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Chapter 10: Problem 17

Find a polynomial equation with real coefficients that has the given roots. $$-3,3$$

Short Answer

Expert verified
The polynomial equation is x^2 - 9 = 0.

Step by step solution

01

Understand the given roots

The given roots of the polynomial are -3 and 3. These are the values of x that make the polynomial equal to zero.
02

Form the factors

Convert each root into a factor. For the root -3, the factor is (x + 3). For the root 3, the factor is (x - 3).
03

Multiply the factors

Multiply the factors to form the polynomial equation: (x + 3)(x - 3).
04

Simplify the expression

Use the difference of squares formula: (a + b)(a - b) = a^2 - b^2. So, (x + 3)(x - 3) = x^2 - 9.
05

Write the final polynomial equation

The polynomial equation with the given roots (-3, 3) and real coefficients is x^2 - 9 = 0.

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Key Concepts

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

Roots of Polynomials
In algebra, the term 'roots' of a polynomial refers to the values of the variable that make the polynomial equal to zero. Here, the roots given are -3 and 3. To determine the polynomial from these roots, we need to understand that each root corresponds to a factor of the polynomial.

For instance, if -3 is a root, then (x + 3) is a factor. Similarly, if 3 is a root, then (x - 3) is another factor. When we combine these factors, we build the polynomial equation. So, knowing the roots is crucial because they help us form the factors which in turn form the polynomial equation.
Real Coefficients
A polynomial with real coefficients means that all the coefficients in the polynomial are real numbers. In our example, the coefficients are from the polynomial x^2 - 9. This polynomial has a real number coefficient of 1 for x^2 and -9 for the constant term.

Having real coefficients ensures that our polynomial equation is easy to work with and understand. It means the calculations won't involve complex numbers, making equations simpler for most introductory algebra purposes.
Difference of Squares
The difference of squares is a specific algebraic formula used to simplify the multiplication of binomials of the form (a + b)(a - b). This formula states: (a^2 - b^2).

In our example, multiplying the factors (x + 3) and (x - 3) results in (x^2 - 9), because we can apply the difference of squares formula, equating it to: (x^2 - 3^2).

This formula is very useful because it allows us to quickly and accurately simplify polynomial expressions. Recognizing and using the difference of squares can save significant time and effort in algebra.

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Most popular questions from this chapter

For each equation find the value of \(k\) given that 3 satisfies the equation. a) \(x^{4}-3 x^{3}+5 x^{2}-7 x+k=0\) b) \(x^{4}-x^{3}-2 x^{2}+k x-k=0\) c) \(5 x^{3}-k x^{2}-k x-3 k=0\)

Let \(f(x)=x^{4}-1, g(x)=x^{3}-3 x^{2}+5,\) and \(h(x)=4 x^{4}-\) \(3 x^{2}+3 x-1 .\) Find the following function values by using two different methods. See Example \(I\) $$ h\left(\frac{1}{2}\right) $$

In each case, find a polynomial function whose graph behaves in the required manner. Answers may vary. The graph has only two \(x\) -intercepts at \((5,0)\) and \((-6,0)\) It does not cross the \(x\) -axis at either \(x\) -intercept.

State the degree of each polynomial equation. Find all of the real and imaginary roots to each equation. State the multiplicity of a root when it is greater than 1. $$x^{4}+2 x^{3}+x^{2}=0$$

Use the rational root theorem, Descartes' rule of signs, and the theorem on bounds as aids in finding all solutions to each equation. $$x^{4}-5 x^{3}+5 x^{2}+5 x-6=0$$
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