Cambridge Mathematics Y11 Chapter 1 - Methods in Algebra PDF

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CHAPTER ONE

Methods in Algebra Fluency in algebra, and particularly in factoring, is absolutely vital for everything in this course. Much of this chapter will be a review of earlier work, but several topics will probably be quite new, including: • the sum and difference of cubes in Section 1E • three simultaneous equations in three variables in Section 1H.

1 A Arithmetic with Pronumerals A pronumeral is a symbol that stands for a number. The pronumeral may stand for a known number, or for an unknown number, or it may be a variable, standing for any one of a whole set of possible numbers. Pronumerals, being numbers, can therefore take part in all the operations that are possible with numbers, such as addition, subtraction, multiplication and division (except by zero).

Like and Unlike Terms: An algebraic expression consists of pronumerals, numbers and the operations of arithmetic. Here is an example: x2 + 2x + 3x2 − 4x − 3 This particular algebraic expression can be simplified by combining like terms. • The two like terms x2 and 3x2 can be combined to give 4x2 . • Another pair of like terms 2x and −4x can be combined to give −2x. • This yields three unlike terms, 4x2 , −2x and −3, which cannot be combined.

WORKED E XERCISE: Simplify each expression by combining like terms. (a) 7a + 15 − 2a − 20 (b) x2 + 2x + 3x2 − 4x − 3

SOLUTION: (a) 7a + 15 − 2a − 20 = 5a − 5

(b) x2 + 2x + 3x2 − 4x − 3 = 4x2 − 2x − 3

Multiplying and Dividing: To simplify a product like 3y × (−6y), or a quotient like

10x2 y ÷ 5y, work systematically through the signs, then the numerals, and then each pronumeral in turn.

WORKED E XERCISE: Simplify these products and quotients. (a) 3y × (−6y) (b) 4ab × 7bc

(c) 10x2 y ÷ 5y

(a) 3y × (−6y) = −18y 2

(c) 10x2 y ÷ 5y = 2x2

SOLUTION:

ISBN: 9781107679573

(b) 4ab × 7bc = 28ab2 c

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

2

CHAPTER 1: Methods in Algebra

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

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Index Laws: Here are the standard laws for dealing with indices. They will be covered in more detail in Chapter Seven.

THE INDEX LAWS: • To multiply powers of the same base, add the indices: 1

• To divide powers of the same base, subtract the indices: • To raise a power to a power, multiply the indices:

• The power of a product is the product of the powers: • The power of a quotient is the quotient of the powers:

ax ay = ax+ y ax = ax−y ay (ax )n = axn (ab)x = ax bx  x a ax = x b b

In expressions with several factors, work systematically through the signs, then the numerals, and then each pronumeral in turn.

WORKED EXERCISE : Use the index laws above to simplify each expression. (a) 3x4 × 4x3 (c) (3a4 )3

(d) (−5x2 )3 × (2xy )4

(b) (48x7 y 3 ) ÷ (16x5 y 3 )

(e)



2x 3y

4

SOLUTION: 3x4 × 4x3 = 12x7

(a)

(multiplying powers of the same base)

(b) (48x7 y 3 ) ÷ (16x5 y 3 ) = 3x2

(dividing powers of the same base)

(c)

(3a4 )3 = 27a12

(d)

(−5x2 )3 × (2xy )4 = −125x6 × 16x4 y 4

(e)



2x 3y

4

(raising a power to a power) (two powers of products)

= −2000x10 y 4

(multiplying powers of the same base)

=

(a power of a quotient)

4

16x 81y 4

Exercise 1A 1. Simplify: (a) 5x + 3x

(b) 5x − 3x

(c) −5x + 3x

(d) −5x − 3x

2. Simplify: (a) −2a + 3a + 4a

(b) −2a − 3a + 4a

(c) −2a − 3a − 4a

(d) −2a + 3a − 4a

3. Simplify: (a) −2x + x (b) 3y − y

(c) 3a − 7a (d) −8b + 5b

(e) 4x − (−3x) (f) −2ab − ba

(g) −3pq + 7pq (h) −5abc − (−2abc)

4. Simplify: (a) 6x + 3 − 5x (b) −2 + 2y − 1 (c) 3a − 7 − a + 4 (d) 3x − 2y + 5x + 6y

ISBN: 9781107679573

(e) (f) (g) (h)

−8t + 12 − 2t − 17 2a2 + 7a − 5a2 − 3a 9x2 − 7x + 4 − 14x2 − 5x − 7 3a − 4b − 2c + 4a + 2b − c + 2a − b − 2c

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

CHAPTER 1: Methods in Algebra

1A Arithmetic with Pronumerals

5. Simplify: (a) −3a × 2

(b) −4a × (−3a)

(c) a2 × a3

(d) (a2 )3

6. Simplify: (a) −10a ÷ 5

(b) −24a ÷ (−8a)

(c) a7 ÷ a3

(d) 7a2 ÷ 7a

7. Simplify: 5x (a) x 8. Simplify: (a) t2 + t2 9. Simplify: (a) −6x + 3x

(b)

−7x3 x

(c)

−12a2 b −ab

(d)

3

−27x6 y 7 z 2 9x3 y 3 z

(b) t2 − t2

(c) t2 × t2

(d) t2 ÷ t2

(b) −6x − 3x

(c) −6x × 3x

(d) −6x ÷ 3x

DEVELOPMENT

10. If a = −2, find the value of: (a) 3a + 2 (b) a3 − a2 11. If x = 2 and y = −3, find the value of: (a) 3x + 2y (b) y 2 − 5x 12. Subtract: (a) x from 3x 13. Multiply: (a) 5a by 2 (b) 6x by −3

(b) −x from 3x

(c) 3a2 − a + 4 (c) 8x2 − y 3

(d) x2 − 3xy + 2y 2

(c) 2a from −4a

(d) −b from −5b

(c) −3a by a (d) −2a2 by −3ab

14. Divide: (a) −2x by x (b) 3x3 by x2 (c) x3 y 2 by x2 y

(d) a4 + 3a3 + 2a2 − a

(e) 4x2 by −2x3 (f) −3p2 q by 2pq 3

(d) a6 x3 by −a2 x3 (e) 14a5 b4 by −2a4 b (f) −50a2 b5 c8 by −10ab3 c2

15. Find the sum of: (a) x + y + z, 2x + 3y − 2z and 3x − 4y + z (b) 2a − 3b + c, 15a − 21b − 8c and 24b + 7c + 3a (c) 5ab + bc − 3ca, ab − bc + ca and −ab + 2ca + bc (d) x3 − 3x2 y + 3xy 2 , −2x2 y − xy 2 − y 3 and x3 + 4y 3 16. From: (a) 7x2 − 5x + 6 take 5x2 − 3x + 2 (b) 4a − 8b + c take a − 3b + 5c

(c) 3a + b − c − d take 6a − b + c − 3d (d) ab − bc − cd take −ab + bc − 3cd

17. Subtract: (a) x3 − x2 + x + 1 from x3 + x2 − x + 1 (b) 3xy 2 − 3x2 y + x3 − y 3 from x3 + 3x2 y + 3xy 2 + y 3 (c) b3 + c3 − 2abc from a3 + b3 − 3abc (d) x4 + 5 + x − 3x3 from 5x4 − 8x3 − 2x2 + 7

ISBN: 9781107679573

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

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4

CHAPTER 1: Methods in Algebra

18. Simplify: (a) 2a2 b4 × 3a3 b2

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

(b) −6ab5 × 4a3 b3

(c) (−3a3 )2

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(d) (−2a4 b)3

CHALLENGE

19. What must be added to 4x3 − 3x2 + 2 to give 3x3 + 7x − 6? 20. Simplify: 3a × 3a × 3a (a) 3a + 3a + 3a 21. Simplify: (−2x2 )3 (a) −4x

(b)

(b)

3c × 4c2 × 5c3 3c2 + 4c2 + 5c2

(3xy 3 )3 3x2 y 4

(c)

(−ab)3 × (−ab2 )2 −a5 b3

(c)

ab2 × 2b2 c3 × 3c3 a4 a3 b3 + 2a3 b3 + 3a3 b3

(d)

(−2a3 b2 )2 × 16a7 b (2a2 b)5

22. Divide the product of (−3x7 y 5 )4 and (−2xy 6 )3 by (−6x3 y 8 )2 .

1 B Expanding Brackets Expanding brackets is routine in arithmetic. For example, to calculate 7 × 61, 7 × (60 + 1) = 7 × 60 + 7 × 1, which quickly gives the result 7 × 61 = 420 + 7 = 427. The algebraic version of this procedure can be written as:

2

EXPANDING BRACKETS IN ALGEBRA: a(x + y) = ax + ay

WORKED EXERCISE : Expand and simplify each expression. (a) 3x(4x − 7) (b) 5a(3 − b) − 3b(6 − 5a)

SOLUTION: (a) 3x(4x − 7) = 12x2 − 21x

(b) 5a(3 − b) − 3b(6 − 5a) = 15a − 5ab − 18b + 15ab = 15a + 10ab − 18b

Expanding the Product of Two Bracketed Terms: Each pair of brackets should be expanded in turn and the reulting expression should then be simplified.

WORKED EXERCISE : Expand and simplify each expression. (a) (x + 3)(x − 5) (b) (3 + x)(9 + 3x + x2 )

SOLUTION: (a) (x + 3)(x − 5) = x(x − 5) + 3(x − 5) = x2 − 5x + 3x − 15 = x2 − 2x − 15

ISBN: 9781107679573

(b) (3 + x)(9 + 3x + x2 ) = 3(9 + 3x + x2 ) + x(9 + 3x + x2 ) = 27 + 9x + 3x2 + 9x + 3x2 + x3 = 27 + 18x + 6x2 + x3

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

CHAPTER 1: Methods in Algebra

1B Expanding Brackets

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Special Expansions: These three identities are important and must be memorised. Examples of these expansions occur very frequently, and knowing the formulae greatly simplifies the working. They are proven in the exercises.

3

SQUARE OF A SUM:

(A + B)2 = A2 + 2AB + B 2

SQUARE OF A DIFFERENCE:

(A − B)2 = A2 − 2AB + B 2

DIFFERENCE OF SQUARES:

WORKED E XERCISE: (a) (4x + 1)

(A + B)(A − B) = A2 − B 2

Use the three special expansions above to simplify: (b) (s − 3t)2

2

(c) (x + 3y )(x − 3y )

SOLUTION: (a) (4x + 1) 2 = 16x2 + 8x + 1

(the square of a sum)

(b) (s − 3t)2 = s2 − 6st + 9t2

(the square of a difference)

(c) (x + 3y)(x − 3y) = x2 − 9y 2

(the difference of squares)

Exercise 1B 1. Expand: (a) 3(x − 2) (b) 2(x − 3) (c) −3(x − 2) 2. Expand: (a) 3(x + y) (b) −2(p − q) (c) 4(a + 2b) 3. Expand and simplify: (a) 2(x + 1) − x (b) 3a + 5 + 4(a − 2) (c) 2 + 2 (x − 3) (d) −3(a + 2) + 10 4. Expand and simplify: (a) (x + 2)(x + 3) (b) (y + 4)(y + 7) (c) (t + 6)(t − 3) (d) (x − 4)(x + 2) (e) (t − 1)(t − 3)

(e) (f) (g) (h)

(d) −2(x − 3) (e) −3(x + 2) (f) −2(x + 3)

(g) −(x − 2) (h) −(2 − x) (i) −(x + 3)

(d) x(x − 7) (e) −x(x − 3) (f) −a(a + 4 )

(g) 5(a + 3b − 2c) (h) −3(2x − 3y + 5z) (i) xy(2x − 3y)

3 − (x + 1) b + c − (b − c) (2x − 3y) − (3x − 2y) 3(x − 2) − 2(x − 5) (f) (g) (h) (i) (j)

(2a + 3)(a + 5 ) (u − 4)(3u + 2) (4p + 5)(2p − 3) (2b − 7)(b − 3) (5a − 2)(3a + 1)

(i) (j) (k) (l)

−4(a − b) − 3(a + 2b) 4(s − t) − 5(s + t) 2x(x + 6y ) − x(x − 5y ) −7(2a−3b+c)−6(−a+4b−2c) (k) (l) (m) (n) (o)

(6 − c)(c − 3) (2d − 3)(4 + d) (2x + 3)(y − 2) (a − 2)(5b + 4) (3 − 2m)(4 − 3n)

DEVELOPMENT

5. (a) By expanding (A+B)(A+B), prove the special expansion (A+B)2 = A2 +2AB +B 2 . (b) Similarly, prove the special expansions: (i) (A − B)2 = A2 − 2AB + B 2 (ii) (A − B)(A + B) = A2 − B 2

ISBN: 9781107679573

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

r

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CHAPTER 1: Methods in Algebra

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

6. Use the special expansions to expand: (g) (d − 6)(d + 6) (a) (x + y)2 2 (h) (7 + e)(7 − e) (b) (x − y) (c) (x−y)(x+y) (i) (8 + f)2 (j) (9 − g)2 (d) (a + 3) 2 2 (e) (b − 4) (k) (h + 10)(h − 10) (l) (i + 11)2 (f) (c + 5) 2

(m) (n) (o) (p) (q) (r)

(2a + 1) 2 (2b − 3)2 (3c + 2) 2 (2d + 3e)2 (2f + 3g )(2f − 3g ) (3h − 2i)(3h + 2i)

(s) (t) (u) (v) (w) (x)

(5j + 4)2 (4k − 5ℓ)2 (4 + 5m)(4 − 5m) (5 − 3n)2 (7p + 4q)2 (8 − 3r)2

7. Expand: (a) −a(a2 − a − 1) (b) −2x(x3 − 2x2 − 3x + 1)

(c) 3xy(2x2 y − 5x3 ) (d) −2a2 b(a2 b3 − 2a3 b)

8. Simplify:   (a) 14 − 10 − (3x − 7) − 8x

   (b) 4 a − 2(b − c) − a − (b − 2)

9. Expand and simplify:   1 2 (a) t + t

r

   1 1 t − (c) t + t t

  1 2 (b) t − t CHALLENGE

10. Subtract a(b + c − a) from the sum of b(c + a − b) and c(a + b − c). 11. Multiply: (a) a − 2b by a + 2b (b) 2 − 5x by 5 + 4x

(c) 4x + 7 by itself (d) x2 + 3y by x2 − 4y

12. Expand and simplify: (a) (a − b)(a + b) − a(a − 2b) (b) (x + 2)2 − (x + 1) 2 (c) (a − 3)2 − (a − 3)(a + 3)

(e) a + b − c by a − b (f) 9x2 − 3x + 1 by 3x + 1

(d) (p + q)2 − (p − q)2 (e) (2x + 3)(x − 1) − (x − 2)(x + 1) (f) 3(a − 4)(a − 2) − 2(a − 3)(a − 5)

13. Use the special expansions to find the value of: (a) 1022 (b) 9992

(c) 203 × 197

1 C Factoring Factoring is the reverse process of expanding brackets, and will be needed on a routine basis throughout the course. There are four basic methods, but in every situation, common factors should always be taken out first.

THE FOUR BASIC METHODS OF FACTORING: • HIGHEST COMMON FACTOR:

4

• DIFFERENCE OF SQUARES:

• QUADRATICS:

• GROUPING:

Always try this first. This involves two terms.

This involves three terms.

This involves four or more terms.

Factoring should continue until each factor is irreducible, meaning that it cannot be factored further.

ISBN: 9781107679573

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

CHAPTER 1: Methods in Algebra

1C Factoring

7

Factoring by Taking Out the Highest Common Factor: Always look first for any common factors of all the terms, and then take out the highest common factor.

WORKED E XERCISE: Factor each expression by taking out the highest common factor. (a) 4x3 + 3x2 (b) 18a2 b3 − 30b3

SOLUTION: (a) The highest common factor of 4x3 and 3x2 is x2 , so 4x3 + 3x2 = x2 (4x + 3). (b) The highest common factor of 18a2 b3 and 30b3 is 6b3 , so 18a2 b3 − 30b3 = 6b3 (3a2 − 5).

Factoring by Difference of Squares: The expression must have two terms, both of which are squares. Sometimes a common factor must be taken out first.

WORKED E XERCISE: Use the difference of squares to factor each expression. (a) a2 − 36 (b) 80x2 − 5y 2

SOLUTION: (a) a2 − 36 = (a + 6)(a − 6) (b) 80x2 − 5y 2 = 5(16x2 − y 2 ) = 5(4x − y)(4x + y)

(Take out the highest common factor.) (Use the difference of squares.)

Factoring Monic Quadratics: A quadratic is called monic if the coefficient of x2 is 1.

Suppose that we want to factor the monic quadratic expression x2 − 13x + 36. We look for two numbers: • whose sum is −13 (the coefficient of x), and • whose product is +36 (the constant term).

WORKED E XERCISE: Factor these monic quadratics. (a) x2 − 13x + 36

SOLUTION:

(b) a2 + 12a − 28

(a) The numbers with sum −13 and product +36 are −9 and −4, so x2 − 13x + 36 = (x − 9)(x − 4). (b) The numbers with sum +12 and product −28 are +14 and −2, so a2 + 12a − 28 = (a + 14)(a − 2).

Factoring Non-monic Quadratics: In a non-monic quadratic like 2x2 + 11x + 12, where the coefficient of x2 is not 1, we look for two numbers: • whose sum is 11 (the coefficient of x), and • whose product is 12 × 2 = 24 (the constant times the coefficient of x2 ).

ISBN: 9781107679573

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

Cambridge University Press

r

8

CHAPTER 1: Methods in Algebra

CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

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WORKED EXERCISE : Factor these non-monic quadratics. (a) 2x2 + 11x + 12

(b) 6s2 − 11s − 10

SOLUTION:

(a) The numbers with sum 11 and product 12 × 2 = 24 are 8 and 3, so 2x2 + 11x + 12 = (2x2 + 8x) + (3x + 12) (Split 11x into 8x + 3x.) = 2x(x + 4) + 3(x + 4) (Take out the HCF of each group.) = (2x + 3)(x + 4). (x + 4 is a common factor.) (b) The numbers with sum −11 and product −10 × 6 = −60 are −15 and 4, so 6s2 − 11s − 10 = (6s2 − 15s) + (4s − 10) (Split −11s into −15s + 4s.) = 3s(2s − 5) + 2(2s − 5) (Take out the HCF of each group.) = (3s + 2)(2s − 5). (2s − 5 is a common factor.)

Factoring by Grouping: When there are four or more terms, it is sometimes possible to split the expression into groups, factor each group in turn, and then factor the whole expression by taking out a common factor or by some other method.

WORKED EXERCISE : Factor each expression by grouping. (a) 12xy − 9x − 16y + 12

SOLUTION:

(b) s2 − t2 + s − t

(a) 12xy − 9x − 16y + 12 = 3x(4y − 3) − 4(4y − 3) = (3x − 4)(4y − 3) (b) s2 − t2 + s − t = (s + t)(s − t) + (s − t) = (s − t)(s + t + 1)

(Take out the HCF of each pair.) (4y − 3 is a common factor.)

(Factor s2 − t2 using difference of squares.) (s − t is a common factor.)

Exercise 1C 1. Factor, by taking out any common factors: (a) 2x + 8 (e) x2 + 3x (b) 6a − 15 (f) p2 + 2pq (c) ax − ay (g) 3a2 − 6ab (d) 20ab − 15ac (h) 12x2 + 18x

(i) (j) (k) (l)

2. Factor, by grouping in pairs: (a) mp + mq + np + nq (f) ac + bc − ad − bd (b) ax − ay + bx − by (g) pu − qu − pv + qv (c) ax + 3a + 2x + 6 (h) x2 − 3x − xy + 3y (i) 5p − 5q − px + qx (d) a2 + ab + ac + bc (j) 2ax − bx − 2ay + by (e) z 3 − z 2 + z − 1 3. Factor, using the difference of squares: (a) a2 − 1 (e) 25 − y 2 (b) b2 − 4 (f) 1 − n2 2 (g) 49 − x2 (c) c − 9 (d) d2 − 100 (h) 144 − p2

ISBN: 9781107679573

(i) (j) (k) (l)

4c2 − 9 9u2 − 1 25x2 − 16 1 − 49k 2

(k) (l) (m) (n) (o)

20cd − 32c a2 b + b2 a 6a2 + 2a3 7x3 y − 14x2 y 2 ab + ac − b − c x3 + 4x2 − 3x − 12 a3 − 3a2 − 2a + 6 2t3 + 5t2 − 10t − 25 2x3 − 6x2 − ax + 3a (m) (n) (o) (p)

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012

x2 − 4y 2 9a2 − b2 25m2 − 36n2 81a2 b2 − 64

Cambridge University Press

CHAPTER 1: Methods in Algebra

1D Algebraic Fractions

4. Factor each quadratic (a) a2 + 3a + 2 (b) k 2 + 5k + 6 (c) m2 + 7m + 6 (d) x2 + 8x + 15 (e) y 2 + 9y + 20 (f) t2 + 12t + 20

expression. They are all monic quadratics. (g) x2 − 4x + 3 (m) w2 − 2w − 8 (h) c2 − 7c + 10 (n) a2 + 2a − 8 2 (i) a − 7a + 12 (o) p2 − 2p − 15 (j) b2 − 8b + 12 (p) y 2 + 3y − 28 2 (k) t + t − 2 (q) c2 − 12c + 27 (l) u2 − u − 2 (r) u2 − 13u + 42

(s) (t) (u) (v) (w) (x)

9

x2 − x − 90 x2 + 3x − 40 t2 − 4t − 32 p2 + 9p − 36 u2 − 16u − 80 t2 + 23t − 50

DEVELOPMENT

5. Factor each quadratic (a) 3x2 + 4x + 1 (b) 2x2 + 5x + 2 (c) 3x2 + 16x + 5 (d) 3x2 + 8x + 4 (e) 2x2 − 3x + 1 (f) 5x2 − 13x + 6

expression. They are (g) 5x2 − 11x + 6 (h) 6x2 − 11x + 3 (i) 2x2 − x − 3 (j) 2x2 + 3x − 5 (k) 3x2 + 2x − 5 (l) 3x2 + 14x − 5

all non-monic quadratics. (m) 2x2 − 7x − 15 (s) (n) 2x2 + x − 15 (t) 2 (o) 6x + 17x − 3 (u) (p) 6x2 − 7x − 3 (v) 2 (q) 6x + 5x − 6 (w) (r) 5x2 + 23x + 12 (x)

6. Use the techniques of the previous questions to factor each (a) a2 − 25 (i) i2 − 16i − 36 (b) b2 − 25b (j) 5j 2 + 16j − 16 2 (c) c − 25c + 100 (k) 4k 2 − 16k − 9 (d) 2d2 + 25d + 50 (l) 2k 3 − 16k 2 − 3k + 24 3 2 (e) e + 5e + 5e + 25 (m) 2a2 + ab − 4a − 2b (f) 16 − f 2 (n) 6m3 n4 + 9m2 n5 2 3 (g) 16g − g (o) 49p2 − 121q 2 (h) h2 + 16h + 64 (p) t2 − 14t + 40

5x2 + 4x − 12 5x2 − 19x + 12 5x2 − 11x − 12 5x2 + 28x − 12 9x2 − 6x − 8 3x2 + 13x − 30

expression. (q) 3t2 + 2t − 40 (r) 5t2 + 54t + 40 (s) 5t2 + 33t + 40 (t) 5t3 + 10t2 + 15t (u) u2 + 15u − 54 (v) 3x3 − 2x2 y − 15x + 10y (w) 1 − 36a2 (x) 4a2 − 12a + 9

CHALLENGE

7. Write each expression as a product of three factors. (Take out any common factors first.) (a) 3a2 − 12 (e) 25y − y 3 (i) c3 + 9c2 − c − 9 (f) 16 − a4 (b) x4 − y 4 (j) x3 − 8x2 + 7x 3 2 (k) x4 − 3x2 − 4 (c) x − x (g) 4x + 14x − 30 (d) 5x2 − 5x − 30 (h) a4 + a3 + a2 + a (l) ax2 − a −...