x^3 = 0\) are \(x = 0, 3, -3\). Finding roots is crucial because it tells us where the function intersects the x-axis, helping us understand the behavior of the polynomial.

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Most popular questions from this chapter

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