4
SECONDARY EDUCATION
DUAL FOCUS
MATHEMATICS
A
ANDALUSIA
INDEX
Train yourself by solving problems ...........7
Summary • Problems
1. Natural numbers, integers and fractions
9
Summary • 1 Natural numbers • 2 Integers • 3 Fractions • 4 Operations with fractions • 5 Problems with fractions • 6 Powers • The final challenge
2. Decimal numbers ..........................................15
Summary • 1 Types of decimal numbers • 2 Decimal to fraction • 3 Approximate quantities. Errors • 4 Scientific notation • The final challenge
3. Real numbers
23
Summary • 1 Real numbers: the real number line • 2 Sections of the real number line: intervals and halflines • 3 Roots and radicals • The final challenge
4. Polynomials
29
Summary • 1 Polynomials. Operations • 2 Division of a polynomial by (x – a) • 3 Roots of a polynomial • 4 Factorising polynomials • 5 Preparing to solve equations • The final challenge
5. Trigonometry .............................................
37
Summary • 1 Solving first-degree equations • 2 Second-degree equations • 3 Other types of equations • The final challenge
6. Systems of equations and inequations
45
Summary • 1 Systems of linear equations • 2 Solving systems of equations • 3 More complex systems of linear equations • 4 Non-linear systems • 5 Solving problems using systems • 6 Inequations with one unknown • The final challenge
7. Functions. Characteristics ....................
53
Summary • 1 Basic concepts • 2 How functions are presented • 3 Continuous functions. Discontinuities • 4 Variations in a function • 5 Tendency and periodicity • The final challenge
8. Basic functions .........................................
61
Summary • 1 Linear functions • 2 Quadratic functions • 3 Radical functions • 4 Inversely proportional functions • 5 Exponential functions • The final challenge
9. Similarity. Applications
67
Summary • 1 Similarity • 2 Homothety • 3 Rectangles with interesting dimensions • 4 Similarity between triangles • 5 Similarity in right-angled triangles • The final challenge
10. Statistics
77
Summary • 1 Basic concepts • 2 Frequency tables • 3 Statistical parameters: x – and σ • 4 Position parameters • 5 Box plots • 6 Statistical inference • 7 Statistics in the media • The final challenge
11. Bivariate distributions
87
Summary • 1 Bivariate distributions • 2 Correlation value • 3 Using the line of best fit to make estimations • 4 Does correlation imply cause and effect? • 5 Bivariate distributions with a calculator • The final challenge
12. Probability
95
Summary • 1 Random events • 2 Probability of an event • 3 Finding the probability of an event • 4 Compound events. Tree diagrams • 5 Contingency tables • The final challenge
1 NATURAL NUMBERS, INTEGERS AND FRACTIONS
TAKE ACTION!
NATURAL NUMBERS, INTEGERS AND FRACTIONS
Natural
Operations
Absolute value
Operations
Counting
Solving problems
Solving problems
Working with natural numbers, integers, fractions and powers will allow you to face 'The final challenge', in which you will have to find special sets of numbers.
FOCUS ON ENGLISH
Powers with an integer base
Simplifying
Equivalent fractions
Operations
Solving problems
Powers with a rational base
Properties
Operations
Listening
Listen and repeat. The vocabulary is at anayaeducacion.es
Discover
In ancient Mesopotamia, the Babylonians used sexagesimal fractions. The denominators of these fractions were powers with base 60. Centuries later they were replaced by the decimal fractions we use today, when these were found to make performing calculations easier.
Reading
Read the text ‘Integers and fractions through history’ at anayaeducacion.es and answer the questions.
Natural numbers 1
We count with natural numbers. As you know, they are 0, 1, 2, 3, ..., 10, 11, 12, ... 99, 100, 101, ... There are infinitely many of them. The set of all of them is called N. As they are ordered we can represent them on a line:
Natural numbers are also used for ranking. For example, we say that Olivia Gutiérrez is the thirteenth on the class list.
1 Calculate.
Integers 2
The integers are represented on the number line in this way:
• The naturals (zero and the positive integers) were already ordered.
• All natural numbers are greater than negative integers.
• If a and b are natural numbers and a < b then –a > –b
Absolute value of an integer
The absolute value of a number is its magnitude if we disregard its sign.
Rules for operating with integers
• To add positive and negative numbers, we group them together, subtract the results and use the sign of the one with the greater absolute value.
• If a parenthesis is preceded by a minus sign, it can be deleted by changing the sign of all the addends inside it.
• For multiplying whole numbers, remember the 'sign rule':
1 Put the following in order from smallest to largest: –4, 19, 7, 0, –6
2 Calculate. a) | | –3 | | b) | | 5 + (3 – 11) | | c) | 5 + | 3 – 11 | | d) | 30 – (–20 – 9) |
Fractions 3
Fractional numbers to express measurements
Fractional numbers are: ,,,,
In all these fractions, the numerator is smaller than the denominator and are therefore parts of the unit. They are called proper fractions.
Fractional numbers such as 2 3 , 3 7 – and 4 19 , which can be expressed as the sum of an integer and a proper fraction (mixed numbers), are also fractional numbers:
Fractional numbers can be represented on the number line next to the integers:
The fractional numbers, together with the integers, form the set of rational numbers, which is denoted by the letter Q
Simplification of fractions. Equivalent fractions
Remember that to simplify a fraction is to divide its numerator and denominator by the same whole number.
A fraction that cannot be simplified is said to be irreducible.
Two fractions are equivalent if, when simplified, they give rise to the same irreducible fraction.
Operations with fractions 4
1 Express the following as an integer and a fraction a) 9 40 b) 5 86 c) 10 127
2 Find the irreducible fraction. a) 21 18 b) 35
To add (and subtract) fractions with different denominators, we must transform them into equivalent fractions with the same denominator.
The product of two fractions is another fraction whose denominator is the product of their denominators and whose numerator is the product of their numerators:
The quotient of two fractions is the product of the first by the inverse of the second:
Problems with fractions 5
In a class of 36 pupils, 2/3 are boys. 3/4 of the girls do music. What fraction of the class are girls who do music? How many girls is that?
3 2 are boys → 3 1 are girls
Fraction that are girls who do music:
4 3 of 3 1 = 4 3 3 1 12 3 4 1 ·= = of the total class
Number of girls who do music: 4 1 of 36 = 36 : 4 = 9 2
Three friends share a prize. The first one gets 2/5 of the total; the second 5/9 of the remainder, and the third, €92. How much was the prize?
The first gets 5 2 of the prize; the remainder is 5 3 .
The second gets 9 5 of 5 3 = 9 5 · 3 5 = 9 3 = 3 1 of the prize.
The first and the second friend get 5 2 3 1 15 65 15 11 += + = of the prize between them.
The third gets 15 15 15 11 15 4 –= of the prize.
15 4 of the prize is €92; 15 1 is 92 : 4 = €23; 15 15 is 23 · 15 = €345
Therefore, the prize was (92 : 4) · 15 = €345.
1 A piece of land is divided into three parts. Two of them are 2/5 and 1/3 of the total. Which is the largest?
2 In the previous problem, if the land measures 240 m2, what is the surface area of each portion?
3 2/5 of the boys in a class wear glasses. There are 36 people on the class list, 7/12 of whom are girls. How many boys wear glasses?
4 George has spent 2/7 of his allowance on music and 1/5 on books. What fraction of his allowance has he spent? What fraction does he have left?
5 Reading. A dentist spends 1 and 3/4 hours in his practice. If he sees 15 patients in that time, what fraction of an hour can he spend with each one? How many minutes is that?
6 Four-way split: A and B get 2/7 and 13/21 of the total respectively. C receives 7/10 of the remainder. And finally D receives € 390. What was the initial sum of money?
7 A cyclist drops out of the race when she has covered 2/3 of the course. If she had stayed 10 kilometres longer, she would have covered three quarters of the course. How many kilometres was the whole race?
Integer base and positive integer exponent
a aaa a ·· ·…· times n n = 12 3 a is the base; n is the exponent
• If a is positive, an is positive whatever the vale of n
• If a is negative: n n( → even → an positive. For example, (–2)4 = 16. → odd → an negative. For example, (–2)5 = –32.
Properties of powers
1. a m · a n = a m + n
2. (a · b)m = a m · b m
3. (a m)n = a m · n
4. If m > n, a a a n m mn –=
• (–2)3 · (–2)5 = (–2)3 + 5 = (–2)8 • 64 = (2 · 3)4 = 24 · 34 • (–2)5 = (–1 · 2)5 = (–1)5 · 25 = –25 • (53)4 = 53 · 4 = 512 • 10 10 10 10 4 7 74 3 –==
Rational base and integer exponent
a 0 = 1, wherever a ≠ 0.
a –n = a 1 n . Therefore, a –1 = a 1 b a a b –1 = bl
1 Calculate the following powers:
a) –105 b) (–10)5
c) (–10)6 d) –(–10)5 e) (–1)100 f) –106 g) –16 h) –(–1)101
2 Simplify: 5 [( )] () () 3 33 ––· –33 58 4
3 Put the following in order from smallest to largest: 2–3, 2–1, 20, 2–2, 2– 4, (–2)–3, (–2)–1
4 Calculate the following powers:
The final challenge TAKE ACTION!
NATURAL NUMBERS, INTEGERS AND FRACTIONS
Working with natural numbers, integers, fractions and powers will allow you to face 'The final challenge', in which you will have to find special sets of numbers.
Sometimes, in mathematics, we associate human qualities with numbers. We talk about perfect numbers, ambitious, sublime...
What number is perfect for you? If we're talking about exams, it's probably 10. But what is a perfect number?
1 We say that a number is perfect if it is equal to the sum of its proper divisors, that is, the sum of all its divisors except itself.
For example, 28 is perfect because: D(28) = {1, 2, 4, 7, 14, 28}
And it holds that: 28 = 1 + 2 + 4 + 7 + 14
a) What is the first perfect number? Here's a hint: it's a one-digit number.
b) Knowing that all perfect numbers are pairs ending in 6 or 8, what is the next perfect number?
c) Non-perfect numbers are of two types: abundant or deficient. We say that a number is abundant when it is less than the sum of all its proper divisors. Otherwise, it is deficient.
– Classify numbers up to 30 as perfect, abundant or deficient.
– How are prime numbers abundant or deficient? Explain why.
2 A number is ambitious if it can be made perfect by following these steps:
1st We find all the proper divisors of the number and add them together.
2nd We calculate the proper divisors of the resulting number and add them up again.
3rd We continue in this way until we get a perfect number.
a) Is number 24 ambitious, and number 25? Check it out.
b) Are prime numbers ambitious and why?
3 A number is sublime if both the number of its divisors and the sum of its divisors are perfect. Prove that the number 12 is sublime.
Form small groups of classmates and investigate other sets of numbers to which we associate human qualities.
Present the results of your investigations in class and, together, make a poster with all the types of numbers you have found.
4 POLYNOMIALS
POLYNOMIALS
Literal part
Coefficient
Term
Numerical value
Root
Adding and subtracting
Multiplying
Degree
Operating with monomials
Factorising polynomials
A monomial by a polynomial
Dividing
A polynomial by a monomial
Working with polynomials will enable you to tackle ‘The final challenge’, in which we will ask you to invent some problem statements and translate other statements into algebraic language.
FOCUS ON ENGLISH
Listening
Notable products
Two polynomials
Two polynomials
rule
Listen and repeat. The vocabulary is at anayaeducacion.es
Discover
The French mathematician Viète (1540-1603) realised that a new form of notation was needed to improve algebraic language. He proposed using vowels to represent unknowns and consonants to represent known parameters, the opposite of what we do today. In addition, instead of using signs to represent operations, he used abbreviations.
Reading
Read the text ‘Main phases in the history of algebraic language’ at anayaeducacion.es and answer the questions.
Ruffini's
Polynomials. Operations 1
Adding and subtracting polynomials
• To add two polynomials, we group their terms and add similar monomials.
• To subtract two polynomials, we add the first to the opposite of the second.
For example, if
We can also operate directly, removing parentheses and grouping like terms:
The product of a monomial multiplied by a polynomial
To multiply a monomial by a polynomial, we multiply the monomial by each term in the polynomial then add up the results.
We multiply the polynomial M = 2x 3 – 5x 2 + 4x – 3 by 3 and by N = 4x 3: M → 2x 3 – 5x 2 + 4x – 3 M → 2x 3 – 5x 2 + 4x – 3 × 3 → × 3 × N →
We can also multiply directly: 3 · (2x3 – 5x2 + 4x – 3) = 3
1 Remove parentheses and reduce.
a) (x 4 + 2x 3 + 5x 2 – 3x) + (4x 3 – 9x 2 + 7x – 1)
b) (5x 4 – 5x 2 – 3x) – (x 3 + 3x 2 + 6x – 11)
2 Perform.
a) 2 · (3x 2 – 4x) b) –5 · (x 3 – 3x) c) x · (–2x + 3) d) x 2 · (x 2 – x + 1)
3 Find the following products: a) 3x · (2x + y + 1) b) 3a · (a 2 + 2a 4) c) ab 2 · (a – b ) d) –5x 3 · (3x 2 + 7x + 11) e) x 2y · (2x – y + 2) f ) 7x 2y · (3x + y)
4 Calculate the polynomial P in each case. a) 2 · P = 6x 3 – 4x 2 – 8x + 2 b) x · P = x 3 – 3x 2 – 5x
The product of two polynomials
We multiply each monomial of one of the factors by each and every one of the monomials of the other factor. Then, we add up any similar monomials.
Another way to multiply
2x 5 – 8x 3 – 10x 2 ← Product of P and 2x 2
2x 5 + x 4 – 11x 3 – 14x 2 + 7x + 15 ← P · Q
When there are not many terms, we do not need to use the method above. We can simply calculate the product directly: (5x 2 – 2) · (2x – 3) = 10x 3 – 15x 2 – 4x + 6
–14
Dividing polynomials
Dividing polynomials is similar to integer division of natural numbers.
P (x) = 2x 3 – 7x 2 – 11x + 13 Q (x) = 2x + 3 P (x) : Q (x) 2x 3 – 7x 2 – 11x + 13 | 2x + 3 – 2x 3 – 3x 2 x 2 – 5x + 2 – 10x 2 – 11x + 13 10x 2 + 15x 4x + 13 – 4x – 6 7 (2x 3) : (2x ) = x 2 (–10x 2) : (2x ) = –5x (4x ) : (2x ) = 2
Then: 2x 3 – 7x 2 – 11x + 13 = (x 2 – 5x + 2) + 7
5 Given the polynomials P = 5 x 2 – 3, Q = x 2 – 4 x + 1, R = –5 x + 2, calculate.
a) P R b) Q · R c) P · Q
6 Operate and simplify.
a) 3x 2 (2x 3 – 1) + 6 (4x 2 – 3) b) (x – 3)(x 2 + 1) – x 2 (2x 3 + 5x 2)
c) P(x) = 6x3 + 2x2 + 18x + 3 Q(x)= x x 3 – 4x – 5 ← P × 2x 2 + x – 3 ← Q – 3x 3 + 12x + 15 ← Product of P and –3 x 4 – 4x 2 – 5x ← Product of P and x
Remember
Some situations involving polynomials: If when dividing two polynomials P (x) : Q (x) the remainder is 0, we say that the division is exact and that:
• P (x) is divisible by Q (x).
• P (x) is a multiple of Q (x).
7 Perform P(x) : Q(x) in each case and express the result of this signature: P(x) = Q(x) · quotient + remainder a) P(x) = 3x2 – 11x + 5 Q(x) = x + 6
b) P(x) = 6x3 + 2x2 + 18x + 3 Q(x) = 3x + 1
Division of a polynomial
by (x – a)
The division: (2x 4 – 7x 3 + 11x – 2) : (x – 3) can be carried out, synthetically, as follows:
+ 3
quotient: 2 –1 –3 2 → means: 2x 3 – x 2 – 3x + 2
remainder: 4
This method, in which only the coefficients are involved and only the operations that really matter are performed, is called Ruffini's rule after the Italian mathematician who developed it.
Value of a polynomial, P (x), for x = a
Returning to the polynomial that appears as the dividend in the previous example, P (x) = 2x 4 – 7x 3 + 11x – 2, note that the value it takes for x = 3 coincides with the remainder of its division by (x – 3):
P (3) = 2 · 34 – 7 · 33 + 11 · 3 – 2 = 162 – 189 + 33 – 2 = 4
Remainder of the division P (x) : (x – 3) → 4
The coincidence of these values is not accidental, and its justification is simple. If we look at the relationship between the terms of the division, D = d · C + R, we see that:
P (x) = (x – 3) · C (x) + 4 0 P (3) = (3 – 3) · C (3) + 4 = 4
And the above example can be generalised to any division of a polynomial by (x – a), as you can see at left.
The value of a polynomial for x = a coincides with the remainder obtained by dividing by (x – a).
1 Calculate the quotient and remainder in each case:
a) (x 3 – 7x 2 + 9x – 3) : (x – 5)
b) (2x 3 + 7x 2 + 2x + 4) : (x + 3)
Observe
The notation P (a) means: value of the polynomial P for x = a
P (x) x – a Q (x) R
P (x) = (x – a) · Q (x) + R
P (a) = (a – a) · Q (a) + R = = 0 · Q (a) + R = R
2 Let the polynomial M ( x ) = x 4 – 8 x 3 + 15 x 2 + 7 x + 8.
a) Calculate M (4) = 44 – 8 · 43 +15 · 42 + 7 · 4 + 8.
b) Divide, using Ruffini's rule, M (x) : (x – 4).
Roots of a polynomial 3
Let us now consider exact divisions of the type P(x) : (x – a).
Since they are exact, the remainder is zero and, according to the previous section, the value taken by the polynomial P (x) for x = a will also be zero: P (x) = (
We divide the polynomial P (
by (x – 2): P (2) = 23 + 3 · 22 – 6 ·
1 3 – 6 – 8 2 2 10 8 1 5 4 0 → R C (x) = x 2 + 5x + 4 remainder of the division → 0
We then say that the value x = 2 is a root of the polynomial.
If the division P (x) : (x – a) is exact, then the polynomial P (x) cancels out for x = a
We say that a is a root of the polynomial P (x) if P (a) = 0.
Finding the integer roots of a polynomial P (x)
Notice that in the exact division above, the last product, 2 · 4, changed sign, is equal to the last coefficient of the polynomial, –8. Thus, the root, 2, is a divisor of 8.
The division (x 3 – 4x 2 – 11x + 30) : (x – 5) is exact and therefore the value x = 5 is a root of the polynomial x 3 – 4x 2 – 11x + 30. 1 – 4 –11 30 5 5
Applying Ruffini's rule, the last product is 5 – (–6) = –30, the opposite of the last coefficient of the polynomial. Exact division
We see that x = a cannot be an integer root of the polynomial in the example if a is not a divisor of 30.
An integer root of a polynomial with integer coefficients is necessarily a divisor of the independent term, the one that does not carry x A divisor of the independent term of the polynomial may or may not be a root of the polynomial.
Observe
A root, a, of a polynomial allows it to be broken down into two factors:
P (x) = (x – a) · C (x)
1 Find out if any of the values 1, –3, 5, –7 eis the root of the polynomial x 4 – 4x 3 + 2x 2 + 5x – 4.
2 What are the roots of the polynomial ( x – 2)( x + 5)( x – 6)?
3 Write a cubic polynomial whose roots are 2, –2 and 3.
4 Find an integer root of each of these polynomials. If there isn't one, explain why.
A(x) = 4x 3 + 2x 2 + 5x + 7
B(x) = x 3 + 2x 2 + 3x + 1
C(x) = x 4 – 2x 3 – x 2 – 7x + 3
D(x) = x 5 + x 4 + x 3 + x 2 + x + 1
Factorising polynomials
To factor a polynomial is to decompose it into a product of polynomials (factors) of the smallest possible degree.
Take out a common factor
• P (x) = 6x 4 – 9x 3 + 12x 2 – 3x → The common factor is 3x.
P (x) = 6x 4 – 9x 3 + 12x 2 – 3x = 3x (2x 3 – 3x 2 + 4x – 1)
• Q (x) = 45x 5 + 120x 3 + 80x 2 → The common factor is 5x 2 .
Q (x) = 45x 5 + 120x 3 + 80x 2 = 5x 2 (9x 3 + 24x + 16)
Identify notable products
• A (x) = 4x 2 + 4x + 1 → Can be expressed as a square of a sum.
A (x) = 4x 2 + 4x + 1 = (2x)2 + 2 · (2x) · (1) + (1)2 = (2x + 1)2
• B (x) = 9x 2 – 12x + 4 → Can be expressed as a square of a difference.
B (x) = 9x 2 – 12x + 4 = (3x)2 – 2 · (3x) · (2) + (2)2 = (3x – 2)2
• C (x) = 4x 2 – 25 → Can be expressed as a product of a sum and a subtraction.
C (x) = 4x 2 – 25 = (2x)2 – (5)2 = (2x + 5)(2x – 5)
Use Ruffini's rule
A (x) = x 4 – 2x 3– 2x 2– 2x – 3 → We apply Ruffini’s rule, trying with the divisors of 3 (+1, –1, +3, –3).
We see that the division of the polynomial A (x) by (x + 1) is exact, which allows us to perform a preliminary decomposition:
A (x) = x 4 – 2x 3– 2x 2– 2x – 3 = (x + 1) · (x 3 – 3x 2 + x – 3)
We follow the same process with the polynomial C (x) = x 3 – 3x 2 + x – 3, and see that the division by (x – 3) is exact:
C (x) = x 3 – 3x 2 + x – 3 = (x – 3) · (x 2 + 1)
A (x) = (x + 1) · (x 3 – 3x 2 + x – 3) = (x + 1) · (x – 3) · (x 2 + 1)
The last factor, (x 2 + 1), cannot be decomposed, as there is no value of x that cancels it out. In other words, it has no roots.
Remember
I. (a + b)2 = a 2 + b 2 + 2ab
II. (a – b)2 = a 2 + b 2 – 2ab
III. a + b) · (a – b) = a 2 – b 2
Ruffini’s rule
1 Factorise by taking out a common factor and using notable products.
a) x3 + 6x 2 + 9x b) 2x 3 – 4x 2 + 2x c) 3x 4 – 12x 2 d) 8x 5 – 24x 4 + 18x 3
2 Writing. Factorise these polynomials using Ruffini’s rule:
a) x3 – 6x2 + 11x – 6 b) 2x3 + 6x2 – x – 30 c) x3 + 7x2 + 14x + 8 d) 3x5 + x2 – 24x + 36
Preparing to solve equations 5
First-degree expressions
Look how we simplify the following expressions:
a) 3(5x – 7) + 2(x – 1) – 5x + 3 = we remove the parentheses
=15x – 21 + 2x – 2 – 5x + 3 = we group the terms
= 12x – 20
b) 3 2( –4) –2 2––6 5 xy xy + = we multiply by 6 = () () xy xy 6 3 24 2 62 6 6 5 ––––+ eo = we remove the denominators
= 2 · 2(x – y + 4) – 3(2x – y) – 5 = we remove the parentheses
= 4x – 4y + 16 – 6x + 3y – 5 = we group the terms
= –2x – y + 11
Second-degree expressions
Look how we simplify the following expression:
(x + 5)2 – 2(x + 1)(x – 3) = we do the multiplications
= x 2 + 10x + 25 – 2(x 2 – 2x – 3) = we remove the parentheses
= x 2 + 10x + 25 – 2x 2 + 4x + 6 = we group the terms
= –x 2 + 14x + 31
1 Simplify the following expressions:
a) 3(x – 1) + 5(x – 2) – 7x
b) 2(2x – 3) + 1 – (x – 5)
c) 5x + 3(1 – x) – 12 – 2(x – 5)
2 Multiply by the number indicated and simplify: a)
3 Simplify the following expressions:
a) (x – 1)(x + 1) + (x – 2)2 – 3
b) (x + 2)(x – 3) + x – 3
c) (x + 1)2 – 2x(x + 2) + 14
4 Multiply by the number indicated and simplify: a) x(2x + 1) –
TAKE ACTION!
The final challenge
polYNOMIALS
Working with polynomials will enable you to tackle ‘The final challenge’, in which we will ask you to invent some problem statements and translate other statements into algebraic language.
In Mara's village, a flea market is organised every year at the end of December. In its numerous stalls we can find all kinds of handmade articles and organic food products.
1 Read, express algebraically and simplify.
a) This year the number of posts has increased by 8% compared to last year.
b) The number of craft and food stalls, given that there are 6 fewer food stalls than craft stalls.
c) The area of the rectangular market square, given that its dimensions (width and length) add up to 100 m.
d) The area of each stand if they are rectangles with 10 m perimeters.
2 Mara and her family are beekeepers. They have a stall at the market where they sell jars of organic honey and products made from this honey.
To pack the jars of honey, they made a box without a lid out of cardboard 5 dm long and 4 dm wide. To do this, they cut out a square of side x at each corner.
a) Write the area of the base of the box and its volume as a function of x.
b) If the area of the base of the box, as a function of x, were 4x 2 – 20x + 25, how long would the sides of the carton be?
3 At Mara's stall they sell honey in jars of three sizes. The medium jar costs � 2 more than the small one, and the large jar costs � 3 more than the medium one.
a) Express the price of each jar as a function of the price of the small jar.
b) Mara makes a profit of 25% on the sale of each small jar, 20% on the sale of a medium-sized jar and 15% on the sale of a large jar. Express algebraically what she will earn from the sale of the honey.
product
Think about what you would like to sell at a market like the one organised every year in Mara's village. Then, invent problem statements related to your stall and the products you sell so that your classmates can express them algebraically.
NATURAL NUMBERS, INTEGERS AND FRACTIONS
Natural numbers
The numbers we use to count objects are called natural numbers.
N= {0, 1, 2, 3, …, 10, 11, …, 100, 101, …}
Representation
The infinite set of natural numbers can be represented in order on a number line:
1 Count: How many matches are there in a basketball championship in which there are 10 teams and the matches are double-headers?
Integers
Negative integers, together with natural numbers, form the set of integers.
Z= {…, –11, –10, …, –3, –2, –1, 0, 1, 2, 3, …, 10, 11, …}
Representation
The integers can be represented in order on a number line:
Absolute value
The absolute value of a number is its magnitude without its sign.
Graphically, the absolute value of a number is its distance from zero:
Operations
2 Complete: ‘The absolute value of a natural number is equal to...’.
Fractions
Fractions are numerical expressions of fractional numbers.
• Proper fraction: a fraction in which the numerator is smaller than the denominator.
• Mixed number: expression of a fractional number as the sum of an integer and a proper fraction.
Fractional numbers, together with integers, form the set of rational numbers: Q
Representation
Fractional numbers can be represented on the number line together with integers:
Simplification
Simplifying a fraction means dividing the numerator and the denominator by the same number.
We call a fraction that cannot be simplified irreducible.
Equivalent fractions
Two fractions are equivalent if they result in the same irreducible fraction when simplified. 35 –10 and 28 8 –are equivalent
3 Write three fractions equivalent to 15 125 . What is the corresponding irreducible fraction?
POLYNOMIALS 4
A monomial is the product of a number multiplied by one or more letters (variables).
coefficient literal part
1 Complete the statement: ‘A number can be considered a monomial of degree ... because ... = ...’
Polynomials
A polynomial is the sum of two or more monomials. Each of the monomials that form a polynomial is called a term.
Adding and subtracting polynomials
The product of a monomial and a polynomial
Notable products
I. (a + b)2 = a 2 + 2ab + b 2 square of a sum
II. (a – b)2 = a 2 – 2ab + b 2 square of a difference
III. (a + b) · (a – b) = a 2 – b 2 difference of squares
The product of two polynomials
2 What is the degree of the polynomial PQ if P and Q have degree m and n respectively?
Dividing polynomials
This is similar to division with natural numbers.
Dividing polynomials
7x 4 – 11x 3 – 94x + 7 x – 3 – 7x 4 + 21x 3 7x 3 + 10x 2 + 30x – 4 10x 3 – 10x 3 + 30x 2 (7x 4) : x = 7x 3 30x 2 (10x 3) : x = 10x 2 – 30x 2 + 90x (30x 2) : x = 30x
(x)
Ruffini’s rule
Ruffini’s rule can only be used when the divisor is of the type (x – a).
(7x 4 – 11 x 3 – 94 x + 7) : (x – 3)
(x)
3 What is the remainder of the division P(x) : (x – a)?
Roots of a polynomial
If the division of P(x) by x – a is exact, in other words, P(a) = 0, then P(x) can be expressed as a product: P (x) x – a
Q (x) P(x
In this case, we say that a is a root of P(x)
Factorising polynomials
4 Complete: ‘A …root of a polynomial with integer coefficients is necessarily a divisor of ...’.
A polynomial has been factorised when it is broken down into the product of other polynomials (factors) with the lowest degree possible. To factorise polynomials we can:
Take out a common factor
P (x) = 6x 4 – 9x 3 + 12x 2 – 3x
Use Ruffini’s rule
The common factor is 3x.
P (x) = 6x 4 – 9x 3 + 12x 2 – 3x = 3x (2x 3 – 3x 2 + 4x – 1)
Identify notable products
A (x) = 4x 2 + 4x + 1
Can be expressed as a square of a sum.
A (x) = 4x 2 + 4x + 1 = (2x)2 + 2 · (2x) · (1) + (1)2 = (2x – 1)2
5 True or false?: ‘A polynomial without roots cannot be factored’.
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