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Unit 2 – quadratic, polynomial, and radical equations and inequalities. Chapter 6 – Polynomial Functions 6.7 – The Remainder and Factor Theorems. 6.7 – The Remainder and Factor Theorems. In this section we will learn how to: Evaluate functions using synthetic substitution

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unit 2 quadratic polynomial and radical equations and inequalities

Unit 2 – quadratic, polynomial, and radical equations and inequalities

Chapter 6 – Polynomial Functions

6.7 – The Remainder and Factor Theorems

6 7 the remainder and factor theorems
6.7 – The Remainder and Factor Theorems
  • In this section we will learn how to:
    • Evaluate functions using synthetic substitution
    • Determine whether a binomial is a factor of a polynomial by using synthetic substitution
6 7 the remainder and factor theorems1
6.7 – The Remainder and Factor Theorems
  • Synthetic Substitution
    • Synthetic division can be used to find the value of a function. Consider the polynomial function f(a) = 4a2 – 3a + 6. Divide the polynomial by a – 2.
    • Compare the remainder of 16 to f(2).

f(2) = 4(2)2 – 3(2) + 6

= 16 – 6 + 6

= 16 *Notice that the value of f(2) is the same as the remainder when

the polynomial is divided by a – 2. This illustrates the Remainder Theorem

6 7 the remainder and factor theorems2
6.7 – The Remainder and Factor Theorems
  • Remainder Theorem
  • If a polynomial f(x) is divided by x – a, the remainder is the constant f(a), and
    • f(x) = q(x)  (x – a) + f(a)
    • dividend equals quotient times divisor plus remainder
    • where q(x) is a polynomial with degree one less than the degree of f(x)
6 7 the remainder and factor theorems3
6.7 – The Remainder and Factor Theorems
  • When synthetic division is used to evaluate a function, it is called synthetic substitution.
  • It is a convenient way of finding the value of a function, especially when the degree of the polynomial is greater than 2.
6 7 the remainder and factor theorems4
6.7 – The Remainder and Factor Theorems
  • Example 1
    • If f(x) = 3x4 – 2x3 + x2 – 2, find f(4).
6 7 the remainder and factor theorems5
6.7 – The Remainder and Factor Theorems
  • Factors of Polynomials
    • Synthetic division shows that the quotient of x4 + x3 – 17x2 – 20x + 32 and x – 4 is x3 + 5x2 + 3x – 8.
    • When you divide a polynomial by one of its binomial factors, the quotient is called a depressed polynomial. From the results of the division and by using the Remainder Theorem, we can make the following statement.
6 7 the remainder and factor theorems6
6.7 – The Remainder and Factor Theorems
  • x4+ x3 – 17x2 – 20x + 32 = (x3 + 5x2 + 3x – 8)  (x – 4) + 0
  • dividend equals quotient times divisor plus remainder
  • Since the remainder is 0, f(4) = 0. This means that x – 4 is a factor of x4 + x3 – 17x2 – 20x + 32. This illustrated the Factor Theorem, which is a special case of the Remainder Theorem.
6 7 the remainder and factor theorems7
6.7 – The Remainder and Factor Theorems
  • Factor Theorem
  • The binomial x – a is a factor of the polynomial f(x) if and only if f(a) = 0.
  • If x – a is a factor of f(x), then f(a) has a factor of (a – a) or 0. Since a factor of f(a) is 0, f(a) = 0.
  • Now assume that f(a) = 0. If f(a) = 0, then the Remainder Theorem states that the remainder is 0 when f(x) is divided by x – a. This means that x – a is a factor of f(x). This proves the Factor Theorem.
6 7 the remainder and factor theorems8
6.7 – The Remainder and Factor Theorems
  • Example 2
    • Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial.
6 7 the remainder and factor theorems9
6.7 – The Remainder and Factor Theorems
  • Example 3
    • The volume of the rectangular prism is given by V(x) = x3 + 7x2 + 2x – 40. Find binomials with integer coefficients for the missing measures.

x – 2

6 7 the remainder and factor theorems10
6.7 – The Remainder and Factor Theorems

HOMEWORK

Page 359

#11 – 27odd, 28, 38 – 41

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