Mathematics: Roots of Polynomials: An Introduction

Britannica Study Guide :: Roots of Polynomials: An Introduction :: Accessible/Printable ... Page 1 of 14

Mathematics: Roots of Polynomials: An Introduction

Contents

1. Multiplying two linear factors 2. Multiplying three linear factors 3. The purpose of factoring 4. Solving quadratic equations 5. Prime factors of polynomials: ambiguity in the word factor 6. Finding one factor from another: polynomial division 7. Finding a factor of a polynomial 8. Strategies for testing possible rational roots 9. Irrational roots Glossary Teacher's Notes Help

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Britannica Study Guide :: Roots of Polynomials: An Introduction :: Accessible/Printable ... Page 2 of 14

1. Multiplying two linear factors

Before you use this study guide, you should be familiar with factoring, solving, and graphing quadratics. You may want to refer to Britannica Study Guides on factoring, Factoring by Finding a Common Factor and Factoring Trinomials. You are going to learn techniques that allow you to factor polynomials of higher degree. As you will discover, these techniques are intimately related to finding the roots of the polynomials. One of the main goals of these techniques is to "reduce" the problem to a point where you can use the techniques that work with quadratics. That is why it is important to know how to factor a quadratic function and to solve a quadratic equation. The techniques we will learn are referred to as the Factor Theorem and the Rational Root Theorem. We begin with a review of multiplying two linear factors. Linear factors are expressions in the form, (ax + b), where x is a variable and a and b are real numbers. They are called linear factors because if you were to graph y = ax + b, you would get a straight line. They are of degree 1, since the highest exponent of the variable x is 1. Two linear factors can be multiplied using the FOIL (First Outer Inner Last) technique. Examples 1. (x + 2)(x ? 3) = x2 ? x ? 6 2. (x ? 6)(x + 5) = x2 ? x ? 30 3. (x ? 5)(x + 5) = x2 ? 25 4. (2x ? 1)(3x ? 4) = 6x2 ? 11x + 4 Click this icon to see an animated presentation of example 1, multiplied using FOIL. Click this icon to see example 2 multiplied by using a cross-problem format. [back to top] [next - Multiplying three linear factors]

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2. Multiplying three linear factors

When you multiply, you should generally work with two factors at a time. When presented with three factors, you can pick any two (since multiplication can be done in any order), and multiply them together using techniques from the previous screen. Example: Multiply (x ? 1)(x + 2)(x ? 3). We choose to multiply the second and the third factor because we already did that multiplication in the last screen. (x + 2)(x ? 3) = x2 ? x ? 6 We then need to multiply this result by the remaining linear factor. (x ? 1)(x2 ? x ? 6) Once you have begun multiplying polynomials that have more than two terms, you may find it useful to organize your work vertically, as shown by clicking on the icon. [back to top] [next - The purpose of factoring]

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3. The purpose of factoring

We can use polynomial equations to describe real-life situations. Finding the roots (x-intercepts) and turning points (peaks and valleys) to these equations yields important and meaningful information. In the case of business or physics situations, the results might indicate at what time an object will hit the ground, or when profits will be maximized. Click the icon to see a sample graph. The polynomial shown above has roots x = ?4, ?1, +5. Notice that roots are also called x-intercepts and zeros. In this guide, we will show that factoring into linear factors allows you to easily determine the roots. In calculus, you discover a simple method to find turning points for any equation. The points on the graph called roots are (?4, 0), (?1, 0), and (5, 0). It is worth emphasizing that these are points on the graph, and they are characterized by an important common feature--they all have 0 for the y-coordinate. [back to top] [next - Solving quadratic equations]

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4. Solving quadratic equations

There are four methods commonly used to solve a quadratic equation. We can solve by guessing and checking, by using the quadratic formula, by factoring the quadratic expression, or by completing the square. We show an example of how some of these techniques relate to each other.

We begin with the function f(x) = 2x2 + 5x ? 12. To graph this quadratic equation, you might want to know the x-intercepts, the y-intercept, and the turning point. Since the x-intercepts occur when f(x) = 0, we can replace f(x) with 0 and solve the equation 0 = 2x2 + 5x ? 12 . To solve this equation, guessing and checking is time-consuming; factoring is difficult when the coefficient of x2 is different from 1; and completing the square is generally difficult. Therefore we use the quadratic formula to solve this equation. Practice using the quadratic formula (which you can see when you click the first icon), and then check your answer by clicking the second icon.

As you can see, the solutions to the equation are x = 1.5 and ?4. This means that f(1.5) = 0, and f(?4) = 0, which means that the x-intercepts are at ?4 and 1.5, and that (1.5, 0), and (?4, 0) are points on the graph. Substituting zero for x in the equation results in f(0) = ?12. Thus the y-intercept is ?12 and the point (0, ?12) is on the graph. The graph of a quadratic equation is a special curve known as a parabola. All parabolas have the same general "U" shape, involving one turning point, known as its vertex. In analytic geometry you will discover that its vertex occurs when x = ?b/2a, from which the corresponding value of y can be determined and the point added to the graph. Click the icon to see the graph.

Since the number inside the square root of the quadratic formula (the determinant b2 ? 4ac) was a perfect square (121), the quadratic is factorable.

When you factor 2x2 + 5x ? 12, you get (2x ? 3)(x + 4). See the Britannica Study Guide Factoring Trinomials if you need to review this. This means that the equation above, which we solved by looking for x-intercepts, can be rewritten as follows: 2x2 + 5x ? 12 = 0 (2x ? 3)(x + 4) = 0 Using the zero product property, we can say that this means that solving the equations

2x ? 3 = 0 and x + 4 = 0

will give us solutions to the equation. The solutions to these are x = 1.5, and x = ?4, which match the previous roots.

Factoring enables us to find the x-intercepts. Sometimes factoring is faster than using the quadratic formula, and sometimes it is slower. Unfortunately, something like the quadratic formula does not exist for all polynomial functions. Factoring, and guessing and checking, are the only methods available until you learn calculus. The following screens review techniques that can help you factor and find roots.

[back to top] [next - Prime factors of polynomials: ambiguity in the word factor]

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