Math 017 Materials With Exercises

( x y)(b c) xb xc yc yb yc yb xb xc yc xc yb xb xc xb yc yb55 46 53 2 2 24 a) F 10b) 18y 24yc) x 2x d) 2c 22de) a a b a b94 2 3 2 54 2 3f) 2xy xy x y g) xz yz xw yw h) 10a 4b a a b753 23 2i) ac bc gc ad bd gd j) x y x y xy y x x x 1k) 2xy 4y 3x 6 l) 2ac c 2abc bc 4a 2 4ab 2b225 a) ( x y)5 5x 5y b) ( 4x 1) 4x1c) a( c 1) ac a33 3 e) ( x 1)(y 2) xy y 2x 2d) 2y(a 2b d) 2ya 4yb 2yd2 42 4f) ( x x 2x) x x 2x6 a) 3a 3b 3b 3a( a b)3 a 3b3 b3a 3(answers vary)b) 2 z yz yz 2z z(2 y) z 2 zy 2z zy (answers vary)7 . mn mp( n p)m pm nm mn p mp nm8 ( 1)( x y z)( x y z)(1) 1(x y z)( x y z)1 x y z x y z x y z ( x z y)9 a) 2 b) 3 c) m10 a) 5(x y)b) 7(1 7a)c) ( 3 2c)2e) 11 t(14t)211 a) (2 a )c d) y ( 8xy 5 1)2 f) 2(lw lh wh)g) x (4x 5x 5)xy b) xy( x y)c) xy ( a b 1)12 a) 1(3 x) b) 1( b 1)13 a) 5 (2 3a)a b) 2e) d2 (8 ad)a c) x y z 1 a 2 13 211t t 4 c) 5 (3x1) 2x d) 4y5 (2xy1)c f) 3 2 x1x y( 13xy)g) ( 2 a)h) (1 s)ya b2 ( x 2 y 2zb) 1 1 3x 5c) )10.6(6x 10y1d) (5a 6b) 5 62 (2x 2 17311f) 0.1(20a 3b1)g) (11xy 2x ) h) 0.02(20m 50n)113( a b)(6 x b) ( a b)[4 3( a b)]14 a) )e) )15 a) )16 a) 2xy((1y 2xy)173 3 44 2 242 4 b) a b ( b 5a1)c) 17xy ( x y 2xy 3)d) m n(1 5n m )34 5e) 4ac( 4 2c 3cd) 3 a(1a a 18 a) 2(2 )2222 f) 7ab (5b 2ab 3)g) ( x 2y)(2 z ) h) ( c d)[1 4a(c d)]x 4 2 1 1 1 b) x 1 c) 2x d) 4x 2 x x 2 x 4 19 a) Numerator: One term, t. (can be viewed as two factors, 1 and t) Denominator: two terms: 2t (with factors 2 and t) and─ty (with factors ─1, t and y). Therefore t is a common factor of ALL terms (numerator and denominator.) We can dividethe numerator and the denominator by t.b) Numerator: two terms, x (with factors 1 and x) and xy (with factors x and y) Denominator: one term, 2ax (with factors2, a and x). ALL terms in the numerator AND in the denominator have a common factor, x. We can therefore divide boththe numerator and denominator by x.Answers to Exercises 123