C H A P T ER Polynomial Equations The Fundamental Theorem of Algebra, which states that any polynomial equation of degree n must have exactly n complex roots or solutions, was first proposed by Peter Rothe in 1608. Over the next two centuries, many mathematicians attempted proofs of the Theorem, but it was not until 1814 that Swiss mathematician Jean-Robert Argand published what is believed to be the first complete and formal proof. You will determine all the roots of a polynomial equation using several methods..1 Higher Order Polynomials and Factoring Roots of Polynomial Equations p. 365.2 Greater Than or Less Than Solving Polynomial Inequalities p. 373.3 It s Fundamental The Fundamental Theorem of Algebra p. 383.5 Remains of a Polynomial The Remainder Theorem p. 397.6 More Factors of a Polynomial The Factor Theorem p. 403.7 Let s Be Rational Rational Root Theorem p. 409.8 Getting to the Root of it All Solving Polynomial Equations p. 417.4 When Division is Substitution Polynomial and Synthetic Division p. 387 Chapter l Polynomial Equations 363
364 Chapter l Polynomial Equations
.1 Higher Order Polynomials and Factoring Roots of Polynomial Equations Objective In this lesson you will: l Calculate the roots of polynomial equations using factoring. Problem 1 Solving Polynomial Equations using Factoring Previously, you solved quadratic equations using factoring. For example, solve x 2 7x 0 using factoring. (x 2)(x 5) 0 x 2 0 or x 5 0 x 2 or x 5 Check: ( 2) 2 7( 2) 4 14 0 ( 5) 2 7( 5) 25 35 0 Factoring can also be used to solve some higher order polynomial equations. 1. Solve each polynomial equation using factoring. Then, check your solution(s). a. x 3 3x 2 x 3 Lesson.1 l Roots of Polynomial Equations 365
b. 0 x 4 5x 2 6 c. x 4 29x 2 0 0 d. x 4 4x 3 x 2 4x 0 e. x 4 1 366 Chapter l Polynomial Equations
2. What types of quadratic equations can be solved using factoring? 3. If a quadratic equation cannot be solved using factoring, what other methods can be used to solve the equation? 4. What types of polynomial equations can be solved using factoring? 5. If a polynomial equation cannot be solved using factoring, what other methods can be used to solve the equation? Formulas have been developed to solve any cubic equation and any quartic equations; however, the formulas are complex and difficult to use. Other methods for solving polynomial equations will be explored later in this course. Lesson.1 l Roots of Polynomial Equations 367
Problem 2 Solving Polynomial Equations Solve each polynomial equation. Then, check your solution(s). 1. x 2 7x 4 0 368 Chapter l Polynomial Equations
2. x 3 3x 2 x Lesson.1 l Roots of Polynomial Equations 369
3. 0 x 3 5x 2 6x 30 4. x 4 7x 2 18 0 370 Chapter l Polynomial Equations
5. 3x 3 6x 2 9x 0 Lesson.1 l Roots of Polynomial Equations 371
6. x 5 x 4 3x 3 3x 2 4x 4 0 Be prepared to share your methods and solutions. 372 Chapter l Polynomial Equations
.2 Greater Than or Less Than Solving Polynomial Inequalities Objective In this lesson you will: l Determine solutions to polynomial inequalities algebraically and graphically. Problem 1 Solving Polynomial Inequalities Algebraically 1. The solutions to the quadratic inequality x 2 4x 21 0 can be determined algebraically by the following steps. a. Factor the inequality. b. If ab 0, then what must be true about the values of a and b? Lesson.2 l Solving Polynomial Inequalities 373
c. Write two separate compound inequalities to solve x 2 4x 21 0 based on your answer from Question 1, part (b). Then, solve each compound inequality. d. What is the solution(s) to the quadratic inequality x 2 4x 21 0? 374 Chapter l Polynomial Equations
2. The solutions to the polynomial inequality x 3 5x 2 4x 20 0 can be determined algebraically by the following steps. a. Factor the inequality. b. If abc 0, then what must be true about the values of a, b, and c? Lesson.2 l Solving Polynomial Inequalities 375
c. Write compound inequalities to solve x 3 5x 2 4x 20 0 based on your answer from Question 2, part (b). Then solve each compound inequality. d. What is the solution(s) to the polynomial inequality x 3 5x 2 4x 20 0? 376 Chapter l Polynomial Equations
3. Solve x 4 6x 2 5 0. Lesson.2 l Solving Polynomial Inequalities 377
As the degree of a polynomial inequality increases, the number of possible cases increases. So, this method is useful for smaller degree polynomial inequalities, but less useful for higher degree polynomial inequalities. Problem 2 Solving Polynomial Inequalities Graphically A graphical method can also be used to determine the solutions to a polynomial inequality. 1. The solutions to the quadratic inequality x 2 4x 21 0 can be determined graphically by the following steps. a. Solve the corresponding quadratic equation x 2 4x 21 0. b. Graph y x 2 4x 21. c. Determine the values of x for which the values of y are less than or equal to zero using the graph. 378 Chapter l Polynomial Equations
d. How does the solution in Problem 2, Question 1 compare to the solution in Problem 1, Question 1? 2. The solutions to the polynomial inequality 2x 3 4x 2 12x 24 0 can be determined graphically by the following steps. a. Solve the corresponding polynomial equation 2x 3 4x 2 12x 24 0. b. Graph y 2x 3 4x 2 12x 24. Lesson.2 l Solving Polynomial Inequalities 379
c. Why is the scale on the y-axis unimportant? d. What is the solution to the inequality 2x 3 4x 2 12x 24 0? 380 Chapter l Polynomial Equations
3. Solve x 5 4x 3 x 2 4 0. Be prepared to share your methods and solutions. Lesson.2 l Solving Polynomial Inequalities 381
382 Chapter l Polynomial Equations
.3 It s Fundamental The Fundamental Theorem of Algebra Objectives In this lesson you will: l State the Fundamental Theorem of Algebra. l Use the Fundamental Theorem of Algebra to determine roots and zeros of polynomial equations and functions. l Determine the number and characteristics of the roots and zeros of polynomial equations and functions. Key Terms l Fundamental Theorem of Algebra l complex roots l double roots l triple roots Problem 1 Fundamental Theorem of Algebra Take Note The Fundamental Theorem of Algebra was first proposed in the early 1600s, but would not be proven until almost two centuries later. Both simple and elegant, it states that any polynomial equation of degree n must have exactly n complex roots or solutions; also, every polynomial function of degree n must have exactly n complex zeros. However, any solution or zero may be a multiple solution or zero. The proof of this theorem is beyond the content of this course, but determining all the roots of a polynomial equation is not. When a root or zero appears more than once, it is said to have a multiplicity. For example, double roots have a multiplicity of 2, and triple roots have a multiplicity of 3. 1. x 2 1 0 For each of the following polynomial equations, state the number of roots, and calculate all of the roots by any method (factoring; successive approximation; or graphing, tracing, and zooming). 2. x 2 1 0 Lesson.3 l The Fundamental Theorem of Algebra 383
3. 7x 3 5x 2 6x 0 4. x 4 5x 2 4 0 5. x 4 x 2 3 0 384 Chapter l Polynomial Equations
Problem 2 State the number of zeros and calculate all of the zeros by any method (factoring; successive approximation; or graphing, tracing, and zooming) for each of the following polynomial functions. 1. f(x) x 4 2. f(x) x 5 13x 3 36x Lesson.3 l The Fundamental Theorem of Algebra 385
3. f(x) x 5 5x 4 x 5 4. f(x) x 3 7x 2 2x 1 Calculating roots and zeros can be very time consuming; this is especially true for irrational or complex roots. In the next lessons, you will concentrate on identifying and calculating rational roots and zeros. Be prepared to share your methods and solutions. 386 Chapter l Polynomial Equations
.4 When Division is Substitution Polynomial and Synthetic Division Objectives In this lesson you will: l Divide polynomials. l Use synthetic division. Key Term l synthetic division Problem 1 The Fundamental Theorem of Algebra tells you that every polynomial equation of degree n must have n roots. This means that every polynomial can be written as the product of n factors of the form ax b. For example, 2x 2 3x 9 (2x 3)(x 3) 0. Once you have factored the polynomial completely, you can calculate the roots by setting each factor equal to 0. You also know that a factor of a number must divide into the number evenly with a remainder of zero. Factors of polynomials must also divide a polynomial evenly without a remainder. The division algorithm for dividing one polynomial by another is very similar to the long division algorithm for whole numbers. Lesson.4 l Polynomial and Synthetic Division 387
The following table illustrates the division algorithms for both whole numbers and polynomials. 3 25 ) 775 ( Divide : 25 ) 2x 77 ) x 3 ) 2x 2 3x 9 ( Divide: x ) 2x ) 2 3 25 ) 775 (Multiply: 3(25)) 75 3 25 ) 775 (Subtract) 75 2 3 25 ) 775 75 (Bring down) 25 3 1 25 ) 775 ( Divide : 2 ) 2 ) 75 25 31 25 ) 775 (Multiply: 1(25)) 75 25 25 (Subtract) 0 2x x 3 ) 2x 2 3x 9 (Multiply: 2x(x 3)) 2x 2 6x 2x x 3 ) 2x 2 3x 9 (Subtract) 2x 2 6x 3x 2x x 3 ) 2x 2 3x 9 2x 2 6x (Bring down) 3x 9 2x 3 x 3 ) 2x 2 3x 9 ( Divide : x ) 3x ) 2x 2 6x 3x 9 2x 3 x 3 ) 2x 2 3x 9 (Multiply: 3(x 3)) 2x 2 6x 3x 9 3x 9 (Subtract) 0 1. This algorithm of Divide Multiply Subtract Bring down Repeat is familiar and straightforward. Determine the quotient for each of the following polynomial division problems. a. x ) 4x 3 0x 2 7x 388 Chapter l Polynomial Equations
b. x 4 ) x 3 2x 2 5x 16 c. x 1 ) x 3 0x 2 0x 1 d. 2x 3 ) 4x 4 0x 3 5x 2 7x 9 Lesson.4 l Polynomial and Synthetic Division 389
e. 3x 2 ) 9x 4 3x 3 4x 2 7x 2 2. Answer the following questions about Question 1, parts (a) through (e). a. What do you notice in parts (a), (c), and (d)? Why was this necessary? b. When there was a remainder, was the divisor a factor of the dividend? Explain. c. Describe the remainder when you divide a polynomial by a factor. A whole number like 775 that is divided evenly by a factor like 25 can be written as a product of factors, 775 25(31). When a polynomial equation a n x n a n 1 x n 1 a 2 x 2 a 1 x a 0 x 0, or a polynomial function f( x) a n x n a n 1 x n 1 a 2 x 2 a 1 x a 0 x 0 is divided evenly by a factor, you can write each as a product of two polynomials: a n x n a n 1 x n 1 a 2 x 2 a 1 x a 0 x 0 (x r)(b n 1 x n 1 b n 2 x n 2 b 2 x 2 b 1 x b 0 ) f a n x n a n 1 x n 1 a 2 x 2 a 1 x a 0 x 0 or (x r)(b n 1 x n 1 b n 2 x n 2 b 2 x 2 b 1 x b 0 ), f(x) (x r )q(x), where q(x) is the quotient polynomial. 390 Chapter l Polynomial Equations
3. Rewrite the dividend as the product of the divisor and the quotient for each division problem in Question 1 that has a remainder of 0. a. 4x 3 0x 2 7x b. x 3 0x 2 0x 1 c. 9x 4 3x 3 4x 2 7x 2 When you divide a whole number by a number that is not a factor, you can say that it does not divide evenly and has a remainder. You write the remainder as a fraction. For instance: 8 12 ) 1 96 5 1 12 8 5 12 When one polynomial does not divide evenly into another, you can write the answer in a similar form: a n x n a n 1 x n 1 a 2 x 2 a 1 x a 0 x 0 x r b n 1 x n 1 b n 2 x n 2 b 2 x 2 b 1 x b 0 R x r, f( x) or with functions r( x) x r q(x) x r, where q(x) is the quotient and r(x) is the remainder. 4. Rewrite the dividend as the product of the divisor and the quotient plus the remainder for each division problem in Question 1 that has a remainder other than 0. a. x 2x 2 5x 6 b. 4x 4 0x 3 5x 2 7x 9 Lesson.4 l Polynomial and Synthetic Division 391
5. Perform the indicated division and write the answer as a product of the divisor and the quotient plus the remainder for each problem. a. Divide f(x) 4x 4 3x 3 5x 2 2x 1 by x 2 b. Divide f(x) 3x 4 2x 3 0x 2 2x 1 by x 3 392 Chapter l Polynomial Equations
Problem 2 Although dividing polynomials is a straightforward method for determining factors, it can become very time consuming. In 1809, Paolo Ruffini introduced a shortcut for long division of a polynomial by a linear factor (x r), which is called synthetic division. Synthetic division makes this division more efficient. It uses only the coefficients of the terms. The following example compares long division of polynomials with synthetic division. Long Division Synthetic Division 2x 3 x 3 ) 2x 2 3x 9 2x 2 6x 3x 9 3x 9 0 3 2 3 9 2 3 0 3 2 3 9 2 Bring down the 2 3 2 3 9 2 6 Multiply 3 by 2 and place in the next column 3 2 3 9 6 2 3 Add the values in the second column 3 2 3 9 6 9 2 3 0 Repeat the process until complete The quotient is 2x 3 and the remainder is 0. Notice that the opposite sign of r is used, and that every power must have a place holder as in long division. Lesson.4 l Polynomial and Synthetic Division 393
1. Here are two examples of synthetic division. Perform the following steps for each problem: i. Write the dividend. ii. Write the divisor. iii. Write the quotient. iv. Write the dividend as the product of the divisor and the quotient plus the remainder. a. 2 1 0 4 3 6 2 4 0 6 1 2 0 3 0 i. ii. iii. iv. b. 3 2 4 4 3 6 6 30 78 243 2 26 81 249 i. ii. iii. iv. 2. Use synthetic division to perform the following divisions. Write the dividend as the product of the divisor and the quotient plus the remainder. a. x 4 ) x 3 2x 2 5x 6 b. f(x) 3x 3 4x 2 8 divided by x 2 394 Chapter l Polynomial Equations
Take Note Synthetic division works only for (x r) divisors. So you need to rewrite 2x 3 to the equivalent root x 3 2 before doing the synthetic division. 2x 3 0 x 3 2 x 3 2 0 c. 2x5 5x 4 2x 2 6x 7 2x 3 d. g(x) 4x 4 2x 3 4x 2 2x 11 r(x) 2x 1 Calculate g(x) r(x). Be prepared to share your methods and solutions. Lesson.4 l Polynomial and Synthetic Division 395
396 Chapter l Polynomial Equations
.5 Remains of a Polynomial The Remainder Theorem Objectives In this lesson you will: l Use synthetic substitution. l Use the Remainder Theorem to evaluate polynomial equations and functions. Key Terms l synthetic substitution l Remainder Theorem Problem 1 Remainder Theorem From the Fundamental Theorem of Algebra, you know that any polynomial equation of degree n has n roots, and with the more efficient method of synthetic division, calculating them becomes less tedious. You are also able to rewrite any polynomial equation or function as the product of a divisor and a quotient plus the remainder, f(x) d(x)q(x) r(x). If you are trying to find roots or zeros, however, rewriting the equation or function as the product of a linear divisor by a polynomial quotient plus a whole number remainder is more useful. Algebraically, this form is written f(x) (x r)q(x) R. Rewrite each of the following polynomials as the product of (x 3) and the quotient plus the remainder. 1. x 3 27 2. f(x) x 4 2x 3 3x 2 Lesson.5 l The Remainder Theorem 397
3. Evaluate each of the polynomials in Questions 1 and 2 for x 3. Compare these values to your answers to Questions 1 and 2. 4. You can write any polynomial as the product of a linear factor and a quotient polynomial plus a whole number remainder as follows: f(x) (x r)q(x) R. Calculate f(r). What can you conclude about f(r) and the whole number remainder R? Explain. This result of using synthetic division to evaluate a polynomial function for a specific value r is called synthetic substitution because f(r) R, the remainder. This result also provides the basis for the Remainder Theorem, which states that when any polynomial equation or function is divided by a linear factor (x r), the remainder is the value of the equation or function when x r. 5. Determine the remainder of each equation or function by evaluating it at the given value. a. 3x 6 2x 3 3x 2 divided by x 1. b. f(x) 3x 5 4x 4 2x 3 x 2 5x 3 divided by x 2 c. x 3 2x 2 x 2 2x 1 d. g(x) 4x 3 x 2; 2x 3 398 Chapter l Polynomial Equations
Problem 2 Signs and Signs 1. Evaluate the polynomial 4x 3 5x 2 for each value using synthetic substitution. a. x 5 b. x 0.25 c. x 0.7 d. x 1 Lesson.5 l The Remainder Theorem 399
2. Evaluate the polynomial x 4 x 3 39x 2 9x 270 for each value using synthetic substitution. a. x 6 b. x 4 c. x 2 d. x 4 e. x 7 400 Chapter l Polynomial Equations
3. In Questions 1 and 2, the sign of values of the polynomials changed between each value of x. You know that a polynomial equation is defined for every value of x. What can you conclude about at least one value of the polynomial between each of these values of x? Be prepared to share your methods and solutions. Lesson.5 l The Remainder Theorem 401
402 Chapter l Polynomial Equations
.6 More Factors of a Polynomial The Factor Theorem Objectives In this lesson you will: l Use the Factor Theorem to determine if a polynomial is a factor of another polynomial. l Use the Factor Theorem to calculate factors of polynomial equations and functions. l Determine the products and sums of roots of polynomial equations. Key Term l Factor Theorem Problem 1 Factor Theorem 1. When you divide a whole number by another whole number and the remainder is zero, what conclusion can you draw about these two numbers? 2. Using the Remainder Theorem, you divide a polynomial by a linear polynomial. If the remainder is zero, what can you conclude about these two polynomials? This result is the Factor Theorem, which states that a polynomial has a linear polynomial as a factor if and only if the remainder is zero; f(x) has x r as a factor if and only if f(r) 0. 3. For each of the following, determine if the second polynomial is a factor of the first. a. 2x 3 3x 2 3x 2; x 1 Lesson.6 l The Factor Theorem 403
b. x 4 3x 3 5x 2; x 2 c. 5x 4 3x 2 2; x 3 4. For each of the following, determine if the given linear polynomial is a factor of the polynomial function. a. f(x) x 7 3x 2; x 1 b. f(x) x 7 3x 2; x 1 c. g(x) 4x 3 2x 2 6x 5; 2x 1 You can also use the Factor Theorem to completely factor higher-order polynomials: 2x 3 3x 2 3x 2 (x 1)(2x 2 x 2) x b b 2 4ac 1 (1) 2 4(2)(2) 1 15 1 2a 2(2) 4 4 15 4 i 2x 3 3x 2 3x 2 (x 1)(2x 2 x 2) (x 1) ( x 1 4 15 4 i ) ( x 1 4 15 4 i ) 404 Chapter l Polynomial Equations
5. Determine if the second polynomial expression is a factor of the first polynomial for each problem. Then, completely factor the first polynomial expression. a. x 3 1; x 1 b. x 4 3x 2 28; x 2, x 2 c. x 4 x 3 x 2 x 2; x 1 Lesson.6 l The Factor Theorem 405
Using the Fundamental Theorem of Algebra, the Remainder Theorem, and the Factor Theorem, every polynomial equation or function of degree n can be rewritten as the product of n linear factors of the form x r, where r is a complex number. In future lessons, you will use this information to develop a method to calculate all of the rational roots or zeros of any polynomial equation or function. Problem 2 Reversing the Process 1. Determine an equation that would have each of the following sets of roots. a. x 1, 2, 3 b. x 2, 5 2 2. After completing Question 3 of Problem 1, a student said that anytime there is one complex root, it seems that there actually must be two, the root and its conjugate. Is this student correct? Explain. 3. Determine the equation that would have each of the following sets of roots. a. x i, 3 4 b. x 2 3 3 i, 2 2 406 Chapter l Polynomial Equations
c. x 1 i, 1 2, 3 2 Problem 3 Products and Sums of Roots 1. Calculate the product and sum of each set of roots. a. The roots of the equation x 2 4x 1 0 are x 2 5, 2 5. Product: Sum: b. The roots of the equation x 3 2x 2 5x 6 0 are x 1, 2, 3. Product: Sum: c. The roots of the equation 2x 3 5x 2 8x 20 0 are x 2, 5 2. Product: Sum: d. The roots of the equation 4x 3 3x 2 4x 3 0 are x i, 3 4. Product: Sum: Lesson.6 l The Factor Theorem 407
e. The roots of the equation 36x 3 24x 2 43x 86 0 are x 2 3 3 i, 2. 2 Product: Sum: f. The roots of the equation 4x 4 12x 3 13x 2 2x 6 0 are x 1 i, 1 2, 3 2. Product: Sum: 2. Compare the sums of the roots to the first two coefficients of the polynomial equations. What conclusion can you draw? 3. Compare the products of the roots to the first and last coefficients of the odd degree polynomial equations. What conclusion can you draw? 4. Compare the products of the roots to the first and last coefficients of the even degree polynomial equations. What conclusion can you draw? Be prepared to share your methods and solutions. 408 Chapter l Polynomial Equations
.7 Let s Be Rational Rational Root Theorem Objective In this lesson you will: l Use the Rational Root Theorem to determine the rational roots of polynomial equations. Key Term l Rational Root Theorem Problem 1 Rational Root Theorem The Fundamental Theorem of Algebra states that every polynomial of degree n has n roots. So, every polynomial can be written as the product of n linear factors of the form (b n x c n ) where each factor represents a root or solution x c n. b n a n x n a n 1 x n 1... a 2 x 2 a 1 x a 0 x 0 (b n x c n )(b n 1 x c n 1 )...(b 1 x c 1 ) 0 Applying the Remainder Theorem and the Factor Theorem together, a n x n... Q, where Q is a polynomial of degree n 1 and there is no remainder. Also, the product of the roots is a 0 a for even degree polynomials and a 0 n a for odd degree polynomials. n The Rational Root Theorem states that a rational root of a polynomial a n x n a n 1 x n 1... a 2 x 2 a 1 x a 0 x 0 must be of the form p q, where p is a factor of the constant term and q is a factor of the leading coefficient. 1. List all possible rational roots for each polynomial. a. x 2 7x b. x 3 3x 2 x 3 Lesson.7 l Rational Root Theorem 409
c. 2x 4 5x 2 6 d. 4x 3 5x 7 2. The product of the roots is by definition a rational number. a. The irrational roots must be pairs in the form x d e _ f, where d, e, and f are rational numbers. Why? b. The complex roots must also be pairs of conjugates of the form x g hi. Why? The Rational Root Theorem is useful to solve higher order polynomial equations by reducing the number of possible solutions. 4 Chapter l Polynomial Equations
Problem 2 Determining Rational Roots 1. Solve the polynomial equation 2x 2 x 6 0 by performing the following steps. a. List the possible rational roots of the equation. Then, use synthetic division to determine one of the rational roots. b. Rewrite the equation in factored form. c. What are the solutions to the equation? Lesson.7 l Rational Root Theorem 411
To determine all the roots or solutions of a polynomial equation: l Determine the possible rational roots. l Use synthetic division to determine one of the roots. l Rewrite the original polynomial as a product. l Determine the possible rational roots of the quotient. l Repeat the process until all the rational roots are determined. l Factor the remaining polynomial to determine any irrational or complex roots. l Recall that some roots may have a multiplicity. 2. Determine all roots for x 4 7x 2 18 0. a. Determine the possible roots. b. Use synthetic division to determine one of the roots. 412 Chapter l Polynomial Equations
c. Rewrite the original polynomial as a product. d. Determine the possible rational roots of the quotient. e. Use synthetic division to determine one of the roots. f. Rewrite the original polynomial as a product. Lesson.7 l Rational Root Theorem 413
g. Determine the possible rational roots of the quotient. h. Determine the remaining roots. i. Rewrite the original polynomial as a product. 3. Determine all roots for each equation. a. x 4 5x 2 6x 2 0 414 Chapter l Polynomial Equations
b. 4x 3 5x 2 0 Be prepared to share your methods and solutions. Lesson.7 l Rational Root Theorem 415
416 Chapter l Polynomial Equations
.8 Getting to the Root of it All Solving Polynomial Equations Objective In this lesson you will: l Determine all roots of polynomial equations. Problem 1 Determine all roots of each polynomial equation. Then, classify each root as rational, irrational, imaginary, or complex. Lesson.8 l Solving Polynomial Equations 417
1. 2x 4 7x 3 11x 2 28x 12 0 418 Chapter l Polynomial Equations
2. x 4 x 3 17x 2 21x 4 0 Lesson.8 l Solving Polynomial Equations 419
3. x 4 x 3 2x 2 4x 8 0 420 Chapter l Polynomial Equations
4. x 6 x 5 2x 3 27x 2 35x 0 Be prepared to share your methods and solutions. Lesson.8 l Solving Polynomial Equations 421
422 Chapter l Polynomial Equations