quiz6_v2 - Student’s Name Signature QUIZ 6 Section 3.4...

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Unformatted text preview: Student’s Name: Signature: QUIZ 6' Section 3.4, 3.5 and 3.6 (Version 2) Time: inutes - 3) Question 1: points) [3.4.4bPT]Find the remainder if 2:12" + 22:3 ~ 2:2 — a: —— 6 is divided by w2+1. * 2$2+2X45 I I: -—3 + 9: “fl, [3 _1,_.3x ‘ (“H \ 2xi+2xhxt><~g I —-3——3a: 92% + Ox5+ 2")” [I 22c2+2at—3 MW_X-Q g 92x5 +OxL+2x+O Question Zzpoints) m [3.4.53PT Application of the Intermediate Value Theorem to f (m) = 6x3 —- 5x2 + 23: — 8 on the interval [1,2] implies e _ 3x%+0,( ._ 5 E f does not have a zero in (1,2) £0) (MA .9- f has a zero in (1,2) ‘ 3X _ 3) 4(1) Law [3 no conclusion about a zero of f in (1,2) A (Lifidfld‘ $0): Q-fi'tél—Sz-E <0 mom a— ssams Ur 1(a): Ejf— azi' 2.:2 4%. Question 3: points) 1+ (g __ a0 + L} _ g > O [3.5.2aPT] Select the real and complex zeros of f = m2 + 9 5 none of these Ag+ w z c 3—i,3+i #0:} z a: 'z I —3i,3z‘ fig): 02:) KO +9-O G) X e 8.9!. E —3,3 ._> x : t 51 42rd wig: UR Jam qmm Equla; DiSuiminanl: D: lJ‘b— how. : .. 5Q<O —>NEXTPAGE (a_ 4 L4) _ 9) |V - 2 - »°- awrhx(mlrw&) VS: )(i: Jo+ W: -O'+ Gt = zero; - g1 02c: 32. x62: - 2)? (cumin mrgvgafi 0;? XL), l Question 4: ( points) [3.6.1bPT] elect the polynomial with real coefficients of degree 5 having zeros of 1, 2 -—4z', -3+ i. 5 None of these (2:—1)(a:+2+4i)(x+2—~4£)(a:+3+i)(a:+3-i) (2——1)(a:,—2+4é)(a:—2—4i)(x+3+a:)(a:+3—i) ' (a:——1)(a:~—2+4é)(m—~—2——4i)(x—3+i)(zv~3~i) (z—1)(:z:+2+4i)(m+2-4i)(m—3+i)(m—3—i) nnln X L05 dear“ 5 .——) 5 (nka 2.91m; aivm Zeros: 4 g) _ h; - 5-H f7 rfimaim‘rf) ZemS: ' 4- '— W raprfl flak} ‘QC‘UQ—‘hfi. rvwg’¥ (gnAIQiCL Q’Dwe jacfm‘s ...
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