Slide 6- 1Copyright © 2012 Pearson Education, Inc.

Copyright © 2012 Pearson Education, Inc.

6.1 Introduction to Polynomial

Factorizations and Equations

■ Graphical Solutions

■ The Principle of Zero Products

■ Terms with Common Factors

■ Factoring by Grouping

■ Factoring and Equations

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Graphical Solutions

Whenever two polynomials are set equal to each other, the result is a polynomial equation.

In this section we learn how to solve polynomial equations both graphically and algebraically.

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ExampleSolve: x2 = 4x.SolutionIntersect Method: We can find the real-number solutions of a polynomial equation by finding the points of intersection of two graphs.

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continuedZero Method: Rewrite the equation so that one side is 0 and then find the x-intercepts of one graph or the zeros of the function.Rewrite the equation so that one side is 0: x2 = 4xx2 – 4x = 0

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ExampleFind the zeros of the function given by f(x) = x3 – 2x2 – 5x + 6.

SolutionGraph the equation, choosing a window that shows the x-intercepts of the graph. This may require several attempts.

To find the zeros use the ZERO option from the CALC menu.

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continued f(x) = x3 – 2x2 – 5x + 6

To find the zeros use the ZERO option from the CALC menu.

Use the same procedure for the other two zeros.

The zeros are 2, 1 and 3.

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The Principle of Zero Products

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ExampleSolve: (x – 4)(x + 3) = 0.SolutionAccording to the principle of zero products, at least one factor must be 0.x – 4 = 0 or x + 3 = 0 x = 4 or x = 3For 4: For 3:(x – 4)(x + 3) = 0 (x – 4)(x + 3) = 0(4 – 4)(4 + 3) = 0 (3 – 4)(3 + 3) = 0 0(7) = 0 0(7) = 0 0 = 0 TRUE 0 = 0 TRUE

(3, 0)

(4, 0)

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Terms with Common Factors

When factoring a polynomial, we look for factors common to every term and then use the distributive law.Multiply Factor

4x(x2 + 3x 4) 4x3 + 12x2 16x = 4xx2 + 4x3x 4x4 = 4xx2 + 4x3x 4x4 = 4x3 + 12x2 16x = 4x(x2 + 3x 4)

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Example

Factor: 28x6 + 32x3.SolutionThe prime factorization of 28x6 is

2 2 7 x x x x x xThe prime factorization of 32x3 is

2 2 2 2 2 x x x The largest common factor is 2 2 x x x or 4x3.28x6 + 32x3 = 4x3 7x3 + 4x3 8

= 4x3(7x3 + 8)

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Example Factor: 12x5 21x4 + 24x3

SolutionThe prime factorization of 12x5 is

2 2 3 x x x x xThe prime factorization of 21x4 is

3 7 x x x xThe prime factorization of 24x3 is

2 2 2 3 x x x The largest common factor is 3 x x x or 3x3.12x5 21x4 + 24x3 = 3x3 4x2 3x3 7x + 3x3 8

= 3x3(4x2 7x + 8)

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Tips for Factoring1. Factor out the largest common factor, if one exists. 2. The common factor multiplies a polynomial with the same number of terms as the original polynomial.3. Factoring can always be checked by multiplying. Multiplication should yield the original polynomial.

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Factoring by Grouping

Sometimes algebraic expressions contain a common factor with two or more terms.Example Factor x2(x + 2) + 3(x + 2).

Solution The binomial (x + 2) is a factor of both x2(x + 2) and 3(x + 2). Thus, x + 2 is a common factor.x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3

= (x + 2)(x2 + 3)

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Example

Write an equivalent expression by factoring.a) 3x3 + 9x2 + x + 3b) 9x4 + 6x 27x3 18Solutiona) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3)

= 3x2(x + 3) + 1(x + 3) = (x + 3)(3x2 + 1)

Don’t forget to include the 1.

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Example continued

b) 9x4 + 6x 27x3 18= (9x4 + 6x) + (27x3 18)= 3x(3x3 + 2) + (9)(3x3 + 2)= (3x3 + 2)(3x 9)= (3x3 + 2)3(x 3)= 3(3x3 + 2)(x 3)

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Factoring out ‒1

b – a = ‒1(a – b) = ‒(a – b)

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Example Factor: ax – bx + by – ay.

Solutionax – bx + by – ay = (ax – bx) + (by – ay)

= x(a – b) + y(b – a)= x(a – b) + y(‒1)(a – b)= x(a – b) – y(a – b)= (a – b)(x – y)

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Factoring and Equations

Example Solve: 7x2 = 35x.

Solution Use the principle of zero products if there is a 0 on one side of the equation and the other side is in factored form. 7x2 = 35x7x2 – 35x = 0 Subtracting 35x. One side is now 0.

7x(x – 5) = 0 Factoring

7x = 0 or x – 5 = 0 Use the principle of zero products

x = 0 or x = 5We check by substitution or graphically.

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To Use the Principle of Zero Products1. Write an equivalent equation with 0 on one side,

using the addition principle. 2. Factor the nonzero side of the equation.3. Set each factor that is not a constant equal to 0.4. Solve the resulting equations.