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Apply algebraic methods in solving problems
Expanding brackets
The distributive law is used to expand brackets:
a(b + c) = a × b + a × c
Expand pairs of brackets using the rule:
(a + b)(c + d) = ac + ad + bc + bd
The following expansions involving two factors
should be familiar.
Example
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To expand an expression with three factors, select
a pair of factors to expand first. This reduces the
expression to the product of two factors.
Example
1. (x + 5)(x + 4)(x + 3)
= (x + 5)(x2 + 3x + 4x + 12)
[expanding (x + 4)(x + 3)]
2
= (x + 5)(x + 7x + 12)
[simplifying]
= x3 + 7x2 + 12x + 5x2 +35x + 60
[(x + 5)(x2 + 7x + 12)]
1. y (2y + 5w)
= y2 × 2y3 + y2 × 5w [distributive law]
[ y n × y m = y n + m]
= 2y5 + 5y2w
2
Externally assessed
4 credits
3
2. –2a(a2 – 7a + 8)
= –2a × a2 + –2a × –7a + –2a × 8
[a(b + c + d) = ab + ac + ad]
3
= –2a + 14a2 – 16a
3. (3x + 5)(4x – 7)
= 12x2 – 21x + 20x – 35
[(a + b)(c + d) = ac + ad + bc + bd]
2
= 12x – x –35
[simplifying]
4. (5x – 2)2 + x(4 – x)
= (5x – 2)(5x – 2) + x(4 – x)
= 25x2 – 10x – 10x + 4 + 4x – x2
[distributive law]
2
[collecting like terms]
= 24x – 16x + 4
Perfect squares can be expanded more quickly using
the following formula.
(a ± b)2 = a2 ± 2ab + b2
e.g. (x + 5)2 = x2 + 2 × x × 5 + 52
= x3 + 12x2 + 47x + 60
[collecting like terms]
2
2. –4p(8 – p)
= –4p(8 – p)(8 – p)
= –4p(64 – 16p + p2)
[expanding and simplifying (8 – p)2]
= –256p + 64p2 – 4p3
3. (x + 5)3 = (x + 5)(x + 5)2
= (x + 5)(x2 + 10x + 25)
= x3 + 10x2 + 25x + 5x2 + 50x + 125
= x3 + 15x2 + 75x + 125
Note: A simple check on an expansion can be done
by substituting x = 1 into both the original
expression and its expansion. If the answers
don’t match, you have made an error.
For example, substituting x = 1 in (x + 5)(x + 4)(x + 3)
gives (1 + 5)(1 + 4)(1 + 3) = 120.
Substituting x = 1 in x3 + 12x2 + 47x + 60 gives
1 + 12 + 47 + 60 which is also 120.
The two results agree, so the expansion is most likely
to be correct.
= x2 + 10x + 25
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
AS 91261
MATHEMATICS AND STATISTICS 2.6
2
AS 91261
Ans. p. 59
Achievement Standard 91261 (Mathematics and Statistics 2.6)
Exercise A: Expanding brackets
j.
4(2x – 3)(x – 4)
1. Expand and simplify.
a. 8x(3x – 4)
b. –6y(1 – x)
2. Expand and simplify the following expressions.
a. x(3x – 4) + 2x(x + 1)
c. 7x4(x3 – 8)
b. 2ab(a + b) – a2(b – a)
d. –8p(p2 – 6p – 4)
c. 9 – x2(9 – x)
e. (7a + 4)(2a –1)
d. (w + 11)(w – 2) + 2w(3w – 5)
f.
(5x + 3y)(xy – 7)
g. (5x + 2y)2
e. 12x – (x + 4)2
h. (2x – 7)2
f.
i.
(3t –7)(3t + 7) – (3t – 7)2
5(x + 1)2
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
Alternatively
x + 2 – 4x 3x + 4
=
[subtracting fractions]
6
2
–3x + 2 3x + 4
=
6
2
–6x + 4 = 18x + 24 [cross-multiplying]
–24x = 20
–20
–5
or
x=
24
6
Use brackets around algebraic expressions in fractions,
so that errors with signs are avoided.
Example
Solve for w:
w + 2 1 – w 3w
–
=
4
2
8
Solution
(w + 2) (1 – w) 3w
–
=
4
2
8
[bracketing
expressions]
8(w + 2) 8(1 – w) 8 × 3w [multiplying
–
=
by LCM of
4
2
8
denominators]
2(w + 2) – 4(1 – w) = 3w
[cancelling]
2w + 4 – 4 + 4w = 3w
[expanding]
6w = 3w
[simplifying]
3w = 0
[subtracting 3w
(both sides)]
w=0
19
Forming linear equations and inequations
Read problems carefully to define the unknown variable
and identify the correct sequence of mathematical
operations (such as adding, dividing, etc.).
You will then need to solve your equation. Make sure
you check your answer in the context of the question
to ensure it applies to the practical situation.
Example
Miro and Maddie both drive taxis. Miro charges
$6.50 flagfall plus $3.20 per kilometre. Maddie
charges $4 flagfall plus $3.90 per kilometre.
A trip of a certain length costs more with Miro
than with Maddie. How long was this trip? Give
your answer to the nearest tenth of a kilometre.
The SOLV function on a graphics calculator
is useful for checking and solving multi-step
linear equations.
Linear inequations
Linear inequations will also involve multi-step reasoning.
Remember to reverse the inequality sign when
multiplying or dividing by a negative.
Example
3 – 4(x + 2) ) 3(x + 1)
3 – 4x – 8 ) 3x + 3
[expanding]
–7x ) 8
[rearranging]
[dividing by –7 and
x* 8
reversing inequality]
–7
1
x * –1
[simplifying]
7
Note: Alternatively, collect positive numbers of x on
the right side. Rearranging the second line
gives –8 )7x. Dividing by 7 gives –8 )x,
7
8
which can be written in reverse as x * – .
7
Solution
Let the trip length be x kilometres
Miro charges
6.5 + 3.2x
Maddie charges 4 + 3.9x
6.5 + 3.2x > 4 + 3.9x [Miro costs more
than Maddie]
–0.7x + 6.5 > 4
–0.7x > –2.5
–2.5
x<
–0.7
[subtracting 3.9x]
[subtracting 6.5]
[dividing by –0.7 (and
swapping signs)]
x < 3.5714...
The trip was at most 3.5 km (1 d.p.)
[rounding down to
the nearest tenth]
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
AS 91261
Apply algebraic methods in solving problems
22
Achievement Standard 91261 (Mathematics and Statistics 2.6)
AS 91261
4. Robert receives $x per week pocket money.
One week he spends $10 then saves half of the
remaining money. The next week he spends $6
then saves a third of the remaining money. If he
saves more in the second week than the first, solve
the inequation x – 10 < x – 6 to find the greatest
2
3
weekly pocket money Robert could receive.
7. At Occasions reception lounge the charge is $480
booking fee plus $85 per guest. Another reception
lounge Oasis charges $565 booking fee plus $82
per guest.
Set up and solve an inequation to work out the
minimum number of guests for which Oasis would
be the cheaper option. (Your answer should be a
whole number.)
5. Marnie is 32 years younger than her aunt. Marnie
worked out that if she added 13 to her age, then
tripled it, she would be 1 less than double her
aunt’s age.
Solve the equation 3(x + 13) = 2(x + 32) – 1 to
find her aunt’s age.
8. Hone sits three tests. The first result is 15% less
than the second, and the third result is 2.5%
above the average of the first two results. If Hone
must average at least 65% over the three tests,
solve the following inequation
x + (x + 15) + 0.5(2x + 15) + 2.5 * 65
3
to work out the minimum mark Hone needs to
score in the first test.
6. A journey is travelled at a steady speed of 80 km
per hour. If the speed is increased by 10 km per
hour, then the time for the journey is reduced by
half an hour. Solve the following equation to find
the length of the journey.
d
d
1
= +
80 90 2
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
9. Shapes A and B are made from rectangles, with
side lengths given in terms of x.
(x + 5.5) cm
23
The first activity is 1 km further than the second
but only half as long as the third.
If all three activities take a total of 1 hour and
45 minutes, how far did she travel altogether?
A
(x + 2.75) cm
2x cm
B
x cm
2x cm
x cm
For what values of x is the perimeter of A at least
5 cm greater than the perimeter of B?
12. The front of an A-frame building is in the shape of
an isosceles triangle with base width 5 metres and
height 10 metres. The entrance is rectangular and
touches the sides of the building. Let the width
of the entrance be x metres and the height of the
entrance be 5.5 metres. Find the width of the
entrance.
Diagram is
NOT to scale
x
10 m
5.5 m
10. A rectangular enclosure is made up of a rectangular
garden 2x metres wide and (3x + 4) metres long,
surrounded by a path 1 metre wide. If the area of
the path is no more than 200 m2, find the maximum
area the garden can be.
Entrance
5m
11. Sharon records her speed and time for three activities.
Activity 1: A walk to the shops and back at 4 km h–1.
Activity 2: A jog to and from the post-office at 6 km h–1.
Activity 3: A run around the running track at 9 km h–1.
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
AS 91261
Apply algebraic methods in solving problems
32
Achievement Standard 91261 (Mathematics and Statistics 2.6)
AS 91261
The quadratic formula
A general quadratic formula can be used to solve
quadratic equations. This equation is derived by
completing the square.
x=
12 ± 144 – 12
2
x=
12 + 132
12 – 132
or
[± means + or –]
2
2
If ax2 + bx + c = 0 then
x=
–b ± b2 – 4ac
2a
In some cases the quadratic formula can be used as an
alternative to factorisation.
Example
Q. Solve the quadratic equation
2x2 + 7x + 6 = 0
by factorisation and by the quadratic formula.
A. By factorisation:
(2x + 3)(x + 2) = 0
[factorising]
2x + 3 = 0 or x + 2 = 0 [setting factors
to zero]
3
[solving]
x = – or x = –2
2
By formula:
Comparing 2x2 + 7x + 6 = 0 with
ax2 + bx + c = 0 gives a = 2, b = 7, c = 6
Substituting these values in the quadratic
formula gives:
–7 ± 72 – 4 × 2 × 6
2×2
–b ± b2 – 4ac
[x =
]
2a
–7 ± 1
[simplifying]
x=
4
–7 – 1
–7 + 1
or x =
x=
4
4
3
x = –2 or x = –
2
x = 0.255 or 11.745 (3 d.p.)
Note: The negative b-value was bracketed to
avoid substitution errors.
In harder questions, a quadratic equation will be
formed from given information. Solutions will need to
be interpreted.
Example
A rectangular photograph is 5 cm longer than it is
wide. If the area of the photo is 160 cm2, find the
width of the photo.
Solution
Let width of photo be x, then length is x + 5
x+5
x
x=
x(x + 5) = 160
[area rectangle
= width × length]
x2 + 5x = 160
[expanding]
x + 5x – 160 = 0
2
[rearranging]
Comparing with ax2 + bx + c = 0 gives
a = 1, b = 5, c = –160
In other cases, factorisation cannot be used.
x=
–5 ± 52 – 4 × 1 × –160
2×1
Solution
Comparing x2 – 12x + 3 with ax2 + bx + c gives
x=
–5 ± 665
2
a = 1, b = –12, c = 3
The width of the photo is 10.4 cm (1 d.p.)
Example
Solve x2 – 12x + 3 = 0, using the quadratic formula.
[equating coefficients of
x2, x, and constants]
Substituting these values in the formula gives
x=
–(–12) ± (–12)2 – 4 × 1 × 3
2×1
[quadratic
formula]
[simplifying]
x = 10.3938 or –15.3938 (ignore since negative)
Graphical calculators can be used to solve
quadratic equations of the form
ax2 + bx + c = 0. Check how to do it on
your calculator.
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
35
6. Beth enters the number 1 into her calculator,
doubles it then subtracts 5. She then squares the
result of this calculation and divides by 9. She is
surprised to find that the result of these calculations
is equal to the number she first entered. She
wonders if this is true for any other integer.
(2x – 5)2
Use the quadratic formula to solve
=x
9
to find out.
b. Solve this equation to find the length of the
hypotenuse to 1 d.p.
7. A square room is made 3 m longer and 2 m wider.
Its new area is 48.75 m2.
Solve (x + 2)(x + 3) = 48.75 to find its original
area.
x
x
5. Peter is solving the quadratic equation
5x2 – 14x – 3 = 0
He has made at least one error.
Correct his working and find the roots of the
equation.
x=
–14 ± 196 – 4 × 5 × 3
2×5
x=
–4 ± 136
10
3
x = 0.766 or –1.566
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
2
AS 91261
4. A right-angled triangle has hypotenuse
of length (3x + 4) cm.
+4
x
3
Its other two sides are
of lengths (x + 4) cm
and (4x – 1) cm.
4x – 1
a. Show that the relationship between
side lengths can be simplified to
8x2 – 24x + 1 = 0. (Hint: Use Pythagoras’
rule – expand and simplify)
x+4
Apply algebraic methods in solving problems
38
Achievement Standard 91261 (Mathematics and Statistics 2.6)
AS 91261
The nature of the roots of a quadratic
equation
The graph of the quadratic function y = ax + bx + c
is a parabola.
2
Setting y = 0 gives the quadratic equation
ax2 + bx + c = 0 to solve.
A quadratic equation can have two, one or no
solutions, as shown in the graphs below.
y
Some examples are shown below.
Quadratic
equation
Value of
b2 – 4ac
Nature of
roots
4x2 + 12x + 9 = 0
144 – 4 × 4 × 9 = 0
1 repeated
(rational) root
x2 + 9x – 2 = 0
81 – 4 × 1 × –2 = 89
2 irrational
roots
3x2 + 2x + 6 = 0
4 – 4 × 3 × 6 = –68
No real
roots
x2 – 7x – 8 = 0
49 – 4 × 1 × –8 = 81
2 rational
roots
a>0
More complex problems explore the nature of the
roots for quadratics with unknown coefficients.
x
Two solutions
(graph cuts
x-axis twice)
One solution
(graph touches
x-axis once)
For what value(s) of k does the equation
2x2 – kx + k + 2 = 0 have no real roots?
No solutions
(graph doesn’t cut
or touch the x-axis)
x
y
Example
Solution
For the quadratic 2x2 – kx + k + 2 = 0, a = 2,
b = –k and c = k + 2 [comparing with ax2 + bx + c]
Substituting in b2 – 4ac, the discriminant is
(–k)2 – 4 × 2 × (k + 2) = k2 – 8k – 16 [simplifying]
For no real roots, k2 – 8k – 16 < 0.
a<0
The discriminant
Solving k2 – 8k – 16 = 0 (by calculator or formula)
gives k = –1.657 or 9.657.
The formula for the solution to ax2 + bx + c = 0 is
Since y = k2 – 8k – 16 is an upward-pointing parabola,
y
–b ± b2 – 4ac
x=
2a
The expression under the square root sign, b2 – 4ac, is
called the discriminant.
The value of the discriminant determines the number and
nature of the roots (solutions) of a quadratic equation.
Discriminant b2 – 4ac
Nature of roots
Positive
• perfect square
• not a perfect square
Two real roots
• 2 rational roots
• 2 irrational roots
Zero
One (repeated) real root
Negative
No real roots (imaginary
roots)
–1.657
9.657
x
k2 – 8k – 16 is negative between the roots.
Thus the solution is –1.657 < k < 9.657.
Note: This solution is correct to 3 d.p. The exact
8 ± 128
solution to k2 – 8k – 16 = 0 is
2
so the exact solution to k2 – 8k – 16 < 0 is
8 – 128
8 + 128
<k<
which
2
2
simplifies to 4 – 4 2 < k < 4 + 4 2
[since 128 = 64 × 2 = 8 2 ]
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
8. The value, V dollars, of a motorbike t years after it
was purchased is given by the equation:
V = 12 500(0.5)0.2t
47
11. 12 600 people visited a tourist attraction in the
first month that it was open, but the number of
visitors has been falling by 5% per month.
When will the value of the motorbike be $8 000?
A model for the monthly number of visitors
(in thousands) is given by
V = A(1 – 0.05) n – 1
where n is the number of months since opening.
a. What is the value of A?
9. A population of 12 000 000 is growing by 3% per
year. How long does it take for the population to
double?
10. The half-life of uranium-235 is 704 million years.
This means that 50% of the original radioactivity is
present after 704 million years. How long does it take
for the radioactivity of uranium to reduce by 10%?
b. How many people visited in the 3rd month?
Round your answer to the nearest ten.
c. If the number of visitors falls below 10 000 in
a month, then ticket pricing will be reviewed
in the following month’s financial meeting.
If the pattern of monthly visitor numbers
continues as described above, in which month
will the financial meeting to review ticket
pricing take place?
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
AS 91261
Apply algebraic methods in solving problems
48
Achievement Standard 91261 (Mathematics and Statistics 2.6)
AS 91261
Logarithmic equations
Logarithmic equations are converted to exponential
equations using the relationship:
If logb y = x then y = b x
log 52
[dividing by log 3]
log 3
y = 3.597 (3 d.p.) [calculator]
y=
Alternatively, the base-changing formula can be
used to evaluate logarithms to any base:
Example
The logarithmic equation log2 x = 3.5 is
equivalent to the exponential equation x = 23.5
So x = 11.314 (3 d.p.) [evaluating by calculator]
If the base of the logarithm is unknown, then a power
equation must be solved.
When dealing with terms of the form logb x , remember that
• b > 0 (the base of a logarithm is a positive real
number)
• x > 0 (logarithms are defined for positive
numbers only)
Example
1. If logx 3 = 4, then
[rearranging using the rule]
x4 = 3
4
x = 3 [ignore the negative root
(since x > 0)]
x = 1.316 (3 d.p.)
2. logx 16 = 9.7 is equivalent to the equation
x9.7 = 16
To solve this type of equation, raise both
sides to the reciprocal of the power of x.
1
loga x =
log10 52
= 3.597
log10 3
Note: log10 is chosen because it can be evaluated by
calculator using the log key. It is also possible to
use the natural logarithm (loge) using the In key.
For example, log3 52 =
Other logarithmic equations
If two numbers have the same logarithm (to the same
base), then the numbers are equal.
if logb y = logb x then y = x
Example
Q. Solve log (5x – 8) = log (2x + 1)
A. 5x – 8 = 2x + 1 [if logb y = logb x then y = x]
3x = 9
[rearranging]
x=3
The following logarithm laws can be used to rearrange a
logarithmic equation, so that it can be solved more easily.
log x + log y = log xy
x
log x – log y = log
y
n log x = log x n
1
1
]
9.7
[the product of a number
x1 = 16
and its reciprocal is 1]
x = 1.331 (3 d.p.) [by calculator]
(x 9.7)9.7 = 169.7 [the reciprocal of 9.7 is
1
9.7
To find the logarithm of a number to a given base,
form a logarithmic equation and solve it.
Example
Find log3 52
Solution
Let
y = log3 52
[equivalent exponential form]
3y = 52
y
log(3 ) = log 52 [taking log of both sides]
y log 3 = log 52 [logarithm rules]
logb x
logb a
Example
Q. Solve 2 log x – log 4x = log (x – 1), for x > 1
A. log x2 – log 4x = log (x – 1) [n log x = log x n]
x2
= log (x – 1) [log x – log y
log
4x
x
= log ]
y
x
log = log (x – 1)
4
x
=x–1
[if logb y = logb x
4
then y = x]
x = 4x – 4
[multiplying by 4]
3x = 4
4
x=
3
[rearranging]
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
Exercise A: Expanding brackets (page 2)
1.
2.
3.
4.
5.
= a3 + b3
24x – 32x
b.
–6y + 6xy
7x7 – 56x4
d.
–8p3 + 48p2 + 32p
b.
a3 – b3 = (a – b)(a2 + ab + b2)
e.
14a2 + a – 4
f.
5x2y – 35x + 3xy2 – 21y
c.
i.
(p + 2q)(p2 – 2pq + 4q2)
g.
25x +20xy + 4y
h.
4x – 28x + 49
ii.
(10m – n)(100m2 + 10mn + n2)
i.
5x + 10x + 5
j.
8x – 44x + 48
2
2
2
2
2
a.
5x –2x
b.
a b + 2ab + a
c.
9 – 9x2 + x3
d.
7w2 – w – 22
e.
–x + 4x – 16
f.
42t – 98
g.
12x – 6x2
h.
20x
i.
5y2 – 24y – 5
j.
16x2 + 4x – 31
a.
x3 + 12x2 + 41x + 42
b.
x3 – 4x2 – 20x + 48
2
2
2
2
3
1.
c.
6x – 55x – 6x + 135
d.
–18x + 138x – 84x
e.
x3 – 2x2 – 55x + 200
f.
–18x3 – 24x2 – 8x
3
2
3
g.
8x + 12x + 6x + 1
h.
125x – 225x + 135x – 27
–14x2 – 35x + 21
j.
–26x3 – 54x2 – 36x – 8
a.
i.
2
3
(a + b)(a – b) = a2 – ab + ab – b2
iii.
(a + b) = (a + b) (a + b) (a + b)
= (a + b)(a2 + 2ab + b2)
2.
(a – b)3 = a3 – 3a2b + 3ab2 – b3
d.
i.
4x2 + 12xy + 9y2
ii.
x3 + 9x2 + 27x + 27
iii.
8x – 12x + 6x – 1
3
4.
3y
d.
ab–4
e.
7c –3
8
–2b–6
5
1
x4
p3
f.
3.
–2
d.
p –1
3
1
2c –5
5
y2
2a5
3
4y
3
2c2
f.
–w1 or –w
h.
3p4
5
a.
x5
b.
p2
c.
w2
d.
4s 2
e.
3w 2
f.
x5
g.
8y 3
h.
x5
a.
2
b.
3
c.
16
d.
81
e.
8
27
9 or 2 1
4
4
f.
1 000
h.
27
64
a.
g.
4.
a.
7xy(y2 + 3x)
b.
2m2(4m2 – m – 2n)
c.
9y(xy – 9z + 4y)
d.
p6(p – 1)
e.
a4(1 + a4)
f.
2x3(6 + x2)
g.
3ab3(3a3 – 2ab3 + 9b)
h.
x5y8(6x2y2 + 2 – 3y)
a.
(y + 8)(y + 9)
b.
(w – 25)(w – 8)
c.
(x + 6)(x + 2)
d.
(x – 11)(x + 2)
e.
10
e.
(x – 3)(x + 10)
f.
(x – 4)(x – 8)
g.
a.
g.
(x – 8)(x + 3)
h.
(x + 11)(x – 11)
3(x + 7)(x – 2)
b.
2(y – 5)(y – 3)
c.
8(x + 2)(x – 2)
d.
–(x – 9)(x + 3) or (9 – x)(x + 3)
e.
2(7 + 5m)(7 – 5m)
f.
x(x + 2)2
g.
(ab + 2)(b – 3)
h.
(3 – a)(a + 13) or –(a – 3)(a + 13)
a.
(3x + 1)(x + 2)
b.
(2x – 1)(x + 2)
c.
(2y – 3)(2y – 1)
d.
(3p – 4)(p + 1)
e.
(3m – 1)(2m + 3)
f.
(h + 2)(4h + 3)
g.
–(x – 2)(2x + 3) or
(2x + 3)(2 – x)
h.
–(2x – 3)2 or (2x – 3)(3 – 2x)
b.
g.
9
5
–3
7
2
a.
h.
3
Exercise B: Factorising (page 6)
3.
c.
e.
= a3 + 3a2b + 3ab2 + b3
c.
4x–3
c.
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
(a – b)2 = a2 – 2ab + b2
b.
g.
3
b.
(5 – 4p)(25 + 20p + 16p2)
x–5
2
(a + b)(a + b) = a2 + ab + ab + b2
ii.
(2x + 3y)(4x2 – 6xy + 9y2)
iv.
a.
2
i.
3
iii.
Exercise C: Indices (page 9)
= a2 – b2
2.
(a + b)(a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3
a.
= a2 + 2ab + b2
1.
a.
c.
2
–2
1
5.
a.
10 2
b.
10–1
c.
10
d.
100
f.
10 4
10 2
h.
27x – 17
b.
10 3
–1
1 or x 4
–5
2
3
3
–3
6.
c.
5
a4
d.
e.
b2
216a6
324b
a
f.
g.
© ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act.
h.
–
1
1
x4
1
8x6
9x4y2
16
p
32q5
ANSWERS
ANSWERS