Math-3. Lesson 3-1 Finding Zeroes of NOT nice 3rd Degree Polynomials

Transcription

1 Math- Lesson - Finding Zeroes of NOT nice rd Degree Polynomials

2 f ( ) Is this one of the nice rd degree polynomials? a) Sum or difference of two cubes: y 8 5 y 7 b) rd degree with no constant term. c) nice pattern that can be factored by grouping: y y 4 We are left with guessing the first degree polynomial that divides it evenly!

3 Try the bo method ( )( a b c)

4 YES, we can find the quadratic factor! ( )( a b c) ( )( 8)

5 Try: = linear factor would be ( ) y

6 Try: = linear factor would be ( ) y There s got to be an easier way!!!

7 Long Division Look at the left numbers divides 54 4 times = 48 Subtract 48 from 54 Bring the down. 5 times = 60; subtract 60 from Remainder Divisor

8 Your turn: Look at the left numbers 7 divides 44 times 7 = Subtract from 4 Bring the down. 4 times 7 = 8; subtract 8 from Remainder Divisor

9 Steps ) Look at left-most numbers ) What # times left = left? ) Multiply 4) Subtract 5) Bring down. 6) Repeat steps Remainder Divisor

10 Vocabulary Polynomial Long division: One method used to divide polynomials similar to long division for numbers ( ) a b c Divide Evenly: A divisor divides evenly if there is a zero for the remainder.

11 Polynomial Long Division ) Look at left-most numbers ) What # times left = left? ) Multiply? ( ) 4) Subtract ( )

12 Polynomial Long Division ( ) 4) Subtract Careful with the negatives! 5 8 5) Bring down.

13 Polynomial Long Division ( ) 5 8? ) Multiply ( ) 6) Repeat steps -5. ) Look at leftmost numbers ) What # times left = left? 4) Subtract ( )

14 Polynomial Long Division ( ) 5 8 ( ) 4) Subtract Careful of the negatives 8 5) Bring down.

15 Polynomial Long Division ( ) 5) Bring down. 5 8 ( ) 8 8

16 Polynomial Long Division ( ) 5 8 ( 8 ) 8 8 6) Repeat steps -5. ) Look at leftmost numbers ) What # times left = left? 8? ) Multiply 8 8( ) 8 8 4) Subtract ( 8 8)

17 Polynomial Long Division ( ) 5 8 ( ) 8 8 ( 8 8) 0 4) Subtract ( 8 8) Remainder = 0 ( ) divides evenly. ( ) is a factor ( )( 8)

18 Is there an easier way to do this? Yes! st step: Write the polynomial with only its coefficients. nd step: Write the zero of the linear factor. rd step: Bring down the lead coefficient

19 Is there an easier way to do this? Yes! th step: Multiply the zero by the lead coefficient. 5th step: Write the product under the net term to the right. 6 th step: add the second column downward

20 Is there an easier way to do this? Yes! th step: Multiply the zero by the second number 8th step: Write the product under the net term to the right. 9 th step: add the net column downward

21 Is there an easier way to do this? Yes! th step: Multiply the zero by the rd number th step: Write the product under the net term to the right. th step: add the net column downward

22 Is there an easier way to do this? Yes! This last number is the remainder when you divide: by

23 Because the remainder = 0, then ( ) is a factor AND = is a zero of the original polynomial!

24 The Remainder Theorem: When dividing a polynomial epression by a lower degree polynomial epression, If the remainder is zero, then the divisor is a factor of the original polynomial.

25 We call this synthetic division Look at the numbers at the bottom. Are these familiar? ( )( 8) They are the coefficients of the quadratic factor! (wow)

26 ( )( 8) ( )( 6)( )

27 Synthetic Division ) )( ( th degree poly = ( st degree poly)( rd degree poly) 0 Is the rd degree polynomial a nice one?

28 f ( )? ( ) Synthetic Substitution

29 f ( )? ( ) Synthetic Substitution

30 f ( )? ( ) Synthetic Substitution

31 f ( )? ( ) Synthetic Substitution

32 f ( )? ( ) Synthetic Substitution

33 ( )? f ( ) Synthetic Substitution

34 f ( )? ( ) Synthetic Substitution

35 f ( )? ( ) Synthetic Substitution

36 f ( )? ( ) Synthetic Substitution (Since (+) has a remainder of 5, then ( + ) is not a factor).

37 Synthetic Division & Substitution 4 5 ) ( 4 f

38 Good to here.

39 Your turn: Use Synthetic division. 8 8

40 Your turn: Use Synthetic Substitution 4. f ( ) 8 8 f ( 5)?

41 Possible Rational Zeroes f ( ) 4 5 8,,, 6, 9, 8 Factors of the last term divided by the factors of the first term = is a zero of f() ( ) is a factor of f(). Coefficients of quadratic factor Remainder ( 4 5 8) ( )( 8)

42 Polynomial Long Division ( ) 5 8 ( ) 8 8 ( 8 8) 0

43 Polynomial Long Division ( ) 5 8 ( ) 8 8 ( 8 8)

44 Can you follow this? What are the zeroes? 0 ( 5)( )( 7) - Multiply the three binomials (convert to std. form) ,, 7 What do you notice about the first and last terms and the zeroes?

45 0 ( 5)( )( 7) 5-7,, The denominators of the solutions are factors of the lead coefficient. The numerators of the solutions are factors of the constant term.

46 root = zero of a polynomial = inputs that cause outputs to = 0 The Rational Roots Theorem: the possible rational roots of a polynomial are factors of the constant term divided by factors of the lead coefficient. 0 ( 5)( )( 0 ( 5)( 7) ) ,, 5, 7,0, 5, 70,

47 The Rational Roots Theorem: the possible rational roots of a polynomial are factors of the constant divided by factors of the lead coefficient. 0 4,, 5, 7,0, 5, 70, , -, 7,, 5, 7,0, 5,70,,, 5, 7,..., 5, 70 We always check the easy ones st.

48 Your turn: What are the possible rational roots of the following polynomial? y 4 5 8,,, 6, 9,8 If = is a zero, what factor did = come from? ( ) If ( ) is a factor, then the rd degree factors as: y ( )( a b c)