3_4_to_3_6_sol

3_4_to_3_6_sol - REVIEW SECTIONS 3.4 3.5 3.6 Show all work...

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Unformatted text preview: REVIEW SECTIONS 3.4, 3.5, 3.6 Show all work to get full credit. Question 1: (1 point) ' [3.4.4bPT]Find the remainder if 42:5 —— 2x4 - 4 is divided by 21:2 —— x + 1. 1 [I 22:3—92—5 ZX3’X_ Vi c ‘g _%\$ in—N +\ 445' 2.x“?ng *sz‘fo" '4 \/@ 7 1 Q-uﬁawnx?’ 5+” W K -§+%w e -243 +§X"—P" 9-) Question 2: (1 point) 0 «UL: «Hx - a 13:)? [3.5.1aPT] Write 214+ 3i19‘ina+bz'form., @ _W “fax [2- : _ M 0" 1:: 4+3“ “* +3”) ' ‘ '. o, \m—a‘m = (‘ =- :3 1—35 \L a 2‘: ’1. L74 ﬂ : i _1_3i _| —:)3\ “'33: "t C ‘ix “1492 E 4i 4=| ah+L:_1 ‘ . . E: 1 ' * oz 02/ t \ (\%)5 +32 ma Question 3: (1 point) = \ T 0 :-—\ (4)3 [3.4.1aPT]Choose the function that has a factor of (a: — 2) X 6 2 is 2% gm - (-9 ,Z' f(z)=3x3+2x2—7x+2 19m: 1q.+g— n41 Z. - gz'm n, :10 \/® f(x)=3x3—812+39:+2£(1)=Z~t-%’J.Mo+2.:o\/ c f(\$)=3x3+4x2—5x—2 :- ~\$+b fl [3 f(x)=2m3—3m2:73:x—2 Question 4: (1 point) \M \o “ (ii—kn - 34-17 33+? M (NI 313::th : [3.6.1bPT]Select the/Weﬂicients of degree 5 having zeros of l, —2 +333 — i. (a:—1)(m+2+4i)(x+2—4i)(x+3+i)(a:+3—i) None of these [I E (x~\7(%+l~q!) [74+ZT1’}U(X’ -3+£)£><'3‘€ c (a:—1)(a:—2+4i)(:r—2—4i)(x—3+i)(x—-3—i) c (:1:—1)(:1:—2+4i)(:c—2—4i)(:c+3+'i)(a:+3-i) \/@ (a;—1)‘(m+2+4¢)(z+2—4¢)(x—3+i)(x—3—i) A Question 5: (1 point) ' [3.4.4aPT]Find the quotient if 42:5 + 2:34 - 4 is divided by 2:1:2 — a: + 1. E —3——x 341+?” \/ c: 2.1:3—22—-;- 2X24“ Wx5+1xu+ox5f0><fo><4+ Li’fj'l’c‘ *JLL 1 M C 3 2 . \/ 21: +23: +1 G +LH ~Z{3g+o:z Q 2x3+2\$2—1 QM‘LQ.‘ {£2 9 \/ Question6: (1 point) 0 «ca-ﬁrm :07" ‘ 71.) ex 5 [3.4.2aPT]Select the choice containing ALL (according to the Ratio- nal Zeros Theorem) potential rational zeros of the polynomial function f(x)=1514+3z3—8x?i5 Q”:st £2 :3 \$11,115:? [7 :izl, i3, i5, i8, :i:15 [2/ i1, i3,:i:5, £15 £7 i1,i3,i5,i15,i%,i% ' 1 1 1 5 \/@ i1,:l:5,d:5,:t3,:tﬁ,:l:§ Question 7: (1 point) Ugh ['1 I Imam Question 8: (1 point) [3.4.5bMSPT]Select ALL the intervals for which the Intermediate Value Theorem implies that f (1:) = 83:3 + 12.172 - 22: — 3 has a zero. 0 r —- \/ 7 «9 [0’2] £((73)=L03+L\‘€ «an; = \05 V} i©<0 ;Sf(1)>0 In *‘f‘ ' :GH‘i' i r ’ ((sz'S <0 10): 3+§5L<7J5>Q ((a):~<z+\2t?r5=%70\ 1(5); 8. a} Halo -c.-3,)o \ \/@ [—LOI \/ a [0,1] '— l2a 3] Question 9: (1 point) . [3.5.2aPT]Select the real and complex zeros of f(:t) = 1:2 — 42: + 5 Disuimincxfl' 3 none of these a i 4‘, =|b "l (57 _2+i,—2—27 ‘ @;5 3:12: :3 J? 2+i,2-i 5"1 x = = . 5L :2, _ ~ Question 10: (1 point) . Lf‘ - 21x :7 7- a + .L X4, = 2 ” L [3.4.1bPT]Two of the factors of m) = 2x4 + 33:3 — 373:2 + 58.1: — 24 are (23: — 3) and (a: — 2). Another factor is . 2’- :lx +9 ' p \/ E 332—515—4 (2x—5)(x—Z>= ZxZ-[lx -3>< 'l'LO 1 2X ' ® x2+52:-4 ‘ I: (“mm—1) . . q 3__ ,- C (3-4)(a:+1) (uz' Sx—Ho +33:1+58x 3‘! Question 11: (1 point) W 4 \$5 rs [3.4.5aPT]Application of the Intermediate Value Th reg: to f ) = . — 5:124 — 933 + 5:: + 22 on the interval [-1,0] implies Ml“ \_\ 63:5 — — — ﬂ __ ‘ a 'q '7. £1 0 - ‘ 625 a golttghzc’lusion about a zero of in —1,0l————Z'—-3"<——+-—5——:—’— >O f does not have azero in (—1,0) .1 4 T9 «x: +3>5x5 pal"; 4&5? .2“ w _ ._ x + " X " £(o)_ 22X) [3 f has azero 1n (—l,0) ax 93:14 ” 1&5 4 6x0 Question 12: (1 point) \Oxs — q 3x41 +5831?“ [3.4.1cPT]Find the leading coefﬁcient, A, of the polynomial f (3:) = \Ox3 — 35%” 30A ACE _ r1)(m - T2)(\$ — 7‘3) zeros '2’ ‘3’ 01 and = (I'M \/® '2 m; Mwmmaco AWWW C 2 V ’— ,_ (ZXZ’J, \$293924 5 _% job: n —> Aaméwam =4 0 c Question 13: (1 point) . [3.4.1dPT]Find k such that f (2:) = M3 — 3:1:2 — 3:1: — It has a factor of 2 5 /% ‘% f('L),‘gKT‘?”\< 'Ques on14:(lpoint) 1‘) O : ~3kdgg [3.6.1aPT]Select the choice with ALL the remaining zer 2+3i. C noneofthese V6 —1+i,2—3i [I 1-—i,2+3i [3 1+i,—2—3i C l-i,-—2+3i ...
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This note was uploaded on 05/23/2011 for the course MAC 1147 taught by Professor Nuegyen during the Spring '11 term at FSU.

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3_4_to_3_6_sol - REVIEW SECTIONS 3.4 3.5 3.6 Show all work...

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