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Methods in algebra 1

Fluency in algebra, particularly in factoring, is absolutely vital for everything in this course. This chapter is intended as a review of earlier algebraic techniques, and readers should do as much or as little of it as is necessary.

Digital Resources are available for this chapter in the Interactive Textbook and Online Teaching Suite. See the Overview at the front of the textbook for details.

ISBN 978-1-108-46904-3 Photocopying is restricted under law and this material must not be transferred to another party.

© Pender et al. 2019 Cambridge University Press

1A Arithmetic with pronumerals 3

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 and stand for any one of a whole set of possible numbers. Pronumerals, being numbers, can take part in all the operations that are possible with numbers, such as addition, subtraction, multiplication, and division (except by zero).

Like and unlike termsAn algebraic expression consists of pronumerals, numbers and the operations of arithmetic. Here is an example:

x2 + 2x + 3x2 − 4x − 3 = 4x2 − 2x − 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.

Multiplying and dividingTo simplify a product such as 3y × (−6y), or a quotient such as 10x2y ÷ 5y, work systematically through the signs, then the numerals, and then each pronumeral in turn.

Index lawsHere are the standard laws for dealing with indices. They will be covered in more detail in Chapter 7.

1A

Example 1 1A

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

Example 2 1A

Simplify these products and quotients.a 3y × (−6y) b 4ab × 7bc c 10x2y ÷ 5y

SOLUTION

a 3y × (−6y) = −18y2 b 4ab × 7bc = 28ab2c c 10x2y ÷ 5y = 2x2



Chapter 1 Methods in algebra4 1A

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

Exercise 1A

1 Simplify:a 3x + x b 3x − x c −3x + x d −3x − x

2 Simplify:a −2a + 3a + 4a b −2a − 3a + 4a c −2a − 3a − 4a d −2a + 3a − 4a

3 Simplify:a −x + x b 2y − 3y c −3a − 7a d −8b + 5be 4x − (−3x) f −2ab − ba g −3pq + 7pq h −5abc − (−2abc)

4 Simplify:a −3a × 2 b −4a × (−3a) c a2 × a3 d (a2)3

FOUNDATION

1 THE INDEX LAWS

• To multiply powers of the same base, add the indices: ax ay = ax+y

• To divide powers of the same base, subtract the indices: ax

ay = ax−y

• To raise a power to a power, multiply the indices: (ax)n = axn

• The power of a product is the product of the powers: (ab)x = ax bx

• The power of a quotient is the quotient of the powers: (ab)

x

= ax

bx

Example 3 1A

Use the index laws above to simplify each expression.a 3x4 × 4x3 b (20x7y3) ÷ (4x5y3) c (3a4)3

d (−x2)3 × (2xy)4 e (2x3y)

4

SOLUTION

a 3x4 × 4x3 = 12x7 (multiplying powers of the same base)

b (20x7y3) ÷ (4x5y3) = 5x2 (dividing powers of the same base)

c (3a4)3 = 27a12 (raising a power to a power)

d (−x2)3 × (2xy)4 = −x6 × 16x4y4

= −16x10y4

(two powers of products) (multiplying powers of the same base)

e (2x3y)

4

= 16x4

81y4 (a power of a quotient)

• (ab)

x

= ax

bx



1A Arithmetic with pronumerals 5

5 Simplify:a −10a ÷ 5 b −24a ÷ (−8a) c a9 ÷ a3 d 7a2 ÷ 7a

6 Simplify:a t2 + t2 b t2 − t2 c t2 × t2 d t2 ÷ t2

7 Simplify:a −6x + 3x b −6x − 3x c −6x × 3x d −6x ÷ 3x

8 If a = −2, find the value of:a 3a + 2 b a3 − a2 c 3a2 − a + 4 d a4 + 3a3 + 2a2 − a

DEVELOPMENT

9 Simplify:a 6x + 3 − 5x b −2 + 2y − 1c 3a − 7 − a + 4 d 3x − 2y + 5x + 6ye −8t + 12 − 2t − 17 f 2a2 + 7a − 5a2 − 3ag 9x2 − 7x + 4 − 14x2 − 5x − 7 h 3a − 4b − 2c + 4a + 2b − c + 2a − b − 2c

10 Simplify:

a 5xx

b −7m3

−mc −12a2b

abd

−27p6q7r2

9p3q3r

11 Subtract:a x from 3x b −x from 3x c 2a from −4a d −b from −5b

12 Multiply:a 5a by 2 b 6x by −3 c −3a by ad −2a2 by −3ab e 4x2 by −2x3 f −3p2q by 2pq3

13 Divide:a −2x by x b 3x3 by x2 c x3y2 by x2yd a6x3 by −a2x3 e 14a5b4 by −2a4b f −50a2b5c8 by −10ab3c2

14 Simplify:a 2a2b4 × 3a3b2 b −6ab5 × 4a3b3 c (−3a3)2 d (−2a4b)3

15 If x = 2 and y = −3, find the value of:a 3x + 2y b y2 − 5x c 8x2 − y3 d x2 − 3xy + 2y2

CHALLENGE

16 Simplify:

a 3a × 3a × 3a3a + 3a + 3a

b 3c × 4c2 × 5c3

3c2 + 4c2 + 5c2c ab2 × 2b2c3 × 3c3a4

a3b3 + 2a3b3 + 3a3b3

17 Simplify:

a (−2x2)3

−4xb

(3xy3)3

3x2y4c

(−ab)3 × (−ab2)2

−a5b3d

(−2a3b2)2 × 16a7b

(2a2b)5



Chapter 1 Methods in algebra6 1B

Expanding brackets

Expanding brackets is routine in arithmetic. For example, we can calculate 7 × 61 as

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:

There may then be like terms to collect.

Expanding the product of two bracketed termsExpand one pair of brackets, then expand the other pair of brackets. Then collect any like terms.

1B

2 EXPANDING BRACKETS IN ALGEBRA

a(x + y) = ax + ay and (x + y)a = xa + ya

Example 4 1B

Expand and simplify each expression.a 3x(4x − 7) b 5a(3 − b) − 3b(1 − 5a)

SOLUTION

a 3x(4x − 7) = 12x2 − 21x

b 5a(3 − b) − 3b(1 − 5a) = 15a − 5ab − 3b + 15ab = 15a + 10ab − 3b (collecting like terms)

Example 5 1B

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

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

= 27 + 18x + 6x2 + x3



1B Expanding brackets 7

Special expansionsThese three identities are important and must be memorised. Examples of these expansions occur constantly, and knowing the formulae greatly simplifies the working. They are proven in the exercises.

Exercise 1B

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

2 Expand:a 3(x + y) b −2(p − q) c 4(a + 2b)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 Expand and simplify:a 2(x + 1) − x b 3a + 5 + 4(a − 2) c 2 + 2(x − 3)d −3(a + 2) + 10 e 3 − (x + 1) f b + c − (b − c)g (2x − 3y) − (3x − 2y) h 3(x − 2) − 2(x − 5) i 4(2a − 3b) − 3(a + 2b)j 4(s − t) − 6(s + t) k 2x(x + 6y) − x(x − 5y) l 5(2a − 5b) − 6(−a − 4b)

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) f (2a + 3)(a + 5)g (u − 4)(3u + 2) h (4p + 5)(2p − 3) i (2b − 7)(b − 3)j (5a − 2)(3a + 1) k (6 − c)(c − 3) l (2d − 3)(4 + d)

FOUNDATION

3 SPECIAL EXPANSIONS

• Square of a sum: (A + B)2 = A2 + 2AB + B2

• Square of a difference: (A − B)2 = A2 − 2AB + B2

• Difference of squares: (A + B)(A − B) = A2 − B2

Example 6 1B

Use the three special expansions above to simplify:a (x + 4)2 b (s − 3t)2 c (x + 3y)(x − 3y)

SOLUTION

a (x + 4)2 = x2 + 8x + 16 (the square of a sum)

b (s − 3t)2 = s2 − 6st + 9t2 (the square of a difference)

c (x + 3y)(x − 3y) = x2 − 9y2 (the difference of squares)



Chapter 1 Methods in algebra8 1B

DEVELOPMENT

5 a By expanding (A + B)(A + B), prove the special expansion (A + B)2 = A2 + 2AB + B2.b Similarly, prove the special expansions:

i (A − B)2 = A2 − 2AB + B2 ii (A − B)(A + B) = A2 − B2

6 Use the special expansions to expand:a (x + y)2 b (x − y)2 c (x − y)(x + y) d (a + 3)2

e (b − 4)2 f (c + 5)2 g (d − 6)(d + 6) h (7 + e)(7 − e)i (8 + f)2 j (9 − g)2 k (h + 10)(h − 10) l (i + 11)2

m (2a + 1)2 n (2b − 3)2 o (3c + 2)2 p (2d + 3e)2

q (2f + 3g)(2f − 3g) r (3h − 2i)(3h + 2i) s (5j + 4)2 t (4k − 5l )2

u (4 + 5m)(4 − 5m) v (5 − 3n)2 w (7p + 4q)2 x (8 − 3r)2

CHALLENGE

7 Expand and simplify:

a (t + 1t )

2

b (t − 1t )

2

c (t + 1t )(t − 1

t )

8 By writing 102 as (100 + 2), and adopting a similar approach for parts b and c, use the special expansions to find (without using a calculator) the value of:a 1022 b 9992 c 203 × 197

9 Expand and simplify:a (a − b)(a2 + ab + b2) b (x + 2)2 − (x + 1)2

c (a − 3)2 − (a − 3)(a + 3) d (2x + 3)(x − 1) − (x − 2)(x + 1)e (x − 2)3 f (p + q + r)2 − 2(pq + qr + rp)



1C Factoring 9

Factoring

Factoring is the reverse process of expanding brackets, and is needed routinely throughout the course. There are four basic methods, but in every situation common factors should always be taken out first.

Factoring by taking out the highest common factorAlways look first for any common factors of all the terms, and then take out the highest common factor.

Factoring by difference of squaresThe expression must have two terms, both of which are squares. Sometimes a common factor must be taken out first.

1C

4 THE FOUR BASIC METHODS OF FACTORING

• Highest common factor: Always try this first. • Difference of squares: This involves two terms. • Quadratics: This involves three terms. • Grouping: This involves four or more terms.

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

Example 7 1C

Factor each expression by taking out the highest common factor.a 4x3 + 3x2 b 9a2b3 − 15b3

SOLUTION

a The highest common factor of 4x3 and 3x2 is x2, so 4x3 + 3x2 = x2(4x + 3).

b The highest common factor of 9a2b3 and 15b3 is 3b3, so 9a2b3 − 15b3 = 3b3(3a2 − 5).

Example 8 1C

Use the difference of squares to factor each expression.a a2 − 36 b 80x2 − 5y2

SOLUTION

a a2 − 36 = (a + 6)(a − 6)

b 80x2 − 5y2 = 5(16x2 − y2) (take out the highest common factor)= 5(4x − y)(4x + y) (use the difference of squares)



10 1CChapter 1 Methods in algebra

Factoring monic quadraticsA quadratic is called monic if the coefficient of x2 is 1. Suppose that we want to factor the monic quadratic expression x2 − 13x + 36. Look for two numbers:• whose sum is −13 (the coefficient of x), and• whose product is +36 (the constant term).

Factoring non-monic quadraticsIn a non-monic quadratic such as 2x2 + 11x + 12, where the coefficient of x2 is not 1, 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).

Then split the middle term into two terms.

Example 9 1C

Factor these monic quadratics.a x2 − 13x + 36 b a2 + 12a − 28

SOLUTION

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).

Example 10 1C

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)



1C Factoring 11

Factoring by groupingWhen there are four or more terms, it is sometimes possible to factor the expression by grouping. • Split the expression into groups. • Then factor each group in turn. • Then factor the whole expression by taking out a common factor or by some other method.

Exercise 1C

1 Factor each expression by taking out any common factors:a 2x + 8 b 6a − 15 c ax − ay d 20ab − 15ace x2 + 3x f p2 + 2pq g 3a2 − 6ab h 12x2 + 18xi 20cd − 32c j a2b + b2a k 6a2 + 2a3 l 7x3y − 14x2y2

2 Factor each expression by grouping in pairs:a mp + mq + np + nq b ax − ay + bx − by c ax + 3a + 2x + 6d a2 + ab + ac + bc e z3 − z2 + z − 1 f ac + bc − ad − bdg pu − qu − pv + qv h x2 − 3x − xy + 3y i 5p − 5q − px + qxj 2ax − bx − 2ay + by k ab + ac − b − c l x3 + 4x2 − 3x − 12m a3 − 3a2 − 2a + 6 n 2t3 + 5t2 − 10t − 25 o 2x3 − 6x2 − ax + 3a

3 Factor each expression by using the difference of squares:a a2 − 1 b b2 − 4 c c2 − 9 d d2 − 100e 25 − y2 f 1 − n2 g 49 − x2 h 144 − p2

i 4c2 − 9 j 9u2 − 1 k 25x2 − 16 l 1 − 49k2

m x2 − 4y2 n 9a2 − b2 o 25m2 − 36n2 p 81a2b2 − 64

4 Factor each quadratic expression. They are all monic quadratics.a a2 + 3a + 2 b k2 + 5k + 6 c m2 + 7m + 6 d x2 + 8x + 15e y2 + 9y + 20 f t2 + 12t + 20 g x2 − 4x + 3 h c2 − 7c + 10i a2 − 7a + 12 j b2 − 8b + 12 k t2 + t − 2 l u2 − u − 2m w2 − 2w − 8 n a2 + 2a − 8 o p2 − 2p − 15 p y2 + 3y − 28q c2 − 12c + 27 r u2 − 13u + 42 s x2 − x − 90 t x2 + 3x − 40u t2 − 4t − 32 v p2 + 9p − 36 w u2 − 16u − 80 x t2 + 23t − 50

FOUNDATION

Example 11 1C

Factor each expression by grouping.a 12xy − 9x − 16y + 12 b s2 − t2 + s − t

SOLUTION

a 12xy − 9x − 16y + 12 = 3x(4y − 3) − 4(4y − 3) (take out the HCF of each pair) = (3x − 4)(4y − 3) (4y − 3 is a common factor)

b s2 − t2 + s − t = (s + t)(s − t) + (s − t) (factor s2 − t2 using the difference of squares) = (s − t)(s + t + 1) (s − t is a common factor)



12 1CChapter 1 Methods in algebra

DEVELOPMENT

5 Factor each quadratic expression. They are all non-monic quadratics.a 3x2 + 4x + 1 b 2x2 + 5x + 2 c 3x2 + 16x + 5 d 3x2 + 8x + 4e 2x2 − 3x + 1 f 5x2 − 13x + 6 g 5x2 − 11x + 6 h 6x2 − 11x + 3i 2x2 − x − 3 j 2x2 + 3x − 5 k 3x2 + 2x − 5 l 3x2 + 14x − 5m 2x2 − 7x − 15 n 2x2 + x − 15 o 6x2 + 17x − 3 p 6x2 − 7x − 3q 6x2 + 5x − 6 r 5x2 + 23x + 12 s 5x2 + 4x − 12 t 5x2 − 19x + 12u 5x2 − 11x − 12 v 5x2 + 28x − 12 w 9x2 − 6x − 8 x 3x2 + 13x − 30

6 Use the techniques of the previous questions to factor each expression.a a2 − 25 b b2 − 25b c c2 − 25c + 100d 2d2 + 25d + 50 e e3 + 5e2 + 5e + 25 f 16 − f

2

g 16g2 − g3 h h2 + 16h + 64 i i2 − 16i − 36j 5j2 + 16j − 16 k 4k2 − 16k − 9 l 2k3 − 16k2 − 3k + 24m 2a2 + ab − 4a − 2b n 6m3n4 + 9m2n5 o 49p2 − 121q2

p t2 − 14t + 40 q 3t2 + 2t − 40 r 5t2 + 54t + 40s 5t2 + 33t + 40 t 5t3 + 10t2 + 15t u u2 + 15u − 54v 3x3 − 2x2y − 15x + 10y w (p + q)2 − r2 x 4a2 − 12a + 9

CHALLENGE

7 Factor each expression as fully as possible. (Take out any common factors first.)a 3a2 − 12 b x4 − y4 c x3 − x d 5x2 − 5x − 30e 25y − y3 f 16 − a4 g 4x2 + 14x − 30 h a4 + a3 + a2 + a

i c3 + 9c2 − c − 9 j x3 − 8x2 + 7x k x4 − 3x2 − 4 l ax2 − a − 2x2 + 2



1D Algebraic fractions 13

Algebraic fractions

An algebraic fraction is a fraction that contains pronumerals. Algebraic fractions are manipulated in the same way as arithmetic fractions, and factoring may play a major role.

Adding and subtracting algebraic fractionsA common denominator is needed. Finding the lowest common denominator may involve factoring each denominator.

1D

Example 12 1D

Use a common denominator to simplify each algebraic fraction.

a x2

− x3

b 5x6

+ 11x4

c 23x

− 35x

d 1x − 4

− 1x

SOLUTION

a x2

− x3

= 3x6

− 2x6

= x6

b 5x6

+ 11x4

= 10x12

+ 33x12

= 43x12

c 23x

− 35x

= 1015x

− 915x

= 115x

d 1x − 4

− 1x

=x − (x − 4)

x(x − 4)

= 4x(x − 4)

5 ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS

• First factor each denominator. • Then work with the lowest common denominator.

Example 13 1D

Factor the denominators of 2 + x

x2 − x− 5

x − 1 , then simplify the expression.

SOLUTION

2 + x

x2 − x− 5

x − 1= 2 + x

x(x − 1)− 5

x − 1

= 2 + x − 5xx(x − 1)

= 2 − 4xx(x − 1)



14 1DChapter 1 Methods in algebra

Cancelling algebraic fractionsThe key step here is to factor the numerator and denominator completely before cancelling factors.

Multiplying and dividing algebraic fractionsThese processes are done exactly as for arithmetic fractions.

6 CANCELLING ALGEBRAIC FRACTIONS

• First factor the numerator and the denominator. • Then cancel out all common factors.

Example 14 1D

Simplify each algebraic fraction.

a 6x + 86

b x2 − x

x2 − 1

SOLUTION

a 6x + 86

=2(3x + 4)

6

= 3x + 43

( which could be written as x + 43

)

b x2 − x

x2 − 1 =

x(x − 1)(x + 1)(x − 1)

= xx + 1

7 MULTIPLYING ALGEBRAIC FRACTIONS AND DIVIDING BY AN ALGEBRAIC FRACTION

Multiplying algebraic fractions

• First factor all numerators and denominators completely. • Then cancel common factors.

Dividing by an algebraic fraction

• To divide by an algebraic fraction, multiply by its reciprocal. For example: 3x

÷ 4y

= 3x

× y4

.

• The reciprocal of the fraction 4y is

y4

.




Simplifying compound fractionsA compound fraction is a fraction in which either the numerator or the denominator is itself a fraction.

Example 15 1D

Simplify these products and quotients of algebraic fractions.

a 2aa2 − 9

× a − 35a

b 12xx + 1

÷ 6x

x2 + 2x + 1

SOLUTION

a 2a

a2 − 9× a − 3

5a= 2a

(a − 3)(a + 3)× a − 3

5a (factor a2 − 9)

= 25(a + 3)

(cancel a − 3 and a)

b 12xx + 1

÷ 6x

x2 + 2x + 1 = 12x

x + 1× x2 + 2x + 1

6x (multiply by the reciprocal)

= 12xx + 1

×(x + 1)2

6x (factor x2 + 2x + 1)

= 2(x + 1) (cancel x + 1 and 6x)

8 SIMPLIFYING COMPOUND FRACTIONS

• Find the lowest common multiple of the denominators on the top and the bottom. • Multiply top and bottom by this lowest common multiple.

This will clear all the fractions from the top and bottom together.

Example 16 1D

Simplify each compound fraction.

a 12

− 13

14

+ 16

b 1t

+ 1

1t

− 1

SOLUTION

a 12

− 13

14

+ 16

=12

− 13

14

+ 16

× 1212

= 6 − 43 + 2

= 25

b 1t

+ 1

1t

− 1 =

1t

+ 1

1t

− 1× t

t

= 1 + t1 − t



16 1DChapter 1 Methods in algebra

Exercise 1D

1 Simplify:

a xx

b 2xx

c x2x

d a

a2e 3x2

9xyf 12ab

4a2b

2 Simplify:

a x3

× 3x

b a4

÷ a2

c x2 × 3x

d 12b

× b2

e 3x4

× 2

x2f 5

a÷ 10 g 2ab

3× 6

ab2h 8ab

5÷ 4ab

15

3 Write as a single fraction:

a x + x2

b y4

+ y2

c m3

− m9

d n2

+ n5

e x8

− y12

f 2a3

+ 3a2

g 7b10

− 19b30

h xy

30− xy

12

4 Write as a single fraction:

a 1a

+ 1a

b 1x

− 2x

c 1a

+ 12a

d 12x

− 13x

e 34a

+ 43a

f 56x

− 13x

DEVELOPMENT

5 Simplify:

a x + 12

+ x + 23

b 2x − 15

+ 2x + 34

c x + 36

+ x − 312

d x + 22

− x + 33

e 2x + 14

− 2x − 35

f 2x − 13

− 2x + 16

6 Factor where possible and then simplify:

a 2p + 2q

p + qb 3t − 12

2t − 8c x2 + 3x

3x + 9

d aax + ay

e 3a2 − 6ab

2a2b − 4ab2f x2 + 2x

x2 − 4

g a2 − 9

a2 + a − 12h x2 + 2x + 1

x2 − 1i x2 + 10x + 25

x2 + 9x + 20

7 Simplify:

a 1x

+ 1x + 1

b 1x

− 1x + 1

c 1x + 1

+ 1x − 1

d 2x − 3

+ 3x − 2

e 3x + 1

− 2x − 1

f 2x − 2

− 2x + 3

FOUNDATION




8 Simplify:

a 3x + 32x

× x2

x2 − 1b a2 + a − 2

a + 2× a2 − 3a

a2 − 4a + 3

c c2 + 5c + 6

c2 − 16÷ c + 3

c − 4d x2 − x − 20

x2 − 25× x2 − x − 2

x2 + 2x − 8÷ x + 1

x2 + 5x

CHALLENGE

9 Simplify:

a b − aa − b

b 1a − b

− 1b − a

c mm − n

+ nn − m

d x2 − 5x + 62 − x

10 Simplify:

a 1

x2 + x+ 1

x2 − xb 1

x − y+ 2x − y

x2 − y2

c 3

x2 + 2x − 8− 2

x2 + x − 6d 1

x2 − 4x + 3+ 1

x2 − 5x + 6− 1

x2 − 3x + 2

11 Study the worked example on compound fractions and then simplify:

a 1 − 1

2

1 + 12

b 2 + 1

3

5 − 23

c 12

− 15

1 + 110

d 1720

− 34

45

− 310

e 1x

1 + 2x

f t − 1

t

t + 1t

g 11b

+ 1a

h

x

y+ y

xx

y− y

x

i 1 − 1

x + 1

1x

+ 1x + 1

j 3

x + 2− 2

x + 1

5x + 2

− 4x + 1



18 1EChapter 1 Methods in algebra

Solving linear equations

The first principle in solving any equation is to simplify it by doing the same things to both sides. Linear equations can be solved completely this way.

An equation involving algebraic fractions can often be reduced to a linear equation by following these steps.

Changing the subject of a formulaSimilar sequences of operations allow the subject of a formula to be changed from one pronumeral to another.

1E

9 SOLVING LINEAR EQUATIONS

• Any number can be added to or subtracted from both sides. • Both sides can be multiplied or divided by any non-zero number.

Example 17 1E

Solve each equation.

a 6x + 5 = 4x − 9 b 4 − 7x4x − 7

= 1

SOLUTION

a 6x + 5 = 4x − 9− 4x 2x + 5 = −9

− 5 2x = −14

÷ 2 x = −7

b 4 − 7x4x − 7

= 1

× (4x − 7) 4 − 7x = 4x − 7 + 7x 4 = 11x − 7 + 7 11 = 11x ÷ 11 x = 1

Example 18 1E

Change the subject of each formula to x.

a y = 4x − 3 b y = x + 1x + 2

SOLUTION

a y = 4x − 3 + 3 y + 3 = 4x

÷ 4 y + 3

4= x

x =y + 3

4

b y = x + 1x + 2

× (x + 2) xy + 2y = x + 1 Rearranging, xy − x = 1 − 2y Factoring, x(y − 1) = 1 − 2y

÷ ( y − 1) x =1 − 2y

y − 1



1E Solving linear equations 19

Exercise 1E

1 Solve:

a 2x + 1 = 7 b 5p − 2 = −2 c a

2− 1 = 3 d 3 − w = 4

e 3x − 5 = 22 f 4x + 7 = −13 g 1 − x

2= 9 h 11 − 6x = 23

2 Solve:a 3n − 1 = 2n + 3 b 4b + 3 = 2b + 1 c 5x − 2 = 2x + 10d 5 − x = 27 + x e 16 + 9a = 10 − 3a f 13y − 21 = 20y − 35g 13 − 12x = 6 − 3x h 3x + 21 = 18 − 2x

DEVELOPMENT

3 Solve:

a a12

= 23

b y

20= 4

5c 1

x= 3 d 2

a= 5

e 1 = 32y

f 2x + 15

= −3 g 3a − 54

= 4 h 7 − 4x6

= 32

i 20 + aa

= −3 j 9 − 2tt

= 13 k 3x − 1

= −1 l 3x2x − 1

= 53

4 a If v = u + at, find a when t = 4, v = 20 and u = 8.

b Given that v2 = u2 + 2as, find the value of s when u = 6, v = 10 and a = 2.

c Suppose that 1u

+ 1v

= 1t . Find v, given that u = −1 and t = 2.

d If S = −15, n = 10 and a = −24, find ℓ, given that S = n

2(a + ℓ).

e The formula F = 95

C + 32 relates temperatures in degrees Fahrenheit and Celsius. Find the

value of C that corresponds to F = 95.

f Suppose that c and d are related by the formula 1c + 1

= 5d − 1

. Find c when d = 4.

5 Solve each problem by forming, and then solving, a linear equation.a Three less than four times a certain number is equal to 21. Find the number.b Five more than twice a certain number is one more than the number itself. What is the number?c Bill and Derek collect Batman cards. Bill has three times as many cards as Derek, and altogether they

have 68 cards. How many cards does Derek have?d If I paid $1.45 for an apple and an orange, and the apple cost 15 cents more than the orange, how

much did the orange cost?

6 Solve:

a y + y2

= 1 b x3

− x5

= 2 c a10

− a6

= 1 d x6

+ 23

= x2

− 56

e x3

− 2 = x2

− 3 f 1x

− 3 = 12x

g 12x

− 23

= 1 − 13x

h x − 23

= x + 44

i 3x − 2

= 42x + 5

j x + 1x + 2

= x − 3x + 1

FOUNDATION



20 1EChapter 1 Methods in algebra

7 Rearrange each formula so that the pronumeral written in square brackets is the subject.a a = bc − d [b] b t = a + (n − 1)d [n]

c p

q + r= t [r] d u = 1 + 3

v [v]

8 Expand the brackets on both sides of each equation, then solve it.a (x − 3)(x + 6) = (x − 4)(x − 5) b (1 + 2x)(4 + 3x) = (2 − x)(5 − 6x)c (x + 3)2 = (x − 1)2 d (2x − 5)(2x + 5) = (2x − 3)2

CHALLENGE

9 Solve:

a a + 52

− a − 13

= 1 b 34

− x + 112

= 23

− x − 16

c 34

(x − 1) − 12

(3x + 2) = 0 d 4x + 16

− 2x − 115

= 3x − 55

− 6x + 110

10 Rearrange each formula so that the pronumeral written in square brackets is the subject.

a a2

− b3

= a [a] b 1f

+ 2g

= 5h [g] c x = y

y + 2 [y] d a = b + 5

b − 4 [b]

11 Solve each problem by forming, and then solving, a linear equation.a My father is 40 years older than me and he is three times my age. How old am I?b A basketballer has scored 312 points in 15 games. How many points must he average per game in his

next 3 games to take his overall average to 20 points per game?c A cyclist rides for 5 hours at a certain speed and then for 4 hours at a speed 6 km/h greater than his

original speed. If he rides 294 km altogether, what was his first speed?



1F Solving quadratic equations 21

Solving quadratic equations

There are three approaches to solving a quadratic equation:• factoring• completing the square• the quadratic formula.

This section reviews factoring and the quadratic formula. Completing the square is reviewed in Section 1H.

Solving a quadratic by factoringThis method is the simplest, but it only works in special cases.

Solving a quadratic by the formulaThis method works whether the solutions are rational numbers or involve surds. It will be proven in the last challenge question of Exercise 3E.

1F

10 SOLVING A QUADRATIC BY FACTORING

• Get all the terms on the left, then factor the left-hand side. • Then use the principle that if A × B = 0, then A = 0 or B = 0.

Example 19 1F

Solve the quadratic equation 5x2 + 34x − 7 = 0 by factoring.

SOLUTION

5x2 + 34x − 7 = 0

5x2 + 35x − x − 7 = 0 (35 and −1 have sum 34 and product −7 × 5 = −35)

5x(x + 7) − (x + 7) = 0

(5x − 1)(x + 7) = 0 (the LHS is now factored)

5x − 1 = 0 or x + 7 = 0 (one of the factors must be zero)

x = 15 or x = −7 (there are two solutions)

11 THE QUADRATIC FORMULA

• The solutions of ax2 + bx + c = 0 are:

x = −b + √b2 − 4ac2a

or x = −b − √b2 − 4ac2a

.

• Always calculate b2 − 4ac first. (Later, this quantity will be called the discriminant and given the symbol Δ.)



22 1FChapter 1 Methods in algebra

Exercise 1F

1 Solve:a x2 = 9 b y2 = 25 c a2 − 4 = 0

d c2 − 36 = 0 e 1 − t2 = 0 f x2 = 94

g 4x2 − 1 = 0 h 9a2 − 64 = 0 i 25y2 = 16

2 Solve by factoring:a x2 − 5x = 0 b y2 + y = 0 c c2 + 2c = 0 d k2 − 7k = 0e t2 = t f 3a = a2 g 2b2 − b = 0 h 3u2 + u = 0i 4x2 + 3x = 0 j 2a2 = 5a k 3y2 = 2y l 3n + 5n2 = 0

3 Solve by factoring:a x2 + 4x + 3 = 0 b x2 − 3x + 2 = 0 c x2 + 6x + 8 = 0d a2 − 7a + 10 = 0 e t2 − 4t − 12 = 0 f c2 − 10c + 25 = 0g n2 − 9n + 8 = 0 h p2 + 2p − 15 = 0 i a2 − 10a − 24 = 0j y2 + 4y = 5 k p2 = p + 6 l a2 = a + 132m c2 + 18 = 9c n 8t + 20 = t2 o u2 + u = 56p k2 = 24 + 2k q 50 + 27h + h2 = 0 r a2 + 20a = 44

DEVELOPMENT

4 Solve by factoring:a 2x2 + 3x + 1 = 0 b 3a2 − 7a + 2 = 0 c 4y2 − 5y + 1 = 0d 2x2 + 11x + 5 = 0 e 2x2 + x − 3 = 0 f 3n2 − 2n − 5 = 0g 3b2 − 4b − 4 = 0 h 2a2 + 7a − 15 = 0 i 2y2 − y − 15 = 0j 3y2 + 10y = 8 k 5x2 − 26x + 5 = 0 l 4t2 + 9 = 15tm 13t + 6 = 5t2 n 10u2 + 3u − 4 = 0 o 25x2 + 1 = 10xp 6x2 + 13x + 6 = 0 q 12b2 + 3 + 20b = 0 r 6k2 + 13k = 8

FOUNDATION

Example 20 1F

Solve each quadratic equation using the quadratic formula.a 5x2 + 2x − 7 = 0 b 3x2 + 4x − 1 = 0

SOLUTION

a For 5x2 + 2x − 7 = 0, a = 5, b = 2 and c = −7.

Hence b2 − 4ac = 22 + 140 = 144 = 122,

so x = −2 + 1210

or −2 − 1210

= 1 or −125

.

b For 3x2 + 4x − 1 = 0, a = 3, b = 4 and c = −1.

Hence b2 − 4ac = 42 + 12

= 28 = 4 × 7,

so x = −4 + 2√76

or −4 − 2√76

= −2 + √73

or −2 − √73

.



1F Solving quadratic equations 23

5 Solve each equation using the quadratic formula. Give exact answers, followed by approximations correct to four significant figures where appropriate.a x2 − x − 1 = 0 b y2 + y = 3 c a2 + 12 = 7ad u2 + 2u − 2 = 0 e c2 − 6c + 2 = 0 f 4x2 + 4x + 1 = 0g 2a2 + 1 = 4a h 5x2 + 13x − 6 = 0 i 2b2 + 3b = 1j 3c2 = 4c + 3 k 4t2 = 2t + 1 l x2 + x + 1 = 0

6 Solve by factoring:

a x = x + 2x

b a + 10a

= 7

c y + 2y

= 92

d (5b − 3)(3b + 1) = 1

7 Find the exact solutions of:

a x = 1x

+ 2 b 4x − 1x

= x c a = a + 4a − 1

d 5m2

= 2 + 1m

8 a If y = px − ap2, find p, given that a = 2, x = 3 and y = 1.b Given that (x − a)(x − b) = c, find x when a = −2, b = 4 and c = 7.

c Suppose that S = n2

(2a + (n − 1)d). Find the positive value of n that gives S = 80 when a = 4 and d = 6.

9 a Find a positive integer that, when increased by 30, is 12 less than its square.

b Two positive numbers differ by 3 and the sum of their squares is 117. Find the numbers.

c Find the value of x in the diagram opposite.

CHALLENGE

10 Solve each equation.

a 5k + 7k − 1

= 3k + 2 b u + 32u − 7

= 2u − 1u − 3

c y + 1y + 2

= 3 − yy − 4

d 2(k − 1) = 4 − 5kk + 1

e 2a + 3

+ a + 32

= 103

f k + 10k − 5

− 10k

= 116

g 3t

t2 − 6= √3 h 3m + 1

3m − 1− 3m − 1

3m + 1= 2

11 Solve each problem by constructing and then solving a quadratic equation.a A rectangular area can be completely tiled with 200 square tiles. If the side length of each tile was

increased by 1 cm, it would take only 128 tiles to tile the area. Find the side length of each tile.b A photograph is 18 cm by 12 cm. It is to be surrounded by a frame of uniform width whose area is

equal to the area of the photograph. Find the width of the frame.c Two trains each make a journey of 330 km. One of the trains travels 5 km/h faster than the other and

takes 30 minutes less time. Find the speeds of the trains.

(x – 7) cm

(x + 2) cmx cm



24 1GChapter 1 Methods in algebra

Solving simultaneous equations

There are two algebraic approaches to solving simultaneous equations — substitution and elimination. They can be applied to both linear and non-linear simultaneous equations.

Solution by substitutionThis method can be applied whenever one of the equations can be solved for one of the variables.

Solution by eliminationThis method, when it can be used, is more elegant, and usually involves less algebraic manipulation.

1G

12 SOLVING SIMULTANEOUS EQUATIONS BY SUBSTITUTION

• Solve one of the equations for one of the variables. • Then substitute it into the other equation.

Example 21 1G

Solve each pair of simultaneous equations by substitution.

a 3x − 2y = 29 (1) 4x + y = 24 (2)

b y = x2 (1) y = x + 2 (2)

SOLUTION

a Solving (2) for y, y = 24 − 4x. (2A)Substituting (2A) into (1), 3x − 2(24 − 4x) = 29 x = 7.Substituting x = 7 into (1), 21 − 2y = 29

y = −4.

Hence x = 7 and y = −4. (This should be checked in the original equations.)

b Substituting (1) into (2), x2 = x + 2 x2 − x − 2 = 0

(x − 2)(x + 1) = 0 x = 2 or −1.

From (1), when x = 2, y = 4, and when x = −1, y = 1. Hence x = 2 and y = 4, or x = −1 and y = 1. (Check in the original equations.)

13 SOLVING SIMULTANEOUS EQUATIONS BY ELIMINATION

• Take suitable multiples of the equations so that one variable is eliminated when the equations are added or subtracted.

1Methods in algebra



1G Solving simultaneous equations 25

Exercise 1G

1 Solve by substituting the first equation into the second:a y = x and 2x + y = 9 b y = 2x and 3x − y = 2c y = x − 1 and 2x + y = 5 d a = 2b + 1 and a − 3b = 3e p = 2 − q and p − q = 4 f v = 1 − 3u and 2u + v = 0

2 Solve by either adding or subtracting the two equations:a x + y = 5 and x − y = 1 b 3x − 2y = 7 and x + 2y = −3c 2x + y = 9 and x + y = 5 d a + 3b = 8 and a + 2b = 5e 4c − d = 6 and 2c − d = 2 f p − 2q = 4 and 3p − 2q = 0

DEVELOPMENT

3 Solve by substitution:a y = 2x and 3x + 2y = 14 b y = −3x and 2x + 5y = 13c y = 4 − x and x + 3y = 8 d x = 5y + 4 and 3x − y = 26e 2x + y = 10 and 7x + 8y = 53 f 2x − y = 9 and 3x − 7y = 19g 4x − 5y = 2 and x + 10y = 41 h 2x + 3y = 47 and 4x − y = 45

4 Solve by elimination:a 2x + y = 1 and x − y = −4 b 2x + 3y = 16 and 2x + 7y = 24c 3x + 2y = −6 and x − 2y = −10 d 5x − 3y = 28 and 2x − 3y = 22e 3x + 2y = 7 and 5x + y = 7 f 3x + 2y = 0 and 2x − y = 56g 15x + 2y = 27 and 3x + 7y = 45 h 7x − 3y = 41 and 3x − y = 17i 2x + 3y = 28 and 3x + 2y = 27 j 3x − 2y = 11 and 4x + 3y = 43

FOUNDATION

Example 22 1G

Solve each pair of simultaneous equations by elimination.a 3x − 2y = 29 (1)

4x + 5y = 8 (2)b x2 + y2 = 53 (1)

x2 − y2 = 45 (2)

SOLUTION

a Taking 4 × (1) and 3 × (2),

12x − 8y = 116 (1A) 12x + 15y = 24. (2A)

Subtracting (1A) from (2A),

23y = −92

y = −4.

Substituting into (1).

3x + 8 = 29

x = 7. Hence x = 7 and y = −4.

b Adding (1) and (2),

2x2 = 98

x2 = 49. Subtracting (2) from (1),

2y2 = 8

y2 = 4.

Hence x = 7 and y = 2, or x = 7 and y = −2, or x = −7 and y = 2, or x = −7 and y = −2.

÷ 23



26 1GChapter 1 Methods in algebra

5 Solve by substitution:a y = 2 − x and y = x2 b y = 2x − 3 and y = x2 − 4x + 5c y = 3x2 and y = 4x − x2 d x − y = 5 and y = x2 − 11e x − y = 2 and xy = 15 f 3x + y = 9 and xy = 6

6 Solve each problem by forming and then solving a pair of simultaneous equations.a Find two numbers that differ by 16 and have a sum of 90.b I paid 75 cents for a pen and a pencil. If the pen cost four times as much as the pencil, find the cost

of each item.c If 7 apples and 2 oranges cost $4, and 5 apples and 4 oranges cost $4.40, find the cost of each apple

and orange.d Twice as many adults as children attended a certain concert. If adult tickets cost $8 each, child tickets

cost $3 each, and the total takings were $418, find the numbers of adults and children who attended.e A man is 3 times as old as his son. In 12 years’ time he will be twice as old as his son. How old is

each of them now?f At a meeting of the members of a certain club, a proposal was voted on. If 357 members voted and

the proposal was carried by a majority of 21, how many voted for and how many voted against?

CHALLENGE

7 Solve simultaneously:a x + y = 15 and x2 + y2 = 125 b x − y = 3 and x2 + y2 = 185c 2x + y = 5 and 4x2 + y2 = 17 d x + y = 9 and x2 + xy + y2 = 61e x + 2y = 5 and 2xy − x2 = 3 f 3x + 2y = 16 and xy = 10

8 Set up a pair of simultaneous equations to solve each problem.a Kathy paid $320 in cash for a CD player. If she paid in $20 and $10 notes, and there were 23 notes

altogether, how many of each type were there?b Two people are 16 km apart on a straight road. They start walking at the same time. If they walk

towards each other, they will meet in 2 hours, but if they walk in the same direction (so that the distance between them is decreasing), they will meet in 8 hours. Find their walking speeds.



1H Completing the square 27

Completing the square

Completing the square can be done with all quadratic equations, whereas factoring is only possible in special cases.

The review in this section is mostly restricted to monic quadratics, in which the coefficient of x2 is 1. Chapter 3 will deal with non-monic quadratics. Chapter 3 will also require completing the square in a quadratic function, which is only slightly different from completing the square in a quadratic equation.

Perfect squaresThe expansion of the quadratic (x + 5)2 is

(x + 5)2 = x2 + 10x + 25.

Notice that the coefficient of x is twice 5, and the constant is the square of 5.

Reversing the process, the constant term in a perfect square can be found by taking half the coefficient of x and squaring the result.

Solving a quadratic equation by completing the squareThis process always works.

1H

14 COMPLETING THE SQUARE IN AN EXPRESSION x2 + bx + ⋯

Halve the coefficient b of x, and square the result.

Example 23 1H

Complete the square for each expression.a x2 + 16x + ⋯ b x2 − 3x + ⋯

SOLUTION

a The coefficient of x is 16, half of 16 is 8, and 82 = 64, so x2 + 16x + 64 = (x + 8)2.

b The coefficient of x is −3, half of −3 is −112 , and (−11

2)2 = 21

4 ,

so x2 − 3x + 214

= (x − 112)

2.

15 SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE

Complete the square in the quadratic by adding the same to both sides.



28 1HChapter 1 Methods in algebra

The word ‘Algebra’Al-Khwarizmi was a famous and influential Persian mathematician who worked in Baghdad during the early ninth century when the Baghdad Caliphate excelled in science and mathematics. The Arabic word ‘algebra’ comes from al-jabr, a word from the title of his most important work, and means ‘the restoration of broken parts’ — a reference to the balancing of terms on both sides of an equation. Al-Khwarizmi’s own name came into modern European languages as ‘algorithm’.

Example 24 1H

Solve each quadratic equation by completing the square.a t2 + 8t = 20 b x2 − x − 1 = 0

SOLUTION

a t2 + 8t = 20 + 16 t2 + 8t + 16 = 36

(t + 4)2 = 36

t + 4 = 6 or t + 4 = −6 t = 2 or −10

b x2 − x − 1 = 0+ 1 x2 − x = 1

+ 14 x2 − x + 1

4= 11

4 (x − 1

2)2

= 54

x − 1

2= 1

2√5 or x − 1

2= −

12√5

x = 12

+ 12√5 or 1

2− 1

2√5



1H Completing the square 29

Exercise 1H

1 What constant must be added to each expression in order to create a perfect square?a x2 + 2x b y2 − 6y c a2 + 10a d m2 − 18me c2 + 3c f x2 − x g b2 + 5b h t2 − 9t

2 Factor:a x2 + 4x + 4 b y2 + 2y + 1 c p2 + 14p + 49 d m2 − 12m + 36e t2 − 16t + 64 f x2 + 20x + 100 g u2 − 40u + 400 h a2 − 24a + 144

3 Copy and complete:a x2 + 6x + … = (x + … )2 b y2 + 8y + … = (y + … )2

c a2 − 20a + … = (a − … )2 d b2 − 100b + … = (b − … )2

e u2 + u + … = (u + … )2 f t2 − 7t + … = (t − … )2

g m2 + 50m + … = (m + … )2 h c2 − 13c + … = (c − … )2

DEVELOPMENT

4 Solve each quadratic equation by completing the square.a x2 − 2x = 3 b x2 − 6x = 0 c a2 + 6a + 8 = 0d x2 + 4x + 1 = 0 e x2 − 10x + 20 = 0 f y2 + 3y = 10g b2 − 5b − 14 = 0 h y2 − y + 2 = 0 i a2 + 7a + 7 = 0

CHALLENGE

5 Solve, by dividing both sides by the coefficient of x2 and then completing the square.a 2x2 − 4x − 1 = 0 b 2x2 + 8x + 3 = 0 c 3x2 + 6x + 5 = 0d 4x2 + 4x − 3 = 0 e 4x2 − 2x − 1 = 0 f 2x2 − 10x + 7 = 0

6 a If x2 + y2 + 4x − 2y + 1 = 0, show that (x + 2)2 + (y − 1)2 = 4.b Show that the equation x2 + y2 − 6x − 8y = 0 can be written in the form (x − a)2 + (y − b)2 = c, where a, b and c are constants. Hence find a, b and c.c If x2 + 1 = 10x + 12y, show that (x − 5)2 = 12(y + 2).d Find values for A, B and C if y2 − 6x + 16y + 94 = (y + C)2 − B(x + A).

FOUNDATION



9781108469043c01_01-G_H-Review Page PB 03/12/18 10:29 AM

30 Chapter 1 Methods in algebraRe

view

Chapter review exercise

1 Simplify:

a −8y + 2y b −8y − 2y c −8y × 2y d −8y ÷ 2y

2 Simplify:

a −2a2 − a2 b −2a2 − (−a2) c −2a2 × (−a2) d −2a2 ÷ (−a2)

3 Simplify:

a 3t − 1 − t b −6p + 3q + 10pc 7x − 4y − 6x + 2y d 2a2 + 8a − 13 + 3a2 − 11a − 5

4 Simplify:

a −6k6 × 3k3 b −6k6 ÷ 3k3 c (−6k6)2 d (3k3)3

5 Expand and simplify:

a 4(x + 3) + 5(2x − 3) b 8(a − 2b) − 6(2a − 3b) c −(a − b) − (a + b)d −4x2(x + 3) − 2x2(x − 1) e (n + 7)(2n − 3) f (r + 3)2

g (y − 5)(y + 5) h (3x − 5)(2x − 3) i (t − 8)2

j (2c + 7)(2c − 7) k (4p + 1)2 l (3u − 2)2

6 Factor:

a 18a + 36 b 20b − 36 c 9c2 + 36cd d2 − 36 e e2 + 13e + 36 f f

2 − 12f + 36g 36 − 25g2 h h2 − 9h − 36 i i2 + 5i − 36j 2j2 + 11j + 12 k 3k2 − 7k − 6 l 5l2 − 14l + 8m 4m2 + 4m − 15 n mn + m + pn + p o p3 + 9p2 + 4p + 36p qt − rt − 5q + 5r q u2w + vw − u2x − vx r x2 − y2 + 2x − 2y

Chapter 1 Review

Review activity • Create your own summary of this chapter on paper or in a digital document.

Chapter 1 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF worksheet

version is also available there.



Revi

ew

9781108469043c01_01-G_H-Review Page PB 28/11/18 12:08 PM

Chapter 1 Review 31

7 Simplify:

a x2

+ x4

b x2

− x4

c x2

× x4

d x2

÷ x4

e 3a2b

+ 2a3b

f 3a2b

− 2a3b

g 3a2b

× 2a3b

h 3a2b

÷ 2a3b

i xy

+ yx

j xy

− yx

k xy

× yx

l xy

÷ yx

8 Simplify:

a x + 45

+ x − 53

b 5x + 4

+ 3x − 5

c x + 12

− x − 45

d 2x + 1

− 5x − 4

e x2

− x + 34

f 2x

− 4x + 3

9 Factor each expression where possible, then simplify it.

a 6a + 3b10a + 5b

b 2x − 2y

x2 − y2c x2 + 2x − 3

x2 − 5x + 4

d 2x2 + 3x + 1

2x3 + x2 + 2x + 1e a + b

a2 + 2ab + b2 f 3x2 − 19x − 14

9x2 − 4

10 Solve each linear equation.

a 3x + 5 = 17 b 3(x + 5) = 17 c x + 53

= 17

d x3

+ 5 = 17 e 7a − 4 = 2a + 11 f 7(a − 4) = 2(a + 11)

g a − 47

= a + 112

h a7

− 4 = a2

+ 11

11 Solve each quadratic equation by factoring the left-hand side.

a a2 − 49 = 0 b b2 + 7b = 0 c c2 + 7c + 6 = 0d d2 + 6d − 7 = 0 e e2 − 5e + 6 = 0 f 2f

2 − f − 6 = 0g 2g2 − 13g + 6 = 0 h 3h2 + 2h − 8 = 0

12 Solve using the quadratic formula. Write the solutions in simplest exact form.

a x2 − 4x + 1 = 0 b y2 + 3y − 3 = 0 c t2 + 6t + 4 = 0d 3x2 − 2x − 2 = 0 e 2a2 + 5a − 5 = 0 f 4k2 − 6k − 1 = 0

13 Solve each quadratic by completing the square on the left-hand side.

a x2 + 4x = 6 b y2 − 6y + 3 = 0 c x2 − 2x = 12 d y2 + 10y + 7 = 0



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