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Chapter 12: Problem 3

Find the zeros of the polynomials. $$ x^{2}-x-6 $$

Short Answer

Expert verified
Answer: The zeros of the polynomial equation $$x^2 - x - 6$$ are x = 3 and x = -2.

Step by step solution

01

Given quadratic equation

The given polynomial equation is: $$ x^2 - x - 6 $$
02

Factoring the quadratic equation

Factor the given equation to find the factors of the quadratic equation. We look for two numbers whose sum is -1 and the product is -6. $$ (x - 3)(x + 2) $$
03

Set each factor to zero

Set each factor equal to zero to find the values of x that make the polynomial equal to zero. $$ x - 3 = 0 \quad \text{or} \quad x + 2 = 0 $$
04

Solve for x

Solving the above equations for x, we get the following zeros: $$ x = 3 \quad \text{or} \quad x = -2 $$ Therefore, the zeros of the polynomial $$x^2 - x - 6$$ are x = 3 and x = -2.

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Key Concepts

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

Factoring
When dealing with quadratic equations like \(x^2 - x - 6\), factoring is a valuable technique. It involves re-expressing the equation as a product of two simpler binomials. The goal is to break down the equation into factors that, when multiplied, will give you the original quadratic expression.

Start by identifying two numbers that multiply to the constant term (-6) and add up to the linear coefficient (-1). These numbers are -3 and 2. Consequently, the quadratic equation \(x^2 - x - 6\) can be factored as \((x - 3)(x + 2)\).

This factorization is crucial because it simplifies the process of finding the solutions or "zeros" of the quadratic equation.
Polynomial Zeros
The zeros of a polynomial are the values of \(x\) for which the polynomial is equal to zero. In the context of the equation \(x^2 - x - 6\), we are interested in finding the values of \(x\) that will make the entire expression equal to zero.

Once you have factored the quadratic into \((x - 3)(x + 2)\), set each factor equal to zero. This gives:
  • \(x - 3 = 0\)
  • \(x + 2 = 0\)

Solving these equations provides the zeros of the polynomial, which are the values of \(x\) that satisfy the equation. These zeros are important because they represent the x-intercepts of the graph of the quadratic function.
Solving for x
After setting each factor of the polynomial equal to zero, the next step is solving for \(x\).

Consider each equation separately:
  • For \(x - 3 = 0\), add 3 to both sides to solve for \(x\), yielding \(x = 3\).
  • For \(x + 2 = 0\), subtract 2 from both sides to solve for \(x\), resulting in \(x = -2\).

Hence, the solutions to the equation \(x^2 - x - 6 = 0\) are \(x = 3\) and \(x = -2\). These solutions, or roots, of the equation are the x-values where the parabola represented by the quadratic equation intersects the x-axis. Understanding this process is key for solving various types of polynomial equations.

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Most popular questions from this chapter

For what values of \(a\) does the equation have a solution in \(x\) ? $$ a^{2}+a x^{2}=0 $$

Find two different polynomials of degree 3 with zeros \(1,2,\) and \(3 .\)

Refer to the functions \(f(x)\) and \(g(x),\) where the function $$ g(x)=1+\frac{1}{2} x+\frac{3}{8} x^{2}+\frac{5}{16} x^{3} $$ is used to approximate the values of $$ f(x)=\frac{1}{\sqrt{1-x}} $$ Show that \(f(x)\) is undefined at \(x=1\) and \(x=2\), but that \(g(x)\) is defined at these values. Explain why the algebraic operations used to define \(f\) may lead to undefined values, whereas the operations used to define \(g\) will not.

$$ \begin{aligned} &\text { Find the solutions of }\\\ &\left(x^{2}-a^{2}\right)(x+1)=0, \quad a \text { a constant } \end{aligned} $$

What is the degree of the resulting polynomial? The sum of a degree 8 polynomial and a degree 4 polynomial.
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