Exercises Solve each equation. 1. = 49 2. 6 = 3. 3 = 54 – –8 – 4. 3 48 = 0 5. 16 + 64 = 0 6. 5 + 4 = 0 – – – 29 + 100 = 0 4 73 + 144 = 0 7. 8. – – 5-6 Study Guide and Intervention The Remainder and Factor Theorems Synthetic Substitution Remainder The remainder, when you divide the polynomial f (x ) by (x a), is the constant f (a). – Theorem f(x) = q(x) (x a) + f(a), where q(x) is a polynomial with degree one less than the degree of f(x). – ⋅ Example 1: If = 3 + 2 5 + 2, find f(x) x f(–2). – – Synthetic Substitution Method 1: Method 2 Direct Substitution Replace with x –2. By the Remainder Theorem, should be the f(–2) = 3 + 2– 5 + 2 f(x) x – remainder when you divide the polynomial by + 2. x = 3 + 2– 5+ (–2) 2 f(–2) – 3 2 1 –2 –5 –2 = 48 16 20 2 2 or 8 – – – – 8 10 So = 8 –6 –6 f(–2) 3 3 8 –4 –5 The remainder is 8, so = 8. f(–2) Example 2: If = 5+ 2x 1, find f(x) f(3). – Again, by the Remainder Theorem, should be the remainder when you divide the polynomial by 3. f(3) x – 3 5 0 2 –1
Exercises Solve each equation. 1. = 49 2. – 6 = –8 3. – 3 = 54 4. 3 – 48 = 0 5. – 16 + 64 = 0 6. – 5 + 4 = 0 7. – 29 + 100 = 0 8. 4 – 73 + 144 = 0 5-6 Study Guide and Intervention The Remainder and Factor Theorems Synthetic Substitution Remainder Theorem The remainder, when you divide the polynomial f (x ) by (x – a), is the constant f (a). f(x) = q(x) ⋅ (x – a) + f(a), where q(x) is a polynomial with degree one less than the degree of f(x). Example 1: If f(x) = 3 + 2 – 5 + x – 2, find f(–2). Method 1: Synthetic Substitution By the Remainder Theorem, f(–2) should be the remainder when you divide the polynomial by x + 2. –2 3 2 –5 1 –2 –6 8 –6 10 3 –4 3 –5 8 The remainder is 8, so f(–2) = 8. Method 2 Direct Substitution Replace x with –2. f(x) = 3 + 2– 5 + x – 2 f(–2) = 3 + 2– 5+ (–2) – 2 = 48 – 16 – 20 – 2 – 2 or 8 So f(–2) = 8 Example 2: If f(x) = 5+ 2x – 1, find f(3). Again, by the Remainder Theorem, f(3) should be the remainder when you divide the polynomial by x – 3. 3 5 0 2 –1
chapter_5_study_guide.docx