homework HL no 2 page 11 ex 2,3,4

SAVE OFFLINE

Mathscape 10 ext. - Ch05 Page 164 Thursday, October 13, 2005 4:23 PM

Factorisation and algebraic fractions

5 164

Factorisation and algebraic fractions

This chapter at a glance Stage 5.1/5.2/5.3 After completing this chapter, you should be able to:
















expand binomial products expand perfect squares using the special identities determine whether a given expression is a perfect square complete a perfect square expand expressions using the difference of two squares identity expand expressions that involve a combination of algebraic techniques factorise expressions by taking out the highest common factor factorise expressions using the difference of two squares identity factorise expressions by grouping in pairs factorise monic quadratic trinomials factorise general quadratic trinomials factorise expressions that require a combination of factorisation techniques simplify individual algebraic fractions by factorising then cancelling common factors multiply and divide algebraic fractions by factorising then cancelling common factors add and subtract algebraic fractions by factorising the denominators and finding the lowest common denominator.

Mathscape 10 ext. - Ch05 Page 165 Friday, September 23, 2005 7:50 AM

Chapter

5.1

5:

Factorisation and algebraic fractions

Binomial products review

A binomial product is the product of two binomial expressions. For example, (x + 3)(x − 1), (2a + 5)(a + 2), (4m − 3n)(2m + 5n).

■ The distributive law The distributive law can be used to expand expressions such as a(b + c). When using the distributive law, the term outside the grouping symbols is multiplied by each term inside the grouping symbols. That is, a(b + c) = ab + ac. To expand a binomial product using the distributive law:
multiply the first term in the first factor by each term in the second factor
multiply the second term in the first factor by each term in the second factor. (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd NOTE: The distributive law can be used to expand expressions in which the factors contain more than two terms.

■ The FOIL method The FOIL method is simply the use of the distributive law without writing the first line of working. The acronym FOIL stands for First, Outside, Inside and Last, which is the order in which the terms in the grouping symbols should be multiplied together. This method is also referred to as expansion by inspection. To expand a binomial product using the FOIL method:
multiply the first terms
multiply the outside terms
multiply the inside terms
multiply the last terms
collect any like terms.

(a + b)(c + d) = ac + ad + bc + bd

165

Mathscape 10 ext. - Ch05 Page 166 Friday, September 23, 2005 7:50 AM

166

10

Mathscape

Extension

■ Perfect squares A perfect square is the product of two identical expressions. Some examples of perfect squares are: a2, 9k2, (a + b)2, (3m − 4n)2 To expand a perfect square of the form (a + b)2 or (a − b)2:
square the first term
add or subtract twice the product of the two terms, depending on the sign in the expression
add the square of the last term. (a + b)2 = a2 + 2ab + b2

and

(a − b)2 = a2 − 2ab + b2

NOTE: In the expanded form of the perfect square, the first sign is the same as the sign in the grouping symbols and the last sign is always +. Proofs: (a + b)2 = (a + b)(a + b) (a − b)2 = (a − b)(a − b) = a(a + b) + b(a + b) = a(a − b) − b(a − b) = a2 + ab + ba + b2 = a2 − ab − ba + b2 2 2 = a + 2ab + b = a2 − 2ab + b2

■ Difference of two squares When the sum of two terms is multiplied by their difference, the resulting expression is called a difference of two squares. To expand an expression of the form (a + b)(a − b):
square the first term
subtract the square of the second term. (a + b)(a − b) = a2 − b2 Proof:

EG +S

(a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba − b2 = a2 − b2

Example 1 Expand and simplify: a

(2x + 3)(x + 5)

b (x + 7)2

Solutions a (2x + 3)(x + 5) = 2x(x + 5) + 3(x + 5) = 2x2 + 10x + 3x + 15 = 2x2 + 13x +15

c

(3p− 2q)2

d (m + 4)(m − 4)

b (x + 7)2 = x2 + (2 × x × 7) + 72 = x2 + 14x + 49

Mathscape 10 ext. - Ch05 Page 167 Friday, September 23, 2005 7:50 AM

Chapter

c

EG +S

5:

Factorisation and algebraic fractions

(3p − 2q)2 = (3p)2 − (2 × 3p × 2q) + (2q)2 = 9p2 − 12pq + 4q2

Example 2 Complete each perfect square. a a2 + _____ + 36 = (_____)2

d (m + 4)(m − 4) = m2 − 42 = m2 − 16

b 4e2 + 12e + _____ = (_____)2

Solutions a The first and last terms in the grouping symbols are a 2 = a and 36 = 6 , respectively. The middle term in the trinomial is twice the product of the terms in the grouping symbols. So, middle term = 2 × a × 6 = 12a 2 ∴ a + 12a + 36 = (a + 6)2 b The first term in the grouping symbols is 4e 2 = 2e . The middle term in the trinomial is twice the product of the terms in the grouping symbols. ∴ 12e = 2 × 2e × last term 12e = 4e × last term ∴ the last term is 3. The last term in the grouping symbols is 32 = 9. ∴ 4e2 + 12e + 9 = (2e + 3)2.

EG +S

Example 3 Expand and simplify (x + 5)(x − 5) − (x − 3)2. Solution (x + 5)(x − 5) − (x − 3)2 = x2 − 25 − (x2 − 6x + 9) = x2 − 25 − x2 + 6x − 9 = 6x − 34 Exercise

5.1

1 Expand (x + 5)(x + 3) by using the area diagram.

x

5

x 3

2 Expand each binomial product by using the distributive law. a (x + 2)(y + 3) b (p + 5)(q − 2) c

(u − 4)(v − 6)

167

Mathscape 10 ext. - Ch05 Page 168 Friday, September 23, 2005 7:50 AM

168

Mathscape

10

Extension

3 Expand these binomial products by using the distributive law, then simplify. a (x + 2)(x + 4) b (m + 1)(m + 6) c (u + 8)(u − 5) d (b + 9)(b − 2) e (a − 8)(a + 3) f (t − 1)(t − 2) g (c − 4)(c − 6) h (z − 6)(z − 7) i (d + 12)(d − 5) j (x + 3)(2x + 1) k (3m + 4)(m + 2) l (5a + 2)(a − 4) m (2g − 5)(g − 3) n (3t + 1)(t − 6) o (4n − 7)(n + 5) p (5r + 3)(2r + 7) q (2k + 9)(3k − 5) r (7v − 2)(3v − 8) ■ Consolidation

4 Prove the identity (x + a)(x + b) = x2 + (a + b)x + ab. 5 Expand the following expressions by using the identity in question 4. a (x + 2)(x + 1) b (p + 3)(p + 2) c (a + 7)(a + 3) d (t − 5)(t − 4) e (b − 6)(b − 2) f (c − 3)(c − 4) g (z + 4)(z − 1) h (d + 3)(d − 6) i (s − 2)(s + 10) j (e + 7)(e − 2) k (u − 3)(u + 9) l (k − 11)(k + 4) m (f + 5)(f + 7) n (w − 6)(w + 5) o (r − 8)(r − 4) p (g + 8)(g − 6) q (h + 11)(h − 5) r (v + 3)(v + 8) s (q − 9)(q − 4) t (m − 6)(m + 9) u (i + 3)(i − 12) v (l + 10)(l − 6) w (y − 13)(y + 3) x (j − 8)(j − 7) 6 Expand and simplify 3(2a + 5)(a − 4). 7 Expand: a (a + b)2

b (m + n)2

c

(g − h)2

d (p − q)2

8 Expand: a (a + 3)2 e (u − 5)2

b (p − 4)2 f (k + 10)2

c g

(c + 6)2 (s + 7)2

d (t − 1)2 h (p − 9)2

9 Expand: a (2x + 3)2 d (5c − 2)2 g (3p − q)2 j (4c + 5d)2 m (pq + r)2

b e h k

(3t − 5)2 (7 − 2g)2 (3a + 4b)2 (7g − 3h)2

n (6ab − 7cd)2

10 State whether each expression is a perfect square. a k2 + 4 b (a + 5)2 2 d t + 26t + 169 e a2 b2 2 g n −1 h 4e2 − 6e + 9

c f i l o c f i

(4m + 1)2 (5 + 6r)2 (2e − 11f)2 (12j + 5k)2 2 ⎛ x + 1---⎞ ⎝ x⎠ u2 + 4u + 16 c2 + 12c − 36 x2 + y2 + 2xy

Mathscape 10 ext. - Ch05 Page 169 Friday, September 23, 2005 7:50 AM

Chapter

5:

Factorisation and algebraic fractions

11 Complete these perfect squares. a (m + 5)2 = m2 + 10m + _____ c (_____)2 = y2 + _____ + 36 e (_____)2 = k2 + 8k + _____ g (2b + 3)2 = 4b2 + _____ + 9 i (_____)2 = 25n2 + 60n + _____

b d f h j

(p − 3)2 = p2 − _____ + 9 (_____)2 = a2 − _____ + 49 (_____)2 = u2 − 2u + _____ (_____)2 = 9e2 + _____ + 25 (_____)2 = 49z2 − 42z + _____

12 Use the difference of two squares identity to expand each of the following. a (m + n)(m − n) b (p − 2)(p + 2) c (r + 5)(r − 5) d (3 − g)(3 + g) e (y − 7)(y + 7) f (4 − w)(4 + w) g (t + 1)(t − 1) h (k − 6)(k + 6) i (10 − b)(10 + b) j (2e + 3)(2e − 3) k (5r − 2)(5r + 2) l (3c − 1)(3c + 1) m (4 − 7h)(4 + 7h) n (6 + 5n)(6 − 5n) o (2p − 9q)(2p + 9q) p (10y + 3z)(10y − 3z) q (7s + 6t)(7s − 6t) r (ab − 8c)(ab + 8c) ■ Further applications

13 Expand and simplify the following expressions. a (a + 7)(a +2) + 4 b c 4p + (2p − 3)(p − 2) d e (a + 5)2 − 2(a − 5) f g (m + 4)(m − 4) − (m + 3)2 h i (3k − 4)(k − 2) − (2k − 1)(k + 6) j k (2h + 7)2 − (2h + 3)(2h − 5) l

(t − 6)(t + 6) − 10 3e2 + (e − 4)2 − 6 (x + 4)(x + 6) + (x + 1)2 (y + 7)2 − (y + 7)(y − 7) (5c + 6)2 − (5c − 6)2 (4m − 3)(4m + 3) − (4m − 3)2

14 Expand and simplify: a (x + 2)(x2 + 3x + 4) c (k + 3)(k + 2)(k + 1) e (p + 1)(p + 7)2 g (t + 2)3

(a − 4)(a2 − 2a + 5) (n − 7)(n + 3)(n − 2) (y − 5)(y − 3)2 (e − 3)3

5.2

b d f h

The highest common factor

To factorise an expression means to write the expression as the product of its factors. This is the same as reversing or undoing the expansion process. Expanding a(b + c) = ab + ac Factorising

169

Mathscape 10 ext. - Ch05 Page 170 Friday, September 23, 2005 7:50 AM

170

Mathscape

10

Extension

Many expressions can be factorised in several ways. For example, we can factorise 6a + 12 as 1(6a + 12) or 2(3a + 6) or 3(2a + 4) or 6(a + 2). However, by convention, we take out the highest common factor (HCF) when factorising an expression. That is, we take out the greatest possible factor that is common to every term in the expression. In this example, the HCF of 6a and 12 is 6. Therefore, the correct factorisation of 6a + 12 is 6(a + 2). To factorise an algebraic expression:
write the HCF of the terms outside the grouping symbols
divide each term in the expression by the HCF to find the terms inside the grouping symbols. ab + ac = a(b + c)

and

ab − ac = a(b − c)

NOTE: • If the first term of an expression is negative, then by convention, the HCF is also negative. • Factorisations should be checked by expanding the answers.

EG +S

Example 1 Factorise: a 5p + 20 d 12h2 + 27h

b 8n − 20 e c2d − cd2

c f

m2 + 3m −uv + vw

Solutions a 5p + 20 = 5(p + 4) d 12h2 + 27h = 3h(4h + 9)

b 8n − 20 = 4(2n − 5) e c2d − cd2 = cd(c − d)

c

m2 + 3m = m(m + 3) −uv + vw = −v(u − w)

Exercise

f

5.2

1 Factorise each of these expressions by taking out the highest common factor. a 2c + 6 b 5m + 20 c 21 + 3e d 49 + 7x e 3g − 15 f 6k − 6 g 27 − 3r h 55 − 5t i 2x + 2y j 3m − 12n k 24p + 8q l 12f − 48g m xy + xz n ab − bc o m2 + 7m p c2 − c 2 Factorise by removing the highest common factor. a 4n + 6 b 9b + 15 c 12y − 8 e 21p + 28 f 15g − 18 g 14w + 21 i 44h − 33 j 35d + 49 k 20q − 50 m 25 − 45k n 56 + 63v o 40 − 64a

d h l p

10u − 25 24z − 30 44f − 77 24 − 84s

■ Consolidation

3 Factorise each expression completely. a 3xy + 12yz b 2pq + 10pr

c

5fg − 15gh

d 7cd − 42ce

Mathscape 10 ext. - Ch05 Page 171 Friday, September 23, 2005 7:50 AM

Chapter

e i m q

5:

6ij + 9jk 6b2 + 24b pqr + pqs fg2 − f 2gh

f j n r

Factorisation and algebraic fractions

12mn − 20np 21a2 − 7a cde − def 6jk + 15k2m

4 Factorise: a 4p + 4q + 4r d 3e + 6f + 15g g 30c2 − 12c + 18 j 20y − 25yz + 10y2 m p2q + pq + pq2

g k o s

22uv − 33vw 14q2 − 20q x2y + xy2 16tu2 + 20t2uv

ab + ac − ad 6m2 − 10m + 14n 11ab − a + ab2 7ij + 14i2 − 42ik 9rs − rs2 − r2s

b e h k n

c f i l o

5 Factorise by taking out the greatest negative common factor. a −2k − 8 b −3n − 15 c −10c − 25 e −4x + 4 f −9d + 18 g −16m + 20 i −18 − 27y j −36 + 15e k −32 − 40z m −cd − de n −ij + jk o −a2 − 7a q −8n2 + 20n r −21b − 24b2 s −20f + 36ef

h l p t

32st + 24tu 25u + 30u2 abc − ab2 28g2h2 − 49ghi

x2 − xy + xz 5 − 15v − 30v2 3r2 − 3rs − 9r 24z − 60yz − 96z2 a2bc + ab2c − abc2 d h l p t

−14w − 21 −9g + 21 −44 + 99t −11v + v2 −26c2d − 39cd 2

■ Further applications

6 Factorise by taking out the highest common factor only. a m2 + m3 b x6 + x2 10 3 d y −y e a4 + 3a7 9 5 g 7u + 4u h 5h3 − 4h4 j 22g8 + 11g3 k 16q7 − 24q13 TRY THIS

c f i l

t3 − t8 2g3 − g8 3c10 + 9c5 45z9 − 20z2

Market garden

A market gardener has two vegetable plots, one square and the other a rectangle with one side 3 m shorter than the side of the square, and the other side 4 m longer than the side of the square. Both plots have the same area. 1 Draw a diagram 2 Find the dimensions of each plot.

5.3

Difference of two squares

In exercise 5.1 it was shown that sum by difference products such as (a − b)(a + b) could be expanded, giving a difference of two squares, a2 − b2. This procedure can be used in reverse to factorise a difference of two squares. a2 − b2 = (a − b)(a + b)

171

Mathscape 10 ext. - Ch05 Page 172 Friday, September 23, 2005 7:50 AM

172

EG +S

Mathscape

10

Example 1 Factorise each of the following. a

x2 − 16

Solutions a x2 − 16 = (x − 4)(x + 4)

EG +S

Extension

b 9q2 − 25

c

121u2 − 49v2

b 9q2 − 25 = (3q)2 − 52 = (3q − 5)(3q + 5)

c

121u2 − 49v2 = (11u)2 − (7v)2 = (11u − 7v)(11u + 7v)

Example 2 Factorise completely: a

5y2 − 45

b 7c3 − 28cd2

Solutions In general, when factorising expressions, always remove the highest common factor first. a 5y2 − 45 = 5(y2 − 9) b 7c3 − 28cd 2 = 7c(c2 − 4d 2) = 5(y − 3)(y + 3) = 7c(c − 2d)(c + 2d) Exercise

5.3

1 a Expand (x + 3)(x − 3). Hence, factorise x2 − 9. b Expand (2x + 5)(2x − 5). Hence, factorise 4x2 − 25. 2 Factorise each of these as a difference of two squares. a p2 − q2 b c2 − d 2 c m2 − n2

d u 2 − v2

3 Factorise each expression as a difference of two squares. a x2 − 4 b a2 − 25 c p2 − 1 2 2 e z − 64 f c − 121 g t 2 − 49 i 16 − k2 j 64 − g2 k 100 − m2

d y2 − 36 h b2 − 81 l 144 − u2

4 Factorise: a e2 − 169

b h2 − 256

c

529 − s2

d 361 − j 2

c g k o s

49q2 − 4 1 − 121u2 36e2 − f 2 100p2 − 9q2 4x2 − 9y2z2

d h l p t

■ Consolidation

5 Factorise: a 4a2 − 9 e 36 − 49y2 i 9a2 − b2 m 4g2 − 49h2 q a2b2 − c2

b f j n r

9p2 − 25 25 − 16r2 x2 − 4y2 25m2 − 144n2 p2 − q2r2

16c2 − 1 100 − 81t 2 j 2 − 64k2 36s2 − 121t 2 16e2f 2 − 81g2h2

6 Factorise these expressions completely by first taking out the highest common factor. a 2m2 − 18 b 3a2 − 12 c 11t 2 − 11 d 5y2 − 500 2 2 2 e 75 − 3x f 63 − 7p g 128 − 2e h 54 − 6z2

Mathscape 10 ext. - Ch05 Page 173 Friday, September 23, 2005 7:50 AM

Chapter

5:

i 8n2 − 18 m a3 − a q 8h3 − 32h

Factorisation and algebraic fractions

j 12c2 − 27 n n2 − n4 r 54w3 − 24w

k 45 − 20f 2 o 49d 3 − d s 72s − 32s3

l 18k2 − 50 p 16u − 25u3 t 80j 3 − 125j

Factorise 4a2 − 36 as a difference of two squares. Has it been completely factorised? Why? b Discuss the correct method for factorising this expression, then factorise it completely.

7 a

8 Factorise completely each of these expressions. a 9k2 − 36 b 4c2 − 100 c 25x2 − 100y2

d 4e2 − 144f 2

9 Evaluate 162 − 152 by first expressing it as (16 − 15)(16 + 15). 10 Use the method in question 9 to evaluate: a 172 − 49 b 992 − 1

c

452 − 25

c f i

(x + 3)2 − y2 (p + 2)2 − 4 81(3p + 2q)2 − 64r2

■ Further applications

11 Factorise each expression as a difference of two squares. a (a + b)2 − c2 b (m − n)2 − p2 2 d (j − k) − 36 e (b + 5)2 − 49 2 g 4(m + n) − 9 h 9(c − d)2 − 25

5.4

Grouping in pairs

Expressions that contain four terms can often be factorised by grouping the terms in pairs. To factorise a four-term expression:
group the expression in pairs such that each pair has a common factor
factorise both pairs of expressions
factorise this overall expression by taking out a binomial common factor. ac + bc + ad + bd = c(a + b) + d(a + b) = (a + b)(c + d) NOTE: In some questions it may be necessary to first re-arrange the terms into a different order.

EG +S

Example 1 Factorise: a

pq + 7p + 3q + 21

Solutions a pq + 7p + 3q + 21 = p(q + 7) + 3(q + 7) = (q + 7)(p + 3)

b mn − mp − 4n + 4p

c

xy − yz + x − z

b mn − mp − 4n + 4p = m(n − p) − 4(n − p) = (n − p)(m − 4)

c

xy − yz + x − z = y(x − z) + 1(x − z) = (x − z)(y + 1)

173

Mathscape 10 ext. - Ch05 Page 174 Friday, September 23, 2005 7:50 AM

174

EG +S

Mathscape

10

Extension

Example 2 Factorise e2 + 5g + eg + 5e by grouping in pairs. Solution The expression must be re-arranged since the first pair of terms does not have a common factor. e2 + 5g + eg + 5e = e2 + eg + 5e + 5g = e(e + g) + 5(e + g) = (e + g)(e + 5) Exercise

5.4

1 Factorise each of the following by taking out the binomial common factor. a a(c + d) + b(c + d) b x(x + 3) + 2(x + 3) c 3n(p − 4) + 7(p − 4) d w(x + y) − z(x + y) e t(t − 1) − 9(t − 1) f 4g(a − 8) − 5h(a − 8) 2 2 g p(q + r) + s(r + q) h 5(a + 2) − c(a + 2) i mn (y + 4) − 8(4 + y) j u(d + e) + (d + e) k (p − 2q) + w(p − 2q) l 6h(7i − 2j) − (7i − 2j) ■ Consolidation

2 Factorise each expression by taking out the highest common factor from each pair of terms. a km + kn + 5m + 5n b ac + ad + bc + bd c p2 + pq + 2p + 2q d wx + 3x + 2wy + 6y e ce − 2c + 4de − 8d f 2gh − 2g + 3h − 3 g 4uv + 28u + 5v + 35 h 3x + 6y + 8ax + 16ay i 5mn − 15m + 6pn − 18p j 3gk + 18gh + 4k + 24h k pqr + 6pq + 7r + 42 l mn + np + m + p m a − 5b + ac − 5bc n 3e + f + 3e2 + ef o a3 + a2 + a + 1 3 Factorise each of these by taking out a negative common factor from the second pair of terms. a 2c + 2d − ce − de b pq + ps − qr − rs c p2 − pq − 8p + 8q 2 d xy − 3x − y + 3y e gh + gi − h − i f 3u − 2v − 3uw + 2vw g x3 − 4x2 − xy + 4y h k2 + 7k − 2km − 14m i jkm − 4jk − 2m + 8 j 35np − 14nq − 5p + 2q k 3z2 − 3wz − uz + uw l 10x + 15y − 4xz − 6yz 4 Factorise each expression by first re-arranging the terms into a more suitable order. a xy + wz + xz + wy b pr + 2q + qr + 2p c 3mn + kp + kn + 3mp d 4yz + 15 + 3z + 20y e cd + 7 + 7c + d f 6 + 5ef + 6e + 5f ■ Further applications

5 Explain why a(b − c) = −a(c − b). Hence, factorise each of the following. a 2x − xy + y2 − 2y b yz − 7y + 7z − z2 c ab − ac − bc + b2 3 2 2 2 d a − a b + b − ab e 11m − mn − n + 11n f m − 4n − 20pn + 5mp g 6uv − 6uw − 5w2 + 5vw h 14de − 7ce + c − 2d i 2wx − 2wy + y2 − xy 3 2 2 2 j c − c d − d + cd k 10pq − 25pr + 30qr − 12q l 18rs − 63rt − 28st + 8s2

Mathscape 10 ext. - Ch05 Page 175 Friday, September 23, 2005 7:50 AM

Chapter

5:

Factorisation and algebraic fractions

6 Factorise: a (x + 3)(x + 4) + y(x + 4) c (k + 6)(k − 4) − m(k − 4) e (x − 7)2 − w(x − 7)

b (m − 2)(m − 5) + n(m − 2) d (a + b)2 + c(a + b) f f(g + h) + (g + h)2

7 Factorise: a (2c − d)(c + 3d) − 5(d − 2c)

b (p − q)2 − 3r(q − p)

TRY THIS

Squaring fives

Consider the following: 152 = 10 × 20 + 25 = 225, 252 = 20 × 30 + 25 = 625, 352 = 30 × 40 + 25 = 1225. Why does this work? Use algebra to show why this works. (NOTE: A number mn equals 10 × m + n). Does the rule work for numbers greater than 125, e.g. (125)2?

5.5

Factorising monic quadratic trinomials

Expressions such as 2a + b − 5c, which have three terms, are called trinomials. Trinomials such as x2 + 4x + 7 and 3x2 − 2x + 8 are called quadratic trinomials because the highest power of the variable is 2. If the co-efficient of x2 is 1, the trinomial is said to be monic. Hence, x2 + 4x + 7 is a monic quadratic trinomial. Let x2 + bx + c = (x + p)(x + q) ∴ x2 + bx + c = x2 + px + qx + pq = x2 + (p + q)x + pq Equating the co-efficients of x we have p + q = b and equating the constants we have pq = c. (x + p)(x + q) = x2 + (p + q)x + pq To factorise a monic quadratic trinomial x2 + bx + c:
find by inspection two numbers p, q such that p + q = b and pq = c
factorise the trinomial as (x + p)(x + q). NOTE: In the trinomial x2 + bx + c: 1 if c > 0, then the numbers p, q both have the same sign as b 2 if c < 0, then the numbers p, q have different signs, with the larger number having the same sign as b.

175

Mathscape 10 ext. - Ch05 Page 176 Friday, September 23, 2005 7:50 AM

176

EG +S

Mathscape

10

Extension

Example 1 Factorise: a

x2 + 9x + 20

b x2 − 7x + 12

Solutions a 4 + 5 = 9 and 4 × 5 = 20 ∴ x2 + 9x + 20 = (x + 4)(x + 5) c (−1) + 3 = 2 and (−1) × 3 = −3 ∴ x2 + 2x − 3 = (x − 1)(x + 3)

EG +S

x2 + 2x − 3

c

d x2 − 4x − 12

b (−3) + (−4) = −7 and (−3) × (−4) = 12 ∴ x2 − 7x + 12 = (x − 3)(x − 4) d (−6) + 2 = −4 and (−6) × 2 = −12 ∴ x2 − 4x − 12 = (x − 6)(x + 2)

Example 2 Factorise 3x2 + 21x + 30 Solution 3x2 + 21x + 30 = 3(x2 + 7x + 10) = 3(x + 5)(x + 2) Exercise

5.5

1 Find two integers a and b such that: a a+b=8 b a+b=9 ab = 15 ab = 14 e a + b = −7 f a + b = −3 ab = 10 ab = 2 i a+b=4 j a+b=1 ab = −12 ab = −20

c g k

a + b = 10 ab = 24 a + b = −11 ab = 28 a + b = −2 ab = −35

d h l

a + b = 13 ab = 30 a + b = −17 ab = 72 a + b = −3 ab = −54

2 a Show that (x + a)(x + b) = x2 + (a + b)x + ab. b What can you say about the signs of a and b, where a > b if: i ab > 0 and a + b > 0? ii ab > 0 and a + b < 0? iii ab < 0 and a + b > 0? iv ab < 0 and a + b < 0? ■ Consolidation

3 Factorise each of these monic trinomials. a x2 + 5x + 6 b y2 + 7y + 12 2 e a + 10a + 21 f t 2 + 9t + 18 2 i n + 14n + 45 j d 2 + 12d + 32 2 m e + 15e + 50 n c2 + 17c + 60

c g k o

u2 + 3u + 2 k 2 + 9k + 14 s2 + 11s + 30 r2 + 19r + 88

d h l p

m2 + 10m + 24 p2 + 8p + 7 b2 + 13b + 36 z2 + 20z + 96

4 Factorise each of the following: a m2 − 8m + 15 b q2 − 7q + 10 2 e u − 11u + 18 f e2 − 9e + 8 i h2 − 16h + 63 j v2 − 11v + 24 2 m k − 13k + 30 n j 2 − 18j + 77

c g k o

d 2 − 12d + 20 n2 − 13n + 40 t 2 − 7t + 6 x2 − 15x + 54

d h l p

a2 − 12a + 35 w2 − 9w + 20 s2 − 15s + 44 f 2 − 14f + 48

Mathscape 10 ext. - Ch05 Page 177 Friday, September 23, 2005 7:50 AM

Chapter

5:

Factorisation and algebraic fractions

5 Factorise each of these. a y 2 + 4y − 12 b e v2 − 3v − 10 f i f 2 + 4f − 21 j m z2 − 3z − 40 n q s2 + 3s − 70 r

d 2 + 2d − 15 u2 − 7u − 18 w2 − 2w − 63 i 2 + i − 12 h2 + h − 56

c g k o s

a2 + 3a − 4 m2 − m − 30 k 2 − 6k − 27 r 2 − 2r − 99 b2 − 6b − 40

d h l p t

p2 + 6p − 16 x2 − 4x − 5 c2 + 4c − 32 e2 − 5e − 14 t 2 + 3t − 108

6 Factorise: a n2 − 6n + 8 e q2 + 13q + 22 i g2 − 8g − 20 m e2 + 22e + 120 q z2 − 2z − 80 u w2 − 5w − 36

c2 + 13c + 42 t 2 − 11t + 10 b 2 + 15b + 36 l2 − l − 42 a2 + 4a − 60 k 2 + 15k + 26

c g k o s w

x2 − x − 2 v2 + 14v + 48 r2 − 3r − 54 y2 − 14y + 24 f 2 − 16f + 39 h2 − 19h + 84

d h l p t x

d 2 + 7d − 44 j 2 − 6j − 72 u2 − 13u + 12 p2 + 6p − 55 m2 + 21m + 110 i 2 + i − 132

7 Factorise each of these perfect squares. a p2 + 6p + 9 b c2 + 10c + 25 2 e t + 2t + 1 f r 2 − 12r + 36

c g

g2 − 8g + 16 x2 − 18x + 81

d y 2 − 14y + 49 h j 2 + 22j + 121

b f j n r v

8 Factorise each expression completely by first taking out the highest common factor. a 2m2 − 20m + 32 b 3k 2 + 24k + 36 c 4a2 + 24a − 64 2 2 d 5c − 30c − 135 e 3t − 27t + 24 f 7x2 − 7x − 84 2 2 g 4d − 32d + 60 h 5n + 40n − 100 i 10y2 − 80y − 330 9 Find what: a x + 3 must be multiplied by to give x2 + 9x + 18 b g − 3 must be multiplied by to give g2 + 2g − 15 ■ Further applications

10 Factorise each expression completely. a x4 − 5x2 + 4 b x4 − 29x2 + 100 e x4 + 7x2 − 8 f x4 − 10x2 + 9 4 2 i x − 17x + 16 j x4 − 7x2 + 6

5.6

c x4 − 11x2 + 18 g x4 − 19x2 + 48 k x4 + 3x2 − 28

d x4 + x2 − 20 h x4 − 13x2 + 36 l x4 − 25x2 + 144

Factorising general quadratic trinomials

In quadratic trinomials such as 3x2 + 5x + 2 and 4 + 7x − 5x2, the co-efficient of x2 is not 1. Such trinomials are often referred to as general quadratic trinomials. The identity x2 + (a + b)x + ab = (x + a)(x + b) cannot be used to factorise quadratic trinomials where the leading co-efficient is a number other than 1. In these situations we must resort to a guess and check approach. This is made easier by the use of the cross-method.

177

Mathscape 10 ext. - Ch05 Page 178 Friday, September 23, 2005 7:50 AM

178

Mathscape

10

Extension

To factorise a general quadratic trinomial ax2 + bx + c using the cross method:
draw a cross
on the left-hand side of the cross, write two factors that multiply to give ax2
on the right-hand side of the cross, write two factors that multiply to give c
check that the sum of the cross-products gives the middle term bx
if the cross-products do not give the correct sum, try different combinations until the correct sum is obtained
factorise the trinomial. ax2 + bx + c = (px + r)(qx + s) where • a = pq and c = rs • b = ps + qr

EG +S

Example 1 Factorise 3x2 + 14x + 8 using the cross method. Solution 3x +1

EG +S

(3x × 8) + (x × 1) = 24x + x = 25x

This combination does not give the middle term in the trinomial, so it is not correct.

(3x × 1) + (x × 8) = 3x + 8x = 11x

This combination does not give the middle term in the trinomial, so it is not correct.

(3x × 2) + (x × 4) = 6x + 4x = 10x

This combination does not give the middle term in the trinomial, so it is not correct.

This combination gives the correct middle term in the trinomial.

x

+8

3x

+8

x

+1

3x

+4

x

+2

3x

+2

(3x × 4) + (x × 2) = 12x + 2x = 14x

x

+4

∴ 3x2 + 14x + 8 = (3x + 2)(x + 4)

Example 2 Factorise: a

2x2 − 13x + 21

b 5x2 − 4x − 12

Mathscape 10 ext. - Ch05 Page 179 Friday, September 23, 2005 7:50 AM

Chapter

Solutions a 2x

−7

x b 5x

−3 +6

x

−2

Exercise

5:

Factorisation and algebraic fractions

(2x × −3) + (x × −7) = −6x + (−7x) = −13x

This is equal to the middle term in the trinomial. ∴ 2x2 − 13x + 21 = (2x − 7)(x − 3)

(5x × −2) + (x × 6) = −10x + 6x = −4x

This is equal to the middle term in the trinomial. ∴ 5x2 − 4x − 12 = (5x + 6)(x − 2)

5.6

1 Which diagram correctly shows the factors of 2x2 + 13x + 21? +3 B 2x +7 C 2x +1 A 2x

+1

D 3x

+2

−5

x

−5

+3

D 5x

−3

−2

x

+2

4 Which diagram correctly shows the factors of 2x2 − 3x − 27? −9 B 2x +9 C 2x −3 A 2x

D 2x

+3

x

−9

x

+3

x

+21

+21

x

x

+7

D 2x

2 Which diagram correctly shows the factors of 3x2 − 17x + 10? +5 B 3x −5 C 3x −2 A 3x

x

−2

x

−2

x

3 Which diagram correctly shows the factors of 5x2 + 7x − 6? +6 B 5x −6 C 5x A 5x

x

x

−1

+3

x

x

+1

−3

x

x

+9

■ Consolidation

5 Factorise each of these trinomials. a 2x2 + 7x + 3 b 3x2 + 14x + 8 d 2x2 + 11x + 5 e 5x2 + 22x + 8 2 g 2x + 21x + 54 h 7x2 + 27x + 18 2 j 2x − 5x + 2 k 3x2 − 11x + 6

c f i l

4x2 + 11x + 7 7x2 + 36x + 5 3x2 + 20x + 32 5x2 − 17x + 6

179

Mathscape 10 ext. - Ch05 Page 180 Friday, September 23, 2005 7:50 AM

180

Mathscape

10

Extension

m 2x2 − 13x + 15 p 3x2 − 26x + 35

n 3x2 − 16x + 16 q 2x2 − 19x + 44

o r

5x 2 − 16x + 12 7x2 − 41x + 30

6 Factorise each of these trinomials. a 3x2 + 5x − 2 b 2x2 + x − 10 2 d 3x + 2x − 21 e 2x2 − 5x − 25 2 g 5x − 11x − 36 h 11x2 − 52x − 15 j 6x2 − 17x − 3 k 5x2 + 4x − 12 2 m 7x + 51x − 40 n 7x2 + 36x − 36

c f i l o

2x2 + 7x − 15 3x2 − 22x − 16 4x2 + 27x − 7 4x2 − 17x − 42 4x2 + 9x − 28

7 Factorise: a 2k 2 + 21k + 40 d 3y2 − 17y − 28 g 4b 2 + 31b + 42 j 5h2 + 19h + 12

c f i l

4n2 − 17n + 18 6a2 − 23a − 4 2w2 − 9w − 110 6l 2 − 59l + 45

8 Factorise the following trinomials. b 4n2 − 20n + 21 a 6a2 + 13a + 6 2 d 9p − 9p − 10 e 6c2 + 19c + 15 g 12t 2 − 28t + 15 h 10b2 + 29b − 72 2 j 20y + 33y + 10 k 27w2 − 12w − 55

c f i l

8k 2 + 14k − 9 4e2 + 8e − 45 12m2 − 47m + 40 20q2 + 7q − 6

9 Factorise: a 28 + 3x − x2 d 15 − 28m + 12m2

c f

24 − 14p − 3p2 56 + 19w − 10w2

b e h k

2c2 + 3c − 27 7p2 − 31p + 12 5u2 + 3u − 36 3j 2 + 2j − 40

b 15 − a − 2a2 e 21 + 43g + 20g2

10 Factorise each expression completely by first taking out the highest common factor. a 4k 2 + 18k + 20 b 18p2 − 33p + 9 c 20a2 + 2a − 6 2 2 d 12v − 24v − 63 e 24f − 44f + 12 f 10e2 + 15e − 100 11 Find what: a a + 9 must be multiplied by to give 3a2 + 31a + 36 b 3n − 8 must be multiplied by to give 6n2 + 11n − 72 ■ Further applications

Quadratic trinomials can be factorised using the following identity: ( ax + m ) ( ax + n ) ax 2 + bx + c = ------------------------------------------ , where m + n = b and mn = ac. a For example, to factorise 2x2 + 7x + 6, we first find m, n such that m + n = 7 and mn = 2 × 6, i.e. mn = 12. Therefore, m = 4, n = 3 (or m = 3, n = 4). ( 2x + 4 ) ( 2x + 3 ) So, 2x2 + 7x + 6 = ----------------------------------------2 2 ( x + 2 ) ( 2x + 3 ) = ----------------------------------------2 = (x + 2)(2x + 3) 12 Use this method to factorise the following trinomials. a 3x2 + 10x + 3 b 2x2 − 11x + 12 2 d 4x − 17x − 15 e 7x2 − 15x + 2

c f

3x2 + 4x − 4 5x2 + 22x − 15

Mathscape 10 ext. - Ch05 Page 181 Friday, September 23, 2005 7:50 AM

Chapter

5.7

5:

Factorisation and algebraic fractions

Miscellaneous factorisations

To factorise an expression:
take out the HCF first if there is one
look for a difference of two squares if there are 2 terms
use the sum/product identity or cross method if there are 3 terms
group the terms in pairs if there are 4 terms.

EG +S

Example Factorise: a p2q − pq2 d 40 − s − 6s2 Solutions a p2q − pq2 = pq(p − q) d 40 − s − 6s2 5 −2s

8 +3s = (5 − 2s)(8 + 3s)

Exercise

b 2w2 − 18 e 4t 2 − 36t + 81

c f

4a2 + 20a − 24 7cd + 28ce + 2de + 8e2

b 2w2 − 18 = 2(w2 − 9) = 2(w − 3)(w + 3) e 4t 2 − 36t + 81 2t −9

c

4a2 + 20a − 24 = 4(a2 + 5a − 6) = 4(a + 6)(a − 1) 7cd + 28ce + 2de + 8e2 = 7c(d + 4e) + 2e(d + 4e) = (d + 4e)(7c + 2e)

f

2t −9 = (2t − 9)(2t − 9) = (2t − 9)2

5.7

1 Factorise each of these expressions. a 10e + 30 b a2 − 4 d n2 + 6n + np + 6p e 2x2 + 13x + 20 2 g x −x h t 2 − 6t + 8 j ap − 3a + 5p − 15 k 15y + 35 m g2 − h2 n −7j + 14k p 3b2 + 26b + 16 q 28t 2 − 63tu s 4xy − 8x + 9y − 18 t efg + fgh v 25v2 − 36w2 w 5gm − 30m − 2g + 12

c f i l o r u x

m2 + 7m + 12 25 − k 2 w2 + 4w + 4 2q2 − 7q + 3 u2 − 2u − 15 4c2 − 9 h2 + 8h − 20 2f 2 − 3f − 5

c f

9e2 − 64f 2 60xy − 84y

■ Consolidation

2 Factorise each expression completely. a 27pq + 45p2 b x2 − 4x − 77 d 3b2 − b − 44 e 5g + 15h − 2gi − 6hi

181

Mathscape 10 ext. - Ch05 Page 182 Friday, September 23, 2005 7:50 AM

182

Mathscape

g j m p s v

10

Extension

z2 + 13z − 30 64 − k 2l 2 9h2 − 36 n3 − n 2u2 + 20u + 50 10a2 + 14ab − a

h k n q t w

p2q2 − r2 m3 + 7m2 − mn − 7n 3x2 + 3x − 60 (a + b)2 + c(a + b) −2f 2 + 32 2k 3 + 12k 2 − 54k

i l o r u x

40 + 11k − 2k2 10s2 − 53s + 36 5 − 3j − 20k + 12jk 1 + t + t2 + t3 10a − 25b + 12ac − 30bc 4u2v − 17uv − 42v

c f i

w4 − 26w2 + 25 y2 + 14y − 72 4gx2 − 100g − 75 + 3x2

■ Further applications

3 Factorise each expression completely. a (x + y)2 + z(x + y) b t4 − 1 d 25a2b2 − 100b2c2 e 16 − n4 2 2 g (a + b) − (c + d) h a3 − 9a − a2b + 9b TRY THIS

Difference of two squares

Factorise x4 + x2 + 1. [HINT: Express x4 + x2 + 1 as the difference of two squares.]

5.8

Simplifying algebraic fractions

When simplifying numerical fractions, we divide the numerator and denominator by their highest common factor. Algebraic fractions can be simplified in a similar manner. To simplify algebraic fractions:
factorise both the numerator and the denominator if possible
cancel any common factors.

EG +S

Example Factorise and simplify: 8x + 20 a ----------------------6x 2 + 15x

b

d 2 – 49 -----------------3d – 21

c

b 2 – 11b + 30 -------------------------------b 2 – 3b – 18

b

d 2 – 49 -----------------3d – 21

c

b 2 – 11b + 30 -------------------------------b 2 – 3b – 18

Solutions a

8x + 20 ----------------------6x 2 + 15x 4 ( 2x + 5 ) = --------------------------3x ( 2x + 5 )

(d + 7)(d – 7) = ---------------------------------3(d – 7)

(b – 5)(b – 6) = ---------------------------------(b + 3)(b – 6)

4 = -----3x

d+7 = -----------3

b–5 = -----------b+3

Mathscape 10 ext. - Ch05 Page 183 Friday, September 23, 2005 7:50 AM

Chapter

Exercise

5:

Factorisation and algebraic fractions

5.8

1 Express these algebraic fractions in simplest form. 6k 4t 3a a -----b -----c -----8 20 6 4 5 28c f --------g -----h --------16g 5d 21 25gh 6ab pqr k --------l --------m -----------30h 18a qrs p

u2 ----u

q

s ---2s

2 Factorise and simplify: 3x + 3 a --------------b 3 40t – 50 e -------------------f 20 5t i -----------------------2j 10t – 25t x+y m -----------------n 4x + 4y

r

4k – 8 --------------4 ab + ac -----------------a 14c 2 – 7cd -------------------------21c 2 x + xy ----------------xz + yz

v2w --------2vw

d i n s

c g k o

12m ---------15 ab -----bc 42x -----------49xy

e j o

36r 2 ----------45r

10m + 15 ----------------------5 u ------------------uv – uw 8mn + 16m 2 -----------------------------24mp 12ac – 14bc -----------------------------18ac – 21bc

t

d h l p

15y --------10 pq -----5p 36ef -----------27 fg 60cd 2 -------------35c 2 d

24a – 32 --------------------16 2 x + 6x ----------------x 15g 2 – 20gh ----------------------------55g 2 15c 2 + 6cd -----------------------------218cd + 45c

■ Consolidation

3 Factorise and simplify: m 2 – 49 a -----------------b m–7 9t 2 – 25 e ------------------f 3t – 5 4 Factorise and simplify: x 2 + 4x + 3 a --------------------------x+3 n+8 d ----------------------------n 2 + 3n – 40 z 2 – 81 g -----------------------2 z – z – 90 5 Factorise and simplify: 2x 2 + 17x + 30 a -----------------------------------x+6 d

25k 2 – 64 ----------------------------------5k 2 + 12k – 32

a 2 – 25 -----------------4a + 20 5h 2 – 5 ----------------5h – 5

b e h

b e

c g

x2 – x ------------x2 – 1 6k + 18 ------------------4k 2 – 36

a 2 – 10a + 24 --------------------------------a–6 2 c – 6c + 9 ------------------------5c 2 – 15c s 2 – 11s + 28 ------------------------------s 2 + 4s – 77

2r – 5 -------------------------------------10r 2 – 29r + 10 4h 2 – 19h – 63 ----------------------------------h 2 – 13h + 42

d h

c f i

n + 12 ------------------n 2 – 144 ( p – q )2 -----------------p2 – q2

w 2 – 6w – 40 ------------------------------2w – 20 2 e + 14e + 33 -------------------------------e 2 – 121 2d 2 + 22d + 48 -----------------------------------4d 2 – 36

c

6 p 2 + 17 p + 12 ------------------------------------21 p + 28

f

3v 2 – 7v – 6 -------------------------------4v 2 – 13v + 3

183

Mathscape 10 ext. - Ch05 Page 184 Friday, September 23, 2005 7:50 AM

184

Mathscape

10

Extension

6 Factorise and simplify: px + qx + py + qy a -------------------------------------------x+y d

b

x 2 + xy – 5x – 5y ----------------------------------------x 2 – 8x + 15

e

km – 5k + 3m – 15 -------------------------------------------m 2 – 25 ab – ac – 4b + 4c -----------------------------------------2a 2 – 32

c

u 2 – 49 --------------------------------------------uv – uw – 7v + 7w

f

n 2 + 2n – pn – 2 p ------------------------------------------n 2 + 2n + pn + 2 p

■ Further applications

a–b 7 Show that ------------ = – 1 . Hence, simplify each of these fractions. b–a h–3 8y – 16 e– f a -----------b -----------------c -----------------3–h 2–y 5 f – 5e d

4m – 4n -------------------6n – 6m

e

a 2 – 169 -------------------13 – a

f

c 2 – 12c + 36 -------------------------------12 – 2c

g

s 2 – 17s + 72 ------------------------------24 – 3s

h

g 2 – 12g + 27 -------------------------------9g 2 – g 3

i

48 – 3k 2 ------------------------k 2 – 2k – 8

5.9

Multiplying and dividing algebraic fractions

To multiply or divide algebraic fractions:
change divisions to multiplications and take reciprocals of any fractions that follow division signs
factorise each expression if possible
cancel common factors between the numerator and denominator
multiply the numerators and multiply the denominators.

EG +S

Example 1 z 2 + 2z – 35 3z + 15 ---------------------------× Factorise and simplify ---------------------. 9z 2 + 63z z 2 – 25 Solution z 2 + 2z – 35 3(z + 5) (z + 7)(z – 5) 3z + 15 ------------------------------------------------× = ---------------------- × --------------------------------9z (z + 7) (z + 5)(z – 5) 9z 2 + 63z z 2 – 25 3 1

1 = ----3z

Mathscape 10 ext. - Ch05 Page 185 Friday, September 23, 2005 7:50 AM

Chapter

EG +S

5:

Factorisation and algebraic fractions

Example 2 a 2 + 5a – 24 a 2 + 8a - ÷ -----------------. Factorise and simplify ----------------------------a2 – a – 6 a2 – 4 Solution a 2 + 5a – 24 a 2 + 8a a 2 + 5a – 24 a 2 – 4 ----------------------------- ÷ ----------------- = ----------------------------- × ----------------a2 – a – 6 a2 – 4 a2 – a – 6 a 2 + 8a (a + 8)(a – 3) (a – 2)(a + 2) = ---------------------------------- × ---------------------------------a(a + 8) (a – 3)(a + 2) a–2 = -----------a Exercise

5.9

1 Simplify each of the following. 2 2m x y a --- × --b ---- × ------m 9 2 7 e

m 2 ---- ÷ --6 n

f

5 10y --- ÷ --------x 3x

2 Simplify: 8 a+3 a ------------ × -----------a+3 2 c e

4b ( 2b + 3 ) ( 2b + 3 ) ( b – 4 ) -------------------------- × ------------------------------------6b ( 2b + 3 ) 2 2 ( x – y) 7( x – y) ------------------ ÷ ------------------10 15

15a 24bc --------- × -----------16b 25a

d

g

7r 21r -------- ÷ -------12t 16s

h

b

10q p–1 ------------ × -------------------3( p – 1) 5q

d

27 12 ------------- ÷ --------------------m + 8 2(m + 8)

f

(k + 2)(k – 3) (k – 3)(k + 4) --------------------------------- ÷ --------------------------------9(k + 4) 12 ( k + 1 )

3 Factorise and simplify each of the following. 9 2y + 10 a ------------------ × -----------------b 3y + 15 15 c e

2a 2 + 3a bc 2 -------------------- × ---------------------2 c – 3c 2ab + 3b 10t – 30 8t – 24 -------------------- ÷ ----------------35u 21t

4n + 12 9k – 18 --------------------- × -----------------12k – 24 7n + 21

d

3 4a + 4b ------------------- ÷ -----------------ac + bc 12

f

18m + 12n 27m + 18n ------------------------- ÷ ------------------------20m 2 n 35mn 2

b

s 2 – 2s s + 2 ---------------- × -----------7 s2 – 4

d

x 2 + 8x + 15 9x ------------------------------ × ---------------x+3 x 2 – 25

■ Consolidation

4 Factorise and simplify: a2 – 9 b + 2 a ------------------ × -----------5b + 10 a – 3 c

t + 4 3t 2 – 12t -------------- × -------------------2t – 8 t 2 – 16

15de pq 2 ----------- × --------4 p 2 q 9cd 36ad 42a 2 ------------ ÷ -----------35bc 55ab

c

185

Mathscape 10 ext. - Ch05 Page 186 Friday, September 23, 2005 7:50 AM

186

Mathscape

e g i k

10

Extension

5k + 25 k2 – 4 ----------------------------- × ----------------2 k + 7k + 10 k 2 – 2k n 2 – 9n + 20 n 2 – n – 2 ----------------------------- × --------------------------n 2 – 6n + 5 n 2 – 6n + 8 a 2 + 2ab + b 2 a 2 b – ab 2 -------------------------------- × ---------------------a2 – b2 a2b2 2w 2 + 19w – 33 2w 2 – 242 ------------------------------------- × -----------------------6w – 66 16w 2 – 24w

e2 – 1 2e 2 + 12e + 16 - × ----------------------------------m ----------------------------------2 2 e –e 6e + 18e + 12 5 Factorise and simplify: k 2 – 36 5k + 30 - ÷ -----------------a ---------------35 k 2 – 6k c

h + 3 h 2 + 4h + 3 ------------ ÷ -------------------------18h 9h 2 + 9h

z 2 – 13z + 30 z 2 – 100 ------------------------------÷ ---------------------2 z 2 – 3z ( z + 10 ) 2 4xy + 4y 2 xy – y g ------------------2 ÷ ----------------------( x – y) x2 – y2 2m 2 – 24m m 2 – 144 - ÷ -------------------------i ----------------------------------2 m + 15m + 36 ( m + 3 )2 3q 2 – 13q + 14 3q 2 + 2q – 21 - ÷ -------------------------------k ----------------------------------q 2 – 64 q 2 – 5q – 24 x 2 + 8x + xy + 8y x 2 + 12x + 32 - ÷ -------------------------------m ----------------------------------------x 2 – 4x + xy – 4y 2x 2 – 32 e

f h j

c 2 + 5c + 6 c 2 + 5c -------------------------× ----------------------------2 2 c + 7c c + 8c + 15 2 r – 49 r2 – 9 -----------------------------------------------× 5r 2 – 15r r 2 – 4r – 21 b 2 – 14b + 45 b 3 – 4b 2 -------------------------------- × -------------------b–9 2b 2 – 10b

3a 2 + 22a + 24 a2 – 4 ---------------------------------------------------------------× 3a 2 – 2a – 8 a 2 + 6a d2 – f 2 cd + ce + d 2 + de - × ----------------------------------------n ----------------------------------------2 2 d – df d + df + de + ef l

b

n 2 – 8n n 2 – 64 ----------------- ÷ ----------------4n 4n + 8

d

a2 – 1 a2 + a --------------------- ÷ -------------------------10a + 25 24a 2 + 60a

f h j l n

2t 2 – 6t t 2 + 3t – 18 -------------------- ÷ --------------------------7t 2 – 28t t 2 + 2t – 24 r 2 + 12r + 35 r 2 + r – 42 ------------------------------------------------------------÷ r 2 – 2r – 35 r 2 – 14r + 48 2x 2 + 7x + 5 4x 2 – 25 ----------------------------- ÷ ----------------------x2 – 1 6x 2 – 15x 2u 2 – 18 4u 2 + 48u + 108 ------------------------------------------------------------------÷ 7u 2 v – 21uv 7u 2 + 63u 49c 2 – 121 7c 2 + 3c – 22 -------------------------------÷ -------------------------------2 2c + 17c – 9 c 2 + 11c + 18

■ Further applications

6 Factorise and simplify: 2 – 2a 4 a --------------- × --------------5 3a – 3 c

5x 2 + 5xy 2xy + 2y 2 -------------------------------------------------------÷ 21y 2 z – 42xyz 28x – 14y

b

10n – 20 mn 2 + 2mn ------------------------------- × --------------------------12 – 6n 15mn + 30m

8q 2 – 4 pq 3 p 2 – 6 pq ---------------------------d --------------------------------÷ 27 p 2 q + 9 pq 2 – 36 p – 12q

Mathscape 10 ext. - Ch05 Page 187 Friday, September 23, 2005 7:50 AM

Chapter

5.10

5:

Factorisation and algebraic fractions

Adding and subtracting algebraic fractions

Algebraic fractions, like numerical fractions, can only be added or subtracted if they have a 1 7 common denominator. Consider the addition --- + ------ . The lowest common denominator (LCD) 6 10 is not necessarily found by multiplying the denominators. Rather, it is found by factorising the individual denominators into primes, with each different factor then taken once and multiplied together. 1 7 1 7 For example, --- + ------ = ------------ + -----------6 10 3 × 2 2 × 5 ( 5 × 1 ) + ( 3 × 7 ) • (3 × 2) divides into (3 × 2 × 5) 5 times, 5 × 1 = 5 = ---------------------------------------• (2 × 5) divides into (3 × 2 × 5) 3 times, 3 × 7 = 21 3×2×5 26 = -----30 13 = -----15 The same method is used to add or subtract algebraic fractions. To add or subtract algebraic fractions with binomial or trinomial denominators:
factorise the denominator in each fraction if possible
form the LCD by taking each different factor in the individual denominators once and finding their product
divide the denominators into the LCD then multiply by the numerators
add or subtract the numerators
check whether the resulting fraction can be simplified by factorising and cancelling. NOTE: If one of the denominators is a perfect square, both factors must be included in the LCD.

EG +S

Example 1 5 3 Simplify ------------ + ----------x+4 x–1 Solution 3( x – 1) + 5( x + 4) 5 3 ------------ + ----------- = ---------------------------------------------( x + 4)( x – 1) x+4 x–1 3x – 3 + 5x + 20 = ---------------------------------------( x + 4)( x – 1) 8x + 17 = ---------------------------------( x + 4)( x – 1)

187

Mathscape 10 ext. - Ch05 Page 188 Friday, September 23, 2005 7:50 AM

188

Mathscape

10

Extension

Example 2

EG +S

Simplify: 1 1 a ------------- + -------------------------------2 2 x – 4 x + 13x + 22

3 2 -------------------– ------------------2 2 2x – 5x 4x – 25

b

Solutions 1 1 ------------- + --------------------------------x 2 – 4 x 2 + 13x + 22 1 1 = ---------------------------------- + ------------------------------------( x – 2 ) ( x + 2 ) ( x + 2 ) ( x + 11 ) x + 11 + x – 2 = -----------------------------------------------------( x – 2 ) ( x + 2 ) ( x + 11 )

a

3 2 -------------------– -------------------2x 2 – 5x 4x 2 – 25 3 2 = ----------------------- – ---------------------------------------x ( 2x – 5 ) ( 2x – 5 ) ( 2x + 5 ) 3 ( 2x + 5 ) – 2 ( x ) = -------------------------------------------x ( 2x – 5 ) ( 2x + 5 )

b

2x + 9 = -----------------------------------------------------( x – 2 ) ( x + 2 ) ( x + 11 )

6x + 15 – 2x = -------------------------------------------x ( 2x – 5 ) ( 2x + 5 ) 4x + 15 = -------------------------------------------x ( 2x – 5 ) ( 2x + 5 )

EG +S

Example 3 x 1 Simplify ------------------2- + -----------x+4 ( x + 3) Solution x + 4 + x( x + 3 )2 x 1 ------------------2- + ------------ = ---------------------------------------x+4 ( x + 3 )2( x + 4 ) ( x + 3) x + 4 + x ( x 2 + 6x + 9 ) = ----------------------------------------------------( x + 3 )2( x + 4 ) x + 4 + x 3 + 6x 2 + 9x = -------------------------------------------------( x + 3 )2( x + 4 ) x 3 + 6x 2 + 10x + 4 = -------------------------------------------( x + 3 )2( x + 4 )

Exercise 5.10

1 Simplify each of the following. 11z z 5a 7a a -------- – -----b ------ + -----15 15 8 8 e

2k k ------ – --3 4

f

3f f ------ + --5 8

c

2m m ------- + -----5 10

d

11x 2x --------- – -----14 7

g

5 11 ------ + --------6y 12y

h

5 4 ------ + -----9u 6u

Mathscape 10 ext. - Ch05 Page 189 Friday, September 23, 2005 7:50 AM

Chapter

5:

Factorisation and algebraic fractions

2 Simplify: x+1 x+3 a ------------ + -----------8 4 d

b+3 b+2 ------------ – -----------2 10

b

n+5 n–4 ------------ + -----------7 2

c

3t – 2 t + 6 -------------- + ----------6 4

e

k+7 k–1 ------------ – ----------3 4

f

3z – 5 2z – 3 -------------- – -------------8 12

b

1 1 ------------ + ----------x+1 x–1

c

1 1 ------------ + -----------x+2 x+3

■ Consolidation

3 Simplify: 1 1 a --- + -----------x x+1 d

3 2 --- + -----------x x+5

e

2 3 ------------ + -----------x+3 x+2

f

5 2 ----------- + -----------x–1 x+6

g

1 1 --- – -----------x x+1

h

1 1 ------------ – --x+2 x

i

1 1 ----------- – -----------x–1 x+1

j

4 2 --- – -----------x x+2

k

7 4 ------------ – -----------x+5 x+3

l

5 3 --------------- – --------------2x – 1 3x – 4

b

x 2x ------------ – ----------x+3 x–1

c

x+1 x+2 ------------ + -----------x+3 x+1

4 Simplify: 1 x a ------------ + -----------x+1 x+2

5 Simplify each of the following. 1 1 a -------------------- + -----------x( x + 1) x + 1

b

1 1 ---------------------------------- + -----------( x + 1)( x + 5) x + 5

c

1 1 ---------------------------------- + ---------------------------------( x + 1)( x + 2) ( x + 2)( x + 5)

d

1 1 ---------------------------------- + ---------------------------------( x + 4)( x – 2) ( x + 5)( x – 2)

e

1 1 ------------------- – ---------------------------------x( x – 4) ( x + 6)( x – 4)

f

1 1 ---------------------------------- – ---------------------------------( x – 1)( x + 1) ( x + 1)( x + 2)

g

2 3 ---------------------------------- + ---------------------------------( x + 2)( x + 4) ( x + 1)( x + 4)

h

4 5 ---------------------------------- + ---------------------------------( x – 3)( x + 2) ( x + 2)( x + 5)

i

7 1 -------------------------- – ------------------------------------2x ( 2x – 1 ) ( 2x – 1 ) ( x – 4 )

j

2 4 ------------------------------------- – ---------------------------------( 3x + 2 ) ( x – 2 ) ( x + 6 ) ( x – 2 )

k

2x x --------------------------------- + ---------------------------------( x – 3)( x – 4) ( x – 4)( x + 1)

l

x+2 x+1 ------------------- + -------------------x( x – 1) x( x + 1)

■ Further applications

6 Factorise the denominator in each fraction, then express the fractions with a common denominator and simplify. 1 1 1 1 a --------------- + ----------------b ------------------ + --------------2x + 8 x 2 + 4x 4x + 12 3x + 9

189

Mathscape 10 ext. - Ch05 Page 190 Thursday, October 13, 2005 4:25 PM

190

Mathscape

c

g i k

Extension

3 1 ------------- + ----------2 x – 2 x –4 1 1 ------------- + -------------------------2 2 x – 1 x + 3x + 2 1 1 ----------------------------- + --------------------------------x 2 + 8x + 15 x 2 + 11x + 30 4 3 ---------------- + ----------------------------x 2 – 25 x 2 – 7x + 10 1 6 ------------- + -------------------------------2 2 x – 9 3x – 5x – 12

d f h j l

4 2 ------------- + -----------------2 5x – 15 x –9 1 1 -------------------------+ ---------------2 2 x – 7x + 6 x – 36 1 1 ---------------------------- + ----------------x 2 – 2x – 63 x 2 – 81 2 5 ---------------------------- + -------------------------------x 2 – 7x – 30 x 2 – 14x + 40 2 3 ----------------------------------- + ---------------2 2 2x + 19x + 35 x – 49

7 Factorise the denominator in each fraction, then express the fractions with a common denominator and simplify. 1 1 1 1 - – -----------------a ---------------b --------------- – -----------------2x + 8 5x + 20 x 2 – 6x 3x – 18 2 3 4 1 - – ------------ – -----------------c ---------------d ---------------2 2 x + 4 5x – 40 x – 64 x – 16 1 1 1 1 - – --------------------- – ---------------e ------------------f ---------------------------2 2 2 x – 2x – 35 x – 25 9x – 16 21x – 28 1 1 1 1 – -------------------------------– ----------------------------g -------------------------------h ----------------------7x 2 – 14x x 2 + 4x – 12 x 2 – 13x + 22 x 2 – 14x + 33 3 5 2 2 - – ----------------------------- – ----------------------------i ------------------j ---------------------------x 2 – 5x – 50 x 2 – 4x – 45 x 2 – 144 x 2 + 9x – 36 2x x+2 x–3 4x - – ----------------------------------- – ------------------k ------------------l ----------------------------------2 2 2 2 4x + 37x – 30 x – 100 4x – 49 2x + 11x + 14

0FF

M

I CAAL LL LYY O C U S O N W 0 R K I N G M AATTHHEEMMAATTI C G N I K R O OCUS ON

TAXICAB

W

NUMBERS AND THE SUM OF TWO CUBES

Introduction

F

FOCUS ON WORKING MATHEMATICALLY

e

10

What are ‘taxicab numbers’? How did they come be studied? Who were the mathematicians involved? We can only relate part of the story here. It began during the first world war when the British mathematician GH Hardy (1877–1947) went to visit his protégé and colleague Srinivasa Ramanujan (1877–1920) who lay dying in hospital in London. Hardy had gone out to the hospital in Putney by taxi.

Mathscape 10 ext. - Ch05 Page 191 Thursday, October 13, 2005 4:25 PM

5:

Factorisation and algebraic fractions

191

He was a shy and self conscious man in situations like this, not knowing how to open a conversation easily, despite the fact that he had worked with Ramanujan for years. CP Snow in the foreword to Hardy’s book (A Mathematicians Apology, Cambridge University Press, 2000, page 37) records the conversation. Without a greeting, and certainly as his first remark, Hardy blurted out: ‘I thought the number of my taxicab was 1729. It seemed to me a rather dull number.’ To which Ramanujan replied, ‘No Hardy! No Hardy! It is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways.’ The number 1729 can be written as 13 + 123 and also as 93 + 103 . Today such numbers have become known as ‘taxicab numbers’. Mathematicians define the smallest number expressible as the sum of two cubes in n different ways as the nth taxicab number, denoted by taxicab(n). Thus taxicab(2) = 1729. Taxicab(3) was discovered in 1957 to be 87 539 319 and taxicab(4) in 1991 to be 6 963 472 309 248. Taxicab(5) was discovered in 1997. As you can imagine, computers played a major role in the discovery of taxicab numbers but little else is known about them.

5

E 1 2 3

C HALLENGE

ACTIVITIES

Two cubes have side lengths a and b (whole numbers with a < b). The sum of their volumes is equal to the sum of the lengths of their edges. Find a and b. Given that a3 + b3 = (a + b)(a2 − ab + b2) deduce the factors of a3 − b3 by setting −b for b. Two cubes have side lengths a and b (whole numbers with a > b). The difference in their volumes is equal to the difference of the total lengths of their edges. Investigate whether integer solutions for a and b can be found.

MATHEMAT ICALLY

4

What number is taxicab(1)? Let’s take a closer look at taxicab(2) = 1729. Using a calculator verify that 1729 can be written as the sum of the two cubes 13 + 123 and 93 + 103 . In this chapter you have learnt to factorise algebraic expressions. By multiplying out, verify that the factors of the expression a3 + b3 = (a + b)(a2 − ab + b2). Let a = 9 and b = 10. Use your calculator to verify that the right hand side is 1729. Repeat with a = 1 and b = 12. Equations of the form c = a3 + b3 are called Diophantine equations (named after Diophantus, around 250 AD) where a, b, and c are integers. Solve the Diophantine equation 28 = a3 + b3 for a and b. Why is 28 not a taxicab number?

WORKING

3

ACTIVITIES

ON

1 2

EARNING

FOCUS

2L

FOCUS ON WORKING MATHEMATICALLY

Chapter

Mathscape 10 ext. - Ch05 Page 192 Thursday, October 13, 2005 4:25 PM

192

Mathscape

FOCUS ON WORKING MATHEMATICALLY

E

10

L ET’S

Extension

COMMUNICATE

Reasoning means making logical statements in a sequence. To combine logical statements, we can use ‘linking words’ such as and, if, when, however, because or but to name a few. Write a short paragraph using linking words to communicate the meaning of taxicab numbers. If you have tried this you will immediately see the power of algebra to present meaning in symbolic form.

%R

EFLECTING

There are two ways to describe mathematical thinking on which you should reflect. One is through a search for specific patterns which may suggest a general rule. This type of thinking is inductive. The second concerns the need for proof. In this case the result we suspect to be true is put to the test of deductive reasoning, that is a rigorous chain of argument that leads to an inevitable conclusion. In their book An Introduction to the Theory of Numbers (Oxford University Press, 1954), GH Hardy and his colleague EM Wright proved a theorem to show that taxicab numbers, denoted by taxicab(n), exist for any value of n ≥ 1. What type of mathematical reasoning do you think they used?

Use each of the following in a simple sentence: 1 Binomial factor 2 Difference of two squares 3 The sum of two cubes 4 The difference between inductive and deductive reasoning 5 The Macquarie Learners Dictionary entry for factor:

factor noun 1. one of the things that brings about a result: Hard work was a factor in her success. 2. Specialised one of two or more numbers which, when multiplied together, give the product: Factors of 18 are 3 and 6.

Note the special mathematical meaning of factor and that it applies in algebra as well as in arithmetic.

Mathscape 10 ext. - Ch05 Page 193 Friday, September 23, 2005 7:50 AM

5:

Factorisation and algebraic fractions

1 Expand and simplify: a (y + 4)(y + 5) b (m − 7(m − 3) c (t + 8)(t − 2) d (a − 11)(a + 4) 2 Expand and simplify: a (2p − 3)(3p − 8) b (4 − 3r)(2 + r) 3 Expand these perfect squares. a (m − 7)2 b (2c + 5)2 4 Expand and simplify 2x(9x − 2y)2. 5 Complete these perfect squares. a (____)2 = t 2 + 22t + ____ b (____)2 = n2 − ____ + 81 c (____)2 = 9x2 + ____ + 25 d (____)2 = 16u2 − 88u + ____ 6 State whether each expression is a perfect square. a n2 + 4 b k2 + 6k − 9 2 c a + 2a + 1 d e2 − 36 e q2 − 10q + 100 f 9c2 + 24c + 16 7 Expand and simplify: a (a + 3)(a2 + 5a + 4) b (n + 6)(n + 2) + (3n −4)2 c (a + 10)(a − 10) − (a − 7)2 d (2u + 9)(u − 4) − (u − 3)(u + 6) 8 Factorise: a 7x + 28 b a2 + a c 10m + 15 d 9p − 24q e 12rs + 20st f pqr − pqs g 27y2 − 36y h g2h − gh2 i −ab + 3a j −18c2 − 14cd 9 Factorise: a m 2 − n2 b z2 − 9 2 c 1−p d 49 − r2 e w2 − 529 f 4a2 − 25 2 2 g x − 16y h 81u2 − 100v2 2 2 2 i a b −c j 36p2q2 − 121r2s2

CHAPTER REVIEW

10 Factorise: a x(z + 3) + y(z + 3) b m3 + m 2 + m + 1 c e2 + ef − 2e − 2f d 6jk − 14j − 15k + 35 e 4p − pq + q2 − 4q f 10c − cd − d 2 + 10d 11 Factorise: b b2 − 10b + 21 a x2 + 4x + 3 c e2 − e − 12 d p2 + 3p − 40 2 e a + 10a + 25 f q2 − 18q + 81 12 Factorise: a 3t 2 + 14t + 8 b 2m2 − 13m + 21 2 c 3c + 7c − 6 d 7b2 − 23b − 20 e 6s2 + 31s + 35 f 12d 2 − 19d − 18 13 Factorise: a n2 − 36 b v2 + 8v + 15 c 20k + 35 d 3e2 + 14e + 15 e a2 + 4a + ab + 4b f 9h2 −25 g p2 − 5p − 24 h −21xy − 35yz i 1 + u + u2 + u3 j 16a2 − 121b2 k 5m2 − 14m − 24 l y2 − 9y − 10 m 2cd + 6ce − 5d − 15e n rs2 − r2s + rs o 30g2 + 7g − 15 14 Factorise completely: a 3x2 − 12 b 2a2 + 18a + 36 c n3 − n d 4h2 − 36 e ab2 − 7ab + 12a f 6u2 + 26u + 8 g 12z2 − 75 h x3 − 2x2y − 3x2 + 6xy

193

CHAPTER RE VIEW

Chapter

Mathscape 10 ext. - Ch05 Page 194 Friday, September 23, 2005 7:50 AM

194

Mathscape

VIEW CHAPTER RE

15 Simplify: 3 a -----3d c

25a 2 ----------10a

10

Extension

b

21tu -----------28uv

d

p2q --------2pq

16 Factorise and simplify: a 15e + 25 a --------------------b ----------------5 a 2 + 9a 24m – 16n 9x 2 – 15xy c -------------------------d -------------------------36m – 24n 6x e g i j k

u2 – 4 -------------u–2

f

x 2 + 7x + 10 ----------------------------h x 2 + 5x + 6 3 p 2 – 19 p + 20 -----------------------------------( p – 5 )2 ab – ac – 3b + 3c -----------------------------------------2a 2 – 13a + 21 5w – 10 ------------------l 2–w

18 Simplify: a+8 a–2 a ------------ + -----------4 3 c

1 1 ------------ + --x+3 x

e

x+1 x+5 ------------ + -----------x+2 x+1

b

2h – 3 3h + 7 --------------- – --------------10 5

d

2 x ----------- – -----------x–1 x+1

19 Simplify: 1 1 a --------------- + ------------2 3x + 6 x – 4

16h 2 – 25 ----------------------8h + 10

b

k 2 + 3k – 28 ---------------------------k 2 – 16

c d

1 1 ---------------- + ----------------------------2 2 x – 25 x + 2x – 35 2 3 -------------------------------– -------------------------------2 2 x – 11x + 18 x – 12x + 27 4 x ---------------- – ----------------------------------2 2 4x – 9 2x + 13x + 15

c 2 – 64 -----------------24 – 3c

17 Factorise and simplify: a + 3 4a 2 – 12a a --------------- × ----------------------2a – 6 a2 – 9 c 2 + 9c + 20 ( c – 4 ) 2 × -----------------b ----------------------------c 2 – 16 c 2 + 5c y 2 – 100 y 2 + 14y + 40 - ÷ -------------------------------c ---------------------2y 2 – 20y y 3 + 4y 2 x 2 – 13x + 42 2x 2 – 11x – 6 ÷ -------------------------------d -------------------------------x 2 – 49 x 2 + 7x

CHAPTER REVIEW