Math-3. Lesson 2-8 Factoring NICE 3rd Degree Polynomials

Transcription

1 Math- Lesson -8 Factoring NICE rd Degree Polnomials

2 Find the zeroes of the following rd degree Polnomial 5 Set = Factor out the common factor. 0 ( 5 ) Factor the quadratic 0 ( 1)( ) 0, -1, - Identif the zeroes

3 Nice rd Degree Polnomial (with no constant term) = It has no constant term so it can easil be factored into times a quadratic factor. = ( + 1) If the quadratic factor is nice we can factor that into binomials. = ( + 6)( ) = 0, -6, This is now intercept form so we can read off the -intercepts. What are the?

4 Nice rd Degree Polnomial (with no constant term) 0 What if the quadratic factor is not factorable? = b 6 6 It has no constant term so it can easil be factored into times a quadratic factor. a = (6) (1) = = f = = = 0 ( 6 ) Convert the quadratic factor into verte form and solve. 0 ( ) 5 5 Zeroes: = 0, ± 5

5 Factor the following nice rd degree polnomials then find the zeroes of the polnomial. = ( 5 0 ( 7)( 0, -7, 1) ) 0 = ( 8 + ) = 0 = b a = ( 8) (1) = = f = 8 + = 1 0 = 1 = ± 1

6 Another Nice rd Degree Polnomial a b c d This has the constant term, but it has a ver useful feature: 1 rd 1st What pattern do ou see? 1 th nd 1 If we divide the coefficients of the 1 st and rd terms, and the coefficient of the nd term and the constant, we get the same number.

7 factor b grouping if it has this nice pattern. 1 Group the 1 st and last pair of terms with parentheses. ( ) ( ) Factor out the common term from the first group. ( ) ( ) Factor out the common term from the last group. ( ) ( ) Common factor of ( + )

8 ( ) ( ) These two binomials are now common factors of and Factor out the common binomial term. ( )( ) =, i, i Appl the Comple Conjugates Theorem

9 An easier method is bo factoring (if it has this nice pattern). 1 These terms are the numbers in the bo. Find the common factor of the 1 st row. Fill in the rest of the bo. Rewrite in intercept form. 1 = ( + )( + ) Find the zeroes. = ±i = 0 = ( + )( + ) 0 = + 0 = + =

10 Which of the following rd degree polnomials have the nice pattern we saw on the previous slide? 6 rd th 1 st nd rd th st nd 1 rd th st nd rd th 1 st nd All of them!

11 Find the zeroes using bo factoring = ( + )( 5) 0 = + 5 = = = 5 = ±i = 5 = i, i, 5

12 Find the zeroes using bo factoring = = =

13 What have we learned so far? Nice Common Factor rd degree polnomial: = + + ( ) ( 1)( Nice Factor b bo rd degree polnomial: ) 6

14 Convert to standard form: = ( )( + + 9) = 7 There are NO and NO terms terms The Difference of cubes: factors as the cubed root of each term multiplied b a nd degree polnomial. = 1 = ( 1)(a + b + c) = ( 1)( )

15 The Sum of cubes: factors as the cubed root of each term multiplied b a nd degree polnomial. = + 15 = ( + 5)(a + b + c) 0 = ( + 5)( 5 + 5)

16 Nice Difference of Squares (of higher degree): = 81 = m 81 = (m + 9)(m 9) = ( + 9)( 9) Use m substitution Use m substitution = ( + )( )( + i)( i) = -,, -i, i Find the zeroes. = 6 Let m = Then m =

17 Find the zeroes. = ( 7)( 6 + ) = 7,?,? How do ou find the zeroes of an unfactorable quadratic factor? Convert to verte form then take square roots. = ( 1)( ) = 1,?,? = 0 verte = verte = b a + 1 = = = (1) (1) = 1 0 = = + 1 ± i = + 1 = 1 ± i