HW_sol(19)

HW_sol(19) - ’HW Whmfiisn{fl VCothfion‘ i<§L Student’s Signature MACll40-24 Section 3.4 ‘ What problems are the most difficult for

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2
Background image of page 2
3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ’HW Whmfiisn; '- {fl} VCothfion‘, + i +<§L Student’s Signature MACll40-24 Section 3.4 ‘ What problems are the most difficult for you? Question 1: (1 point) [3.4.1aPT]Choose the function that has a factor of (a: + 2) :. o x :—7. [3 f (m) = 2x3 — 32:2 -— 32: + 2 z-wmmwzwmwwkti‘ ‘U’ ‘5 Q Zero [3 f(m)=3x3+2x2—7x—2 425%; :-q 5i fix \/I f(m)=31:3+4;c2—5mw2 .; ~14 +lb+\o-‘L AE/SUb‘g/Ahfi s C." f(:[t) =31:3 —8x2+3x+2 X:_Q to each CHM, ates“ Question 2: (1 point) Which om. ha), i (v ,2) : C) [3.4.1bPT]0ne of the factors of f (a: = 3x3 — 2:1:2 — 481: +32 is (3:: — 2). ‘ Another factor is "" C'“‘(*"*\ ‘/ XZ-D-\(0 m I! / U (I+8)(I—2)®1sz / 7/ \/I -3K3*ZK"% L // E (m2 + -- (DJ—:2x gxrnwxmi u (x -—4;2 " ""‘ xix: Q2 _\02 : (Mo-wt) Question 3: (1 point) 0 [3.4.1cPT]Find the leading coefficient, A, of the polynomial f(a:) = Au,- ~ mm — mos — 7‘3) with zeros 2, 3, -1, and IN) = -2- «I: —3 mgw-au—oum x I -% flo§=mawam : Ma => U = ~52 "-5 “=3— X'ri' 3 [3 Question 4: (1 point) [3.4.1dPT]Find k such that f (3:) = x3 — 2km2 —— 3kg: — 4 has a. factor of (:r + 2). y -. -7. 174073 —. (.735 VZKC-ZW— ELL-13 - ‘4 E3 % lg HM 10° for.) = -8 -1K(u\ HM -u [j _4 X, 4—7—8 = ‘9 43K arm—H E g .pL—L) = 1K —\'L 7 . O = -7_\( ~\'L Question 5: (1 point) K 3 ‘4’ [3.4.2aPT]Select the choice containing ALL (according to the Ratio- nal Zeros Theorem) potential rational zeros of the polynomial function f(:c) = 21:34 +8.1:2 —- 7x+3 1 1 1 3 1:: :tl,:l:3,:|:§,d:7, aid? [3 :hl, :l:2,:t7, :l:21 [3 2H, :l:3,:|:7,:l:8, :l:21 I: 11, i3,:h7, i21,:l:§,d:§ duo-huff Question 6: (1 point) [3.4.4aPTlFind the quotient if 63:4 + 2;: — 4 is divided by 22:2 + . t 3 \N/. 3$2_§ _,_____s_c_.4L-— E 3 2 g Z¥1+\‘(0qu *O¥3*o *14'1“ 'u “+31: 2: _ _ a 3 :3 3x2 +% C —% +2.1: Question ‘7: (1 point) R \ [3.4.4bPT1Find the remainder if 42:5 + 22:4 + 4 is divided by 2:132 — a: + 1. mmmmlur C 3 — :c \/I 5—1. 2‘3*z‘1vox—l 2547*— x +( q‘§¥2¥q+013*0X7-*OX +‘! r: 3+2: .JWS‘Zy‘l zx’ is: U 233 + 22:2 "' 1 “‘5‘ ' Zlfiioyj- e . LN” i»sz :1“: ____________,__ ,117—wx9r“ 4112+): H r HW Suwanee: +\ Sta-lord g 334 wrhfiun: +\ / Question 4: (1 point) 4.— ;‘ [3.4.1(‘1P’13Find k: such that f = 3:3 —~ 21m:2 ~— 3km? w 4 has a fact-0r of (I? + 2:}. E: 3% E: ...A . 1 46 L' 7 [3 «6 ‘ point) gelect the choice containing ALL (according to the Ratio- nal Zeros. Theorem) potential rational zeros of the polynomial function f(:'17)2 21.114 -=%-8121wv?1x+? a 2:3 4‘ g a; 7 a; g A} “a fluzm \/ I :‘cl,i‘3,i§:i‘§ui§?>i§ 13: nyia J (P ohviclo, a0: 32,: inhau) [3 ilgfizl‘i", 3:21 _ 7 I P ' ' " %‘11,i3,11,i2\~/ ‘* [J i1,:t3,i7,is,:t21 ‘ ‘ +3 ‘ :3 i1,:§:3,i7|,:§:21,:1:§,:t§ J1:H *3E‘3‘FTF‘fiEE V 4 , Question 6: (1 point) ex [3.4.4313'1’ Wind. the quotient if 61? 2:5 - 4 divided by Que2 + I. r: 33:2 a E: 33:2 ‘ E 3:152 + r: w 1+" 2.7; in? xwm 0‘ din‘CS Gt} 3 -2——l—': inbfcaer-‘H: :0 Question 7: (1 point) [3.4.41'1P'11'Fi11d the remainder if 4:125 + 2m“ + 4 is divided by 223:2 —— :1: + 1. U 3 - :1: E3 5 ~— U 3 + :2: E: 2w3+2x2~1 -. am 21:5 -~ 732'1 -~ 91:2 + 5:1: ~+~ 3 on the interval {04% implies U - Ques "I V tion9:( Theorem point) ‘ " ’1‘3Application of the Intermediate Value Theorem to flw) :- f ([095 not have a; zero in (O, l) gamma/fl“ gm: 2-7-4493 “L J no conclusiml about a zero of f in (0.1) has a zero in (0,1) point) "$5.9... ~/ I’T]Select ALL the intervals for which the Intermmfiaie Value implies that f (at) m‘ 83:3 + 205:2 — 2:1: —» 5 has a zero. ["14] X €933 ‘ 8 (‘8~>+ wab 42(1) [4,0] : -—w we} x, [1:23 Q“) r :36!) réfift} - [GM I >g\/ H1 In gm 7410 conclu‘h‘on, aloouf a zero in, [—2,-1] One) [5:23 infirmk [4,0] anc\ {OJ} ...
View Full Document

This note was uploaded on 05/23/2011 for the course MAC 1147 taught by Professor Nuegyen during the Spring '11 term at FSU.

Page1 / 4

HW_sol(19) - ’HW Whmfiisn{fl VCothfion‘ i<§L Student’s Signature MACll40-24 Section 3.4 ‘ What problems are the most difficult for

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online