3.2 Remainder Theorem.pptxdmddmmdmmdmdmdmd

3.2 Remainder Theorem.pptx

1. 3.2 The Remainder Theorem

2. Quotient Dividend Remainder Divisor Long Division

3. Long Division - A Review 8 4 6 3 1 5 4 0 Divisor Remainder Dividend Quotient 7 8 6 5 6 7 6 4 7 3 1 4 3 8 4 96

4. Divide: ( x 2 + 7 x + 2) ÷ ( x + 2) 1. The polynomial must be in descending order of powers. Any missing terms are to be filled with a zero placeholder. x 2 + 7 x + 2 x + 2 2. Only the first term is used when doing the division. Divide x 2 x x 3. Multiply your answer with the entire divisor. x ( x + 2) = x 2 + 2 x x 2 + 2 x 4. Subtract, bring down the next term and repeat the process. 5 x + 5 5 x + 10 -8 = x + 2 multiply Division by a Binomial -( ) -( ) , x  -2

5. 4 x 3 - 11 x 2 + 8 x + 10 x - 2 4 x 2 4 x 3 - 8 x 2 - 3 x 2 - 3 x - 3 x 2 + 6 x 2 x + 2 2 x - 4 14 + 8 x + 10 Division by a Binomial NPV’s

6. x 3 + 0 x 2 - 20 x + 8 x + 4 x 2 x 3 + 4 x 2 - 4 x 2 - 4 x - 4 x 2 - 16 x -4 x - 4 -4 x - 16 24 - 20 x + 8 NPV’s

7. Divide x 3 - 2 x 2 - 33 x + 90 by ( x - 5) using synthetic division . 5 1 -2 -33 90 1. Write only the constant term of the divisor, and the coefficients of the dividend. 2. Bring down the first term of the dividend. 1 3. Multiply 1 by -5 , record the product and subtract . 5 Multiply 3 4. Multiply 3 by -5 , record the product and subtract . 15 -18 5. Multiply -18 by -5 , record the product and subtract. addition -90 Quotient Rem Written as x 2 + 3 x - 18 Using the division statement: P( x ) = ( x - 5)( x 2 + 3 x - 18) Synthetic Division

8. Divide: ( x 4 - 2 x 3 + x 2 + 12 x - 6) ÷ ( x - 2) -2 1 -2 1 12 -6 1 -2 1 -2 14 -28 22 ( x - 2) ( x 3 + x + 14) + 22 Using Synthetic Division x 4 - 2 x 3 + x 2 + 12 x – 6 =

9. Given P ( x ) = x 3 - 4 x 2 + 5 x + 1 , determine the remainder when P ( x ) is divided by x - 1 . -1 1 -4 5 1 1 -1 -3 3 2 -2 3 The remainder is 3. Using f ( x ) = x 3 - 4 x 2 + 5 x + 1, determine P ( 1 ): P ( 1 ) = ( 1 ) 3 - 4( 1 ) 2 + 5( 1 ) + 1 = 1 - 4 + 5 + 1 = 3 NOTE: P (1) gives the same answer as the remainder using synthetic division. Therefore P (1) is equal to the remainder . In other words, when the polynomial x 3 - 4 x 2 + 5 x + 1 is divided by x - 1 , the remainder is P (1). The Remainder Theorem

10. Remainder Theorem: When a polynomial P ( x ) is divided by x - a , the remainder is P( a ). [ think x - a , then x = a ] Determine the remainder when x 3 - 4 x 2 + 5 x - 1 is divided by: a) x - 2 b) x + 1 Calculate P ( 2 ) P ( 2 ) = ( 2 ) 3 - 4( 2 ) 2 + 5( 2 ) - 1 = 8 - 16 + 10 - 1 = 1 The remainder is 1. Calculate P (-1 ) P ( -1 ) = ( -1 ) 3 - 4( -1 ) 2 + 5( -1 ) - 1 = -1 - 4 - 5 - 1 = -11 The r emainder is -11. Point (2, 1) is on the graph of of f ( x ) = x 3 - 4 x 2 + 5 x - 1 Point (-1, -11) is on the graph of of f ( x ) = x 3 - 4 x 2 + 5 x - 1

11. Determine the value of k . When the remainder is 30. is divided by

12. Problem Solving When the polynomial 3 x 3 + ax 2 + bx -9 is divided by x - 2 , the remainder is -5. When the polynomial is divided by x + 1, the remainder is -16. What are the values of a and b ?

13. Assignment Page 124 1, 3c,f, 4a,c, 6a,7b, 8a,c, 9, 11, 14

14. 1. (4 x 3 - 11 x 2 + 8 x + 6) ÷ ( x - 2) -2 4 -11 8 6 4 - 8 -3 6 2 -4 10 P( x ) = ( x - 2)(4 x 2 - 3 x + 2) + 10 2. (2 x 3 - 2 x 2 + 3 x + 3) ÷ ( x - 1) -1 2 -2 3 3 -2 0 -3 2 0 3 6 P( x ) = ( x - 1)(2 x 2 + 3) + 6 Using Long Division,Synthetic Division and Remainder Theorem