1 / 13

ALGEBRA II HONORS/GIFTED @

ALGEBRA II HONORS/GIFTED @. SECTION 5-7 : THE BINOMIAL THEOREM. Work out the problem that corresponds to your group number. 9) What do you notice about the signs before each term?. (x + y) 0 (x – y) 0 (x + y) 1 (x – y) 1 (x + y) 2 (x – y) 2 (x + y) 3 8) (x – y) 3. = 1 = 1

garron
Télécharger la présentation

ALGEBRA II HONORS/GIFTED @

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ALGEBRA II HONORS/GIFTED @ SECTION 5-7 : THE BINOMIAL THEOREM

  2. Work out the problem that corresponds to your group number. 9) What do you notice about the signs before each term? • (x + y)0 • (x – y)0 • (x + y)1 • (x – y)1 • (x + y)2 • (x – y)2 • (x + y)3 • 8) (x – y)3 = 1 = 1 = x + y = x – y = x2 + 2xy + y2 = x2 – 2xy + y2 = x3 + 3x2y + 3xy2 + y3 = x3 – 3x2y + 3xy2 – y3 10) And, those exponents. What about them? 11) Hey, what’s going on with the coefficients?

  3. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 This is Pascal’s Triangle. The triangle’s row are numbered beginning with row 0.

  4. Expand. a) How many terms are in the expansion? 9a) (x + y)4 ANSWER : 5 b) How are the signs of the terms arranged? + + + + + (1) (4) (6) (4) (1) (x)4 (x)3 (x)2 (x)1 (x)0 (y)0 (y)1 (y)2 (y)3 (y)4 ANSWER : all + c) Lay down the coefficients from Pascal’s Triangle. d) Place the terms of the binomial down noting their exponents. e) Simplify. = x4 + 4x3y + 6x2y2 + 4xy3 + y4

  5. f) (x – y)5 ANSWER : x5 - 5x4y + 10x3y2 - 10x2y3 + 5xy4 - y5 g) (x – y)6 ANSWER : x6 – 6x5y + 15x4y2 – 20x3y3 + 15x2y4 – 6xy5 + y6 Find the specified term. 10a) Third term of (x + 2y)6 ANSWER : 60x4y2 b) Fourth term of (2x2 – y)5 ANSWER : -40x4y3

  6. 11)a) Expand using Pascal’s Triangle : (x – 4)5 ANSWER : x5 – 20x4 + 160x3 – 640x2 + 1280x - 1024 b) Now, find : 5C0, 5C1, 5C2, 5C3, 5C4, and 5C5. ANSWERS : 1, 5, 10, 10, 5, 1 c) How do these answers compare with the coefficients found in Pascal’s Triangle? ANSWER : They are the same.

  7. 12) Expand. SOLUTION : +4C0 (3x)4(2)0 -4C1(3x)3(2)1 +4C2(3x)2(2)2 -4C3(3x)1(2)3 +4C4(3x)0(2)4 a) (3x – 2)4 = 81x4 – 216x3 + 216x2 – 96x + 16

  8. b) (4x + y2)3 ANSWER : 64x3 + 48x2y2 + 12xy4 + y6 c) (a + a-1)6 ANSWER : a6 + 6a4 + 15a2 + 20 + 15a-2 + 6a-4 + a-6 d) (1 – i)5 ANSWER : -4 + 4i

  9. 13) Find the specified term. Must add to 9. a) (x + y)9 ; y4 SOLUTION : 9C4 (x)5 (y)4 =126x5y4 Where does the “9” come from? These numbers are always the same. b) (2x – 3y)7 ; y5 ANSWER : 7C5(2x)2(-3y)5= -20412x2y5

  10. c) (a – 2b)8 ; a3 ANSWER : 8C5 (a)3 (-2b)5 = -1792a3b5 d) (a + b)n ; br ANSWER : nCr (a)n-r (b)r

  11. 14) Find the specified term. a) (m + n)20 ; 12th term SOLUTION : 20C11 (m)9 (n)11 = 167960m9n11 The choose number is always 1 less than the term number. The choose number is also known as the “magic number”. b) (r3 – 2s)9 ; 5th term ANSWER : 9C4 (r3)5 (-2s)4 = 2016r15s4

  12. c) (2x – y)12 ; middle term ANSWER : 12C6 (2x)6 (-y)6 = 59136x6y6 d) (a + 2b3)10 ; middle term ANSWER : 10C5 (a)5 (2b)5 = 8064a5b5

More Related