Ws8 Factorisation

Date:

______/______/ 2009

Marks

Worksheet 8: Fundamental Algebra (Substitution and Factorisation) 1

Basic Question Given that x = 5 , y = −4 , z = −2 , h = 0 and k = −1 , evaluate the following: (a) xy ( 2 x − 4k + xhz ) (b)

5 x − 3z xhk − 5 yk

(c)

2k xh − yz

Solution: (a)

xy ( 2 x − 4k + xhz ) = (5)(−4) [ 2(5) − 4(−1) + (5)(0)(−2)] = (−20) [10 + 4] = −280

(c)

(b)

5 x − 3z xhk − 5 yk 5(5) − 3(−2) = (5)(0)(−1) − 5(−4)(−1) 25 + 6 = 0 − 20 11 = −1 20

2k xh − yz 2(−1) = (5)(0) − (−4)(−2) −2 = −8 1 = 4

[Xin Min Sec School/2007] Ans: (a) −280 (b) −1

11 1 (c) 20 4

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2

Intermediate Question Given that a = −1 , b = (a)

1 and c = 2 , find the value of 2

1 1 1 + + a b c

(b)

a2 − b2 1 2 c 3

(b)

a 2 − b2 1 2 c 3

Solution: (a)

1 1 1 + + a b c 1 1 1 = + + −1 1 2 2 1 = −1 + 2 + 2 1 =1 2

1 (−1) −    2 = 1 2 (2) 3

2

2

1− =

1 4

4 3

=

3 4 4 3

=

3 3 × 4 4

=

3 4 [Cedar Girls’ Sec School/2008] 1 3 Ans: (a) 1 (b) 2 4

3

Intermediate Question (a) Factorise the expression p − pq + 3 − 3q completely. Solution:

(a) p − pq + 3 − 3q = p (1 − q ) + 3(1 − q ) = (1 − q )( p + 3) [Xin Min Sec School/2007] Ans: (1 − q )( p + 3)

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(b) Simplify

2y2 + 4y 5y

÷

2

4y + 8 10 y 2

Solution:

2 y2 + 4 y 4 y + 8 ÷ 5 y2 10 y 2

(b)

2 y ( y + 2) 10 y 2 × 5 y2 4( y + 2) =y =

[Cedar Girls’ Sec School /2008] Ans: y

4

Intermediate Question Factorize the following. (a) f (a − b ) + g (b − a ) (b) 2 pq − rs + 2 ps − qr

-( a - b) Solution: (a)

(b)

f ( a − b) + g (b − a ) = f ( a − b ) − g ( a − b) = (a − b)( f − g )

Common Factor

5

2 pq − rs + 2 ps − qr = 2 pq + 2 ps − qr − rs = 2 p(q + s) − r (q + s ) = (q + s)(2 p − r )

[Anglican High School/2007] Ans: (a) ( a − b)( f − g ) (b) ( q + s)(2 p − r )

Common Factor

Advance Question (a) Factorise x3 y − 5 x 2 y 2 − 3 xy 3 + 8 xy 4 . (b) Factorise ab + a 2b 2 − a 3b3 completely. Hence, factorise a n b n + a n +1b n +1 − a n + 2b n + 2 . Solution: (a)

(b)

x 3 y − 5 x 2 y 2 − 3 xy 3 + 8 xy 4 = xy ( x 2 − 5 xy − 3 y 2 + 8 y 3 )

ab + a 2b 2 − a 3b3 = ab (1 + ab − a 2b 2 )

a nb n + a n +1b n +1 − a n + 2b n + 2 = a nb n (1 + ab − a 2b 2 )

Ans: (a) xy ( x − 5 xy − 3 y + 8 y 2

2

3

)

[Anglican High School/2007] (b) ab (1 + ab − a 2b 2 ) , a n b n (1 + ab − a 2b 2 )

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