÷ of Polynomials, Remainder and Factor Theorems

DIVISION OF POLYNOMIALS
REMAINDER AND FACTOR THEOREMS

1. Division of Polynomials, Remainder and Factor Theorems  

2. OBJECTIVES: 1. Divides polynomials using Long Division. 2. Divides polynomials using Synthetic Division. 3. Finds the remainder when the given polynomial is divided by (x – c) using Remainder Theorem. 4. Uses the Factor Theorem to determine whether or not the given binomial (x – c) is a factor of the given polynomial. 5. Applies Factor Theorem and Remainder Theorem in solving problems. (ie. find the value of k...)

3. Division of Polynomials To divide polynomials, we use long division , as follows. DIVISION ALGORITHM: If P(x) and D(x) are polynomials, with D(x) 0, then there exists a unique polynomial Q(x) and R(x) , where R(x) is either 0 or of degree less than the degree of D(x), such that: = Q(x) + or P(x) = D(x) Q(x) + R(x) Dividend = Divisor Quotient + Remainder The polynomials P(x) and D(x) are called the dividend and divisor, respectively; Q(x) is the quotient , and R(x) is the remainder.  

4. A. LONG DIVISION Example 1. Divide (2x 3 – 5x 2 – 8x + 15) by (x – 3). Quotient: 2x 2 + x – 5

5. Example 2. Divide (3x 4 – 12 x + 5) by (x + 1). Solution: Quotient: 3x 3 – 3x 2 + 3x – 15 +  

6. Example 3. Divide (6x 3 + 11x 2 – 31x + 1) by (3x – 2). Solution: Quotient: 2x 2 + 5x – 7 –  

7. Example 4. Divide (3x 4 + x 3 – 17x 2 + 19x – 6 ) by (x 2 –2x + 1). Solution: Quotient: 3x 2 + 7x – 6

8. Example 5. Divide (2x 3 – 3x 2 + 4 x + 5 ) by (x + 2). Solution: Quotient: 2x 2 -7x + 18 -  

9. Example 6. Divide (8x 4 + 6x 2 – 3x + 1 ) by (2x 2 – x + 2). Solution: Quotient: 4 x 2 + 2 x +  

10. Example 7. Divide (– 3x 3 + 7x 2 – 11x – 1 ) by (3x – 1). Solution: Quotient: – x 2 + 2x – 3 –  

11. B . SYNTHETIC DIVISION Example 1. Divide (2x 3 – 5x 2 – 8x + 15) by (x – 3). Solution: 3 2 -5 – 8 15 6 3 -15 2 1 -5 Remainder Quotient: 2 x 2 + x – 5  

12. Example 2. Divide (3x 4 – 12 x + 5) by (x + 1). Solution: -1 3 0 -12 5 -3 3 - 3 15 3 -3 3 -15 20 Remainder Quotient: 3x 3 – 3x 2 +3 x – 15 +  

13. Example 3. Divide (6x 3 + 11x 2 – 31x + 1) by (3x – 2) Solution: 6 11 -31 1 2 4 10 -14 6 15 -21 -13 Remainder 3 2 5 -7 Quotient: 2 x 2 + 5 x – 7 –  

14. Example 4. Divide (2x 3 – 3x 2 + 4x + 5) by (x + 2) Solution: - 2 -3 4 5 -4 14 -36 2 -7 18 -31 Remainder Quotient: 2 x 2 – 7 x + 18 –  

15. Example 5. Divide ( – 3 x 3 + 7 x 2 – 11 x – 1 ) by (3x – 1 ) Solution: – 3 7 –11 – 1 -1 2 – 3 –3 6 –9 – 4 Remainder 3 -1 2 -3 Quotient: – x 2 + 2 x – 3 –  

16. Example 6. Divide ( 3 x 4 + x 3 – 17 x 2 + 19x – 6 ) by (x 2 – 2x + 1 ) Solution: 3 1 -17 19 -6 6 -3 -7 6 14 -12 3 7 -6 0 0 Quotient: 3 x 2 + 7 x – 6

17. Example 7. Divide ( 8 x 4 + 6 x 2 – 3 x + 1 ) by (2x 2 – x + 2 ) Solution: 8 0 6 – 3 1 4 -8 -4 2 8 4 0 -7 1 2 4 2 Quotient: 4 x 2 + 2 x +  

18. The Remainder Theorem: If the polynomial P (x) is divided by (x – c), the remainder R is a constant and is equal to P(c). Hence, R = P(c) Thus, there are two ways to find the remainder when P(x) is divided by (x – c), that is: 1. Use synthetic division, or 2. Calculate P(c). Similarly, there are two ways to find the value of P(c): 1. Substitute c in the polynomial expression P(x), or 2. Use synthetic division. The computation in finding the remainder when the polynomial P(x) is divided by x – c can be easily determined by simply evaluating P (x) for x = c. In other words, simply find P(c).

19. Examples: Find the remainder when the polynomial P(x) is divided by (x – c): P(x) = 2 + 5x – 5 is divided by ( x + 2). Solution 1: By using remainder theorem P(-2) = 2 + 5(-2) – 5 = 2(-8) – 4(4) – 10 – 5 = – 16 – 16 – 15 P(-2) = – 47 Hence, the remainder is – 47. Solution 2: By using synthetic division -2 2 -4 5 -5 -4 16 -42 2 -8 21 -47 Remainder  

20. 2. P(x) = 3 + 4 – 3x + 4 is divided by ( x – 1 ). Solution 1: By using remainder theorem P(1) = 3 + 4 – 3(1) + 4 = 3(1) – 6(1) + 4(1) – 3(1) + 4 = 3 – 6 + 4 – 3 + 4 = 2 Hence, the remainder is 2 . Solution 2: By using synthetic division 1 3 -6 4 -3 4 3 -3 1 -2 3 -3 1 -2 2 Remainder  

21. 3. P(x) = 4 + 4 - 7 x + 6 is divided by (x +3 ). Solution 1: By using remainder theorem P(-3) = 4 + 4 – 7(-3) + 6 = 4(-243) – 2(-27) + 4(9) + 21 + 6 = -972 + 54 + 36 + 21 + 6 = -972 + 117 = -855 Hence, the remainder is -855 . Solution 2: By using synthetic division -3 4 0 -2 4 -7 6 -12 36 -102 294 -861 4 -12 34 -98 287 -855 Remainder  

22. 4. P(x) = 5 + 6 - 7 + 6x – 4 is divided by (x +1 ). Solution : By using remainder theorem P(-1) = 5 + 6 – 7 + 6(-1) – 4 = 5(-1) – 3(1) + 6(-1) – 7(1) – 6 – 4 = -5 – 3 – 6 – 7 – 6 – 4 = -37 Hence, the remainder is –37 . 5. P(x) = 2 + 3 - 8x + 6 is divided by (2x +3 ). Solution : By using remainder theorem P = + 6 = 2 + 6 = + + + 12 + 6 = + + + 18 = + = = Hence, the remainder is  

23. The Factor Theorem: The polynomial P(x) has x – r as a factor if and only if P(r) = 0. Proof: There are two parts of the proof of the Factor theorem, namely: 1. If x – r is a factor of P(x), then P(r) = 0. 2. If P(r) = 0, then (x –r) is a factor of P(x). Example 1: Show that (x – 2) is a factor of P(x) = 2 - 2x + 24. Solution: P(2) = 2 = 2(8) – 9(4) – 4 + 24 = 16 – 36 + 20 = 36 – 36 = 0 Hence, (x – 2 ) is a factor of P(x)  

24. Example 2: Determine whether (x – 3) is a factor of P(x) = 4 + 6x – 9 . Solution: P(3) = 4 = 4 (27) – 3(9) + 18 – 9 = 108 – 27 + 9 = 90 Since the remainder is 90 and not equal to zero, then (x – 3 ) is not a factor of P(x). Example 3: Is (5x + 1) a factor of P(x) = 5 – 29x – 6? Solution: P = 5 - 6 = 5 +6 + - 6 = + + - 6 = - 6 = - 6 = 6 – 6 = 0 Clearly, the remainder is zero, then (5x + 1) is a factor of P(x).  

25. Example 4: Determine whether the given polynomial P(x) is Divisible by the given binomial (x – r) a) P(x) = 2 + 8x – 4 ; (x + 2) Solution: P(-2) = 2 = -16 – 28 – 20 = -64 Since the remainder is – 64, then P(x) is not Divisible by (x + 2) b) P(x) = 2 – 5x + 6 ; (x – 3) Solution: P(3) = 2 = 162 – 27 – 126 – 9 = 162 – 162 = 0 Since the remainder is 0, then P(x) is Divisible by (x – 3 )  

26. Example 5: Find the remainder when P(x) is divided by the D(x) and determine if D(x) is a factor of P(x) . P(x) D(x) Remainder R(x) Factor or Not Factor? 3x 3 – 17x 2 + 18x + 8 x – 2     2. 2x 3 – 3x 2 – 4x + 6 x + 2   3. 2x 4 - 11x 3 +10x 2 + 11x – 12 x – 4     4. 4x 4 + 4x 3 -8x 2 + 9x + 6 x + 3    

27. Applications of Remainder and F actor Theorems: Example 6: Find the value of k so that (x + 2) is a factor of: a) 3 + 3kx + 5 Solution: P(-2) = 0 P(-2) = 3 0 = -24 -8k -6k + 5 14k = -19; k = b) 2k – 6x + 4 Solution: P(-2) = 0 P(-2) = 0 = 32k + 40 + 12 + 16 -32k = 68; k = -  

28. Example 7: Find the value of k so that – 4 is the remainder when 6k - 5x + 4 is divided by x + 2? Solution: P(-2) = -4 P(-2) = 6k -4 = -48k + 12k + 14 -4 = -36k + 14 -36k = -18 k = Example 8: Find the value of k so that x + 2 is a factor of 2k – 6x + 4. Solution: P(-2) = 0 P(-2) = 0 = 32k + 40 + 12 + 16 -32k = 68; k = -  

29. Example 9: Find the values of A & B so that x + 1 is a factor of P(x) = 3A + 4 x + 5 and when divided by x – 2 the remainder is 6. Solution: i ) P(-1) = 0 and ii) P(2) = 6 P(-1) = 3A 0 = -3A + 2B + 1 3A – 2B = 1 (Eq. 1) P(2) = 3A 6 = 24A + 8B + 13 24A + 8B = -7 (Eq.2) 24A + 8B = -7 3A – 2B = 1 12A + 5 = 4 -8 (3A – 2B = 1) 3A– 2 = 1 12A = -1 24A + 8B = -7 3A + = 1 A = -24A + 16B = -8 24B = -15 B =  

30. Example 10: Find the values of A & B so that x – 1 and x + 2 are both factors of P(x) = 4A + 2 x + 6. Solution: i ) P(1) = 0 and ii) P(-2) = 0 P(1) = 4A 0 = 4A - 3B + 8 4A – 3B = -8 (Eq. 1) P(-2) = 4A 0 = -32A - 12B + 2 -32A -12B = -2 (Eq.2) -32A -12B = -2 4A – 3B = -8 8A -11 = -16 8 (4A – 3B = -8) 4A– 3 = -8 8A = -16+11 8A = -5 -32A -12B = -2 4A - = -8 A = 32A -24B = -64 -36B = -66 B =