Test_2_Solutions

Test_2_Solutions - 2 +8x+7 . x-3 x 2-9 x+3 = x 2 +8x+7 ....

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I MATH 110 Test 2 October 8, 2009 Name: ..sO/~L bo Y1 S Class Time: _ Instructions: Show all work. Partial credit will be given. 2. Find the zeros of the following function: [(x) = x 2 - X - 6. (15 points) h/ Sf .{;;l C W .~ po I 'jM 0Vt.;c. . (, f( )() -;:0 (x' - ~) ( X -+- c2 ) !(x)-==O vJ~ X -: S Of X::: -~ So'~ Ct"roS Clre. 3 Ct "1 cl. - ~ - 3. Let p(x) = x 2 + 4x + 2 and q(x) = -2x 2 - 5x - 1 then find the following: a. (p + q)(x) (5 points) b. (pq)(x) (10 points) -=X7. . .,.Lj.x'i~f (-~)('2-_)x-1) ) a.)~i'f)(K):;; p(K)J f-{X ."" - X 2- - >< T (. c:= (X 2- f It X·I ~ ) (- ;). X '2 -~;< -I ) = _ ~ X 4_ 5 X '3 _ )(?. -- ;g- X '3 _ :<0)( '1.- -- 4 >( - LJ X <--) ox - ~ ::::- -:2 x'-t -13)( ?- 2~ I\: ~- )4 X --<- x2+8x+7 ) x-3 . 4. Let p () x = 2 9 and q(x =- then find the follOWing: x - x+3 a. (p + q)(x) (10 points) b. (pq)(x) (10 points) x- 3 j-.- ;(f3 ;::. .'2- f &'x·t ? X - f X.... -lJ
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4b) (pq)(x) = p(x) * q(x) = x
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Unformatted text preview: 2 +8x+7 . x-3 x 2-9 x+3 = x 2 +8x+7 . x-3 (x+3)(x-3) x+3 = x 2 +8x+7 . 1 x+3 x+3 = x 2 +8x+7 x 2 +6x+9 h . f() 3x 2 +9x+20 . h f () R(x) h . h d . f 5 . W rite t e expression x = In t e orm x +-( )' were q IS t e enommator x+2 q x the given expression and G and R are polynomials with deg (R) &lt; deg(q). (20 points) USC ~VI'.5i&quot;vh. Iv c;;d: 3)( + 5 ~ 14;(;&lt;+. .2-) xf~f 3x&quot;2. .+1x ..t.:to -(3X~ i6,X) -----~-'3)( -I J.o _(3;&lt;+0) iLl-6. Find fog for the following functions: 2 a. f(x) = x , g(x) = ax + b (10 points) b. f(x) = 2x 2 , g(x) = x 2 + 2 (5 points) . .-z. ). 2. /) b ';'b~ cA.) Ie j ~-f(~(;()) ::: (tA }cJb) :== a X r &quot;,,~. )( .~ 2... (X2+2).:t. ;: z(xt{ 1-Ltx z .,4) 7. Extra Credit: Derive the exponent rule: x-m = lJx m using the algebraic properties of exponents. (10 points) F...
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Test_2_Solutions - 2 +8x+7 . x-3 x 2-9 x+3 = x 2 +8x+7 ....

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