Question

5) Find the Partial Decomposition of each rational expression. 5x-2 b) (a2-4)(x+1) b) 23+2x2+x

Answer 1

Answer & = s + 5x-2 (2-3) (ara)(22+1) D (An+B) + 2+1 Simplify right hand side 5x-2 = (Ar+B) 6-2) (7172) + C 662 +) 12+2)+ (x-2)(x+2)/2+i) B CORAL 2-2) txt) totalta+2) (5x-2) = (Ax+B) (x-2)(x+2) = D(2241) (2-2) C (x2 + 1) (ata) + - ------ (5x-2) = expand right hand side AM3+ ( x + D x3 + Bxc2 + 2002 C-222D un A t x C + 2D - 4B 426-2D Collect up the likes terms TA+C+ D x² + x2 (B+20-2D+ x (-4A+ C + D) - 4B+ac-2D on comparing both sides

yeti A+C+ D =0 B+ 2 C-2D=0 - 4 A + C + D - 5 -4Brac-2D = -2 __ on solving these equations - A = -1, B = 2 c = z D = 3 Therefore 7 Su-2 (2-4) (2+1) + + 1315 215 Ix-2 at Lara (b) Gta 6e3+2012+2) (2+2) aa2+ 20+1 = (2+2)_Btc 262.4102 -- - term in write the right single fraction = (+2) alator A (atip & Balx+1)+ca ala+12 1 +2 = Ant Ba2+QAX & But ca & A Ant Ba a laut.22 ala+22

(2+2) - (A+B) 2? + (2 A+B+C)* PA. A+B= 0 0 2A+B+C = 1–0 TA=2 - from equation -2 B=-A = B=-2 And from equation ④ 2622-1+0=1 2 + c = 1 Therefore 24+2 2 13 + 2x12 ta 2 +1
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