diff --git a/content/math/algebraic-fractions-formulas/_index.md b/content/math/algebraic-fractions-formulas/_index.md new file mode 100644 index 0000000..5ccaa0d --- /dev/null +++ b/content/math/algebraic-fractions-formulas/_index.md @@ -0,0 +1,437 @@ ++++ +title = "Algebraic fractions & formulas" +description = "Simplify and combine algebraic fractions; rearrange formulas in science contexts." +template = "chapter.html" +weight = 2 +[extra] +subject = "math" +chapter_id = "algebraic-fractions-formulas" +strand = "Number & Algebra" ++++ + +
+ An algebraic fraction is a fraction whose numerator and denominator are + polynomials (or monomials). The same ideas you use for numbers — simplify, find a common + denominator, and combine — apply here, with extra care: you may only cancel + common factors, and you must note values of the variable that would make a + denominator zero. +
+ ++ This chapter builds on expanding and factorising polynomials. If you have not studied those + topics yet, use the callout below as a minimal toolkit. +
+ +Before you start
++ To simplify, factorise the numerator and denominator completely, then divide out any common + factors. +
+ +Rule
++ \[ \frac{ac}{bc} = \frac{a}{b} \quad (b \neq 0,\; c \neq 0) \] +
+Here \(c\) is a common factor of the top and bottom, not a term added to each.
+Example. Simplify \(\dfrac{x^2 - 1}{x^2 + x}\).
++ Factorise: \(x^2 - 1 = (x-1)(x+1)\) and \(x^2 + x = x(x+1)\). Cancel the common factor + \((x+1)\): + \[ \frac{(x-1)(x+1)}{x(x+1)} = \frac{x-1}{x}, \quad x \neq 0,\; x \neq -1. \] + The restrictions matter: at \(x = -1\) the original denominator is zero. +
+ +Example. Simplify \(\dfrac{2x^2 - 8}{x^2 - 4}\).
++ \[ \frac{2(x^2-4)}{x^2-4} = \frac{2(x-2)(x+2)}{(x-2)(x+2)} = 2, \quad x \neq 2,\; x \neq -2. \] +
+ ++ Use a common denominator (the LCM of the denominators), as with numeric + fractions. +
+ +Example. \(\dfrac{2}{x} + \dfrac{3}{x+1}\).
++ LCD is \(x(x+1)\): + \[ \frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1) + 3x}{x(x+1)} = \frac{5x+2}{x(x+1)}, \quad x \neq 0,\; x \neq -1. \] +
+ +Example. \(\dfrac{1}{x-2} - \dfrac{3}{x^2 - 4}\).
++ Note \(x^2 - 4 = (x-2)(x+2)\). With LCD \((x-2)(x+2)\): + \[ \frac{x+2}{(x-2)(x+2)} - \frac{3}{(x-2)(x+2)} = \frac{x-1}{(x-2)(x+2)}, \quad x \neq 2,\; x \neq -2. \] +
+ ++ Multiply numerators and denominators; factorise before cancelling. To divide, multiply by the + reciprocal of the divisor. +
+ +Divide
++ \[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} \] +
+Example. \(\dfrac{x^2 - 9}{2x} \times \dfrac{4}{x+3}\).
++ \[ \frac{(x-3)(x+3)}{2x} \times \frac{4}{x+3} = \frac{4(x-3)}{2x} = \frac{2(x-3)}{x}, \quad x \neq 0,\; x \neq -3. \] +
+ +Example. \(\dfrac{x^2 - 1}{x+1} \div \dfrac{x-1}{2}\).
++ \[ \frac{(x-1)(x+1)}{x+1} \times \frac{2}{x-1} = 2, \quad x \neq -1,\; x \neq 1. \] +
+ ++ These use the same rules; the algebra is longer. Work factorisation and the LCD carefully. +
+ +Example. \(\dfrac{1}{x} + \dfrac{1}{x+1}\) (already seen) gives + \(\dfrac{5x+2}{x(x+1)}\).
+ +Example. Simplify \(\dfrac{x}{x^2 - 5x + 6}\).
++ Factorise \(x^2 - 5x + 6 = (x-2)(x-3)\): + \[ \frac{x}{(x-2)(x-3)}, \quad x \neq 2,\; x \neq 3. \] +
+ ++ A formula links several quantities. Changing the subject means + rewriting the formula so that one chosen letter stands alone on one side — like solving an + equation for that letter. +
++ Undo operations in reverse order to how they build the right-hand side: if the + target variable is multiplied, divide; if it is inside a bracket, expand or divide through; + if it appears in a denominator, multiply both sides by that denominator. +
+ +Example. Make \(t\) the subject of \(v = \dfrac{d}{t}\) (\(t \neq 0\)).
++ Multiply both sides by \(t\): \(vt = d\). Divide by \(v\) (assuming \(v \neq 0\)): + \[ t = \frac{d}{v}. \] +
+ +Example. Make \(A\) the subject of \(P = \dfrac{F}{A}\) (\(A \neq 0\)).
++ Multiply by \(A\): \(PA = F\), so \(A = \dfrac{F}{P}\) (\(P \neq 0\)). +
+ +Example. Make \(R\) the subject of \(V = IR\).
++ Divide by \(I\) (assuming \(I \neq 0\)): \(R = \dfrac{V}{I}\). +
+ +| Formula | +Meaning (typical units) | +
|---|---|
| \(v = d/t\) | +speed = distance ÷ time | +
| \(P = F/A\) | +pressure = force ÷ area | +
| \(V = IR\) | +voltage = current × resistance | +
| \(\rho = m/V\) | +density = mass ÷ volume | +
+ For centuries, algebra was written in words. In the 16th and 17th centuries, mathematicians such + as François Viète began to use letters systematically for both unknowns and + known parameters. That made formulas — general relationships such as + \(v = d/t\) — easy to rearrange without repeating long verbal rules. +
++ Rational expressions (fractions built from polynomials) appear whenever you divide one polynomial + by another. Scientists from Galileo onward used ratios and proportionalities in + motion; later, Newton’s laws and electrical laws (e.g. \(V = IR\)) relied on + changing the subject of a formula to solve for the quantity you can measure. +
++ Today, algebraic fractions show up in simplifying models, in calculus limits, and whenever a + relationship is written as one quantity divided by another. +
+ ++ If \(c \neq 0\), then \(\dfrac{ac}{bc} = \dfrac{a}{b}\) because multiplying numerator and + denominator by the same non-zero number does not change the value of a numeric fraction; the + same idea holds when \(c\) is a polynomial factor shared by top and bottom. +
++ You cannot cancel in \(\dfrac{a+c}{b+c}\) in general — that would mean, for example, + \(\dfrac{1+2}{3+2} = \dfrac{1}{3}\), which is false. Only factors, not + terms, may be cancelled. +
+ ++ A formula like \(v = d/t\) builds \(v\) by first dividing \(d\) by \(t\). To recover \(t\), + undo division by multiplying: \(vt = d\), then undo multiplication by \(v\) (when \(v \neq 0\)) + to get \(t = d/v\). Each step reverses one operation that was applied to the subject you want. +
+ ++ Pause and reason before continuing. Discuss with a partner or write your work in a notebook. +
+ +Checkpoint
++ Simplify \(\dfrac{x^2 - 4}{x + 2}\). What value of \(x\) must be excluded? +
+Checkpoint
++ Can you cancel the \(x\) in \(\dfrac{x + 1}{x}\)? Explain briefly. +
+Checkpoint
++ Write \(\dfrac{3}{x} - \dfrac{1}{2x}\) as a single fraction in simplest form. +
+Checkpoint
++ Make \(d\) the subject of \(v = \dfrac{d}{t}\). State any conditions on \(t\) and \(v\). +
+Checkpoint
++ A wire obeys \(V = IR\). If voltage and resistance are known, which formula gives current? +
+Checkpoint
++ Simplify \(\dfrac{1}{x} + \dfrac{1}{x+1}\). What is the common denominator? +
++ For constant speed, distance \(d\), speed \(v\), and time \(t\) are related by \(v = d/t\). If a + train travels \(240\) km in \(3\) h, then \(v = 240/3 = 80\) km/h. If you know \(v\) and \(t\) + but need distance, rearrange to \(d = vt\). +
+ + + ++ Density \(\rho\) (rho) is mass per unit volume: \(\rho = m/V\). A block of mass \(2.7\) kg with + volume \(0.001\) m³ has \(\rho = 2.7/0.001 = 2700\) kg/m³. To find mass from \(\rho\) and \(V\), + use \(m = \rho V\). +
+ ++ Pressure \(P = F/A\) links force \(F\) and area \(A\). Hydraulic and weather models often solve + for \(F = PA\). In a resistor, \(V = IR\) gives voltage from current and resistance; rearrange + to \(I = V/R\) when you know voltage and resistance. +
+ + + ++ Work each question, then open the solution to check your reasoning. Difficulty increases + through the sets. +
+ +Answer: \(2\), with \(x \neq 0\).
+Steps: Cancel the common factor \(3x\).
+Answer: \(x\).
+Explanation: Cancel one factor \(x\) from numerator and denominator.
+Answer: \(d = vt\) (\(t \neq 0\) when rearranging from the original).
+Steps: Multiply both sides by \(t\).
+Answer: \(R = V/I\) (\(I \neq 0\)).
+Steps: Divide both sides by \(I\).
+Answer: \(x - 3\), with \(x \neq -3\).
+Steps: \(x^2 - 9 = (x-3)(x+3)\); cancel \((x+3)\).
+Answer: \(\dfrac{3}{x}\), \(x \neq 0\).
+Explanation: Same denominator already.
+Answer: \(\dfrac{x-1}{x}\), \(x \neq 0,\; x \neq -1\).
+Steps: Factorise; cancel \((x+1)\).
+Answer: \(m = \rho V\) (\(V \neq 0\) in the original).
+Steps: Multiply both sides by \(V\).
+Answer: \(\dfrac{3x+1}{x(x+1)}\), \(x \neq 0,\; x \neq -1\).
+Steps: LCD \(x(x+1)\); numerator \( (x+1) + 2x = 3x+1\).
+Answer: \(\dfrac{2}{x}\), \(x \neq 0\).
+Steps: Multiply; cancel a factor \(x\).
+Answer: \(\dfrac{x-1}{(x-2)(x+2)}\), \(x \neq 2,\; x \neq -2\).
+Steps: Write \(x^2-4 = (x-2)(x+2)\); common denominator \((x-2)(x+2)\).
+Answer: \(2\), with \(x \neq -1,\; x \neq 1\).
+Steps: Factorise \(x^2-1\); multiply by reciprocal; cancel \((x-1)\) and \((x+1)\).
+Answer: \(t = d/v = 2\) h.
+Steps: \(vt = d \Rightarrow t = d/v\); substitute values.
+Answer: \(2\), \(x \neq 2,\; x \neq -2\).
+Steps: Factorise numerator \(2(x^2-4)\) and denominator \((x-2)(x+2)\); cancel \((x-2)(x+2)\).
+Answer: \(P = 120/0.05 = 2400\) Pa; \(F = PA = 2400 \times 0.05 = 120\) N.
+Explanation: Rearranging checks that the formula is consistent.
++ An algebraic fraction is a fraction whose numerator and denominator are + polynomials (or monomials). The same ideas you use for numbers — simplify, find a common + denominator, and combine — apply here, with extra care: you may only cancel + common factors, and you must note values of the variable that would make a + denominator zero. +
+ ++ This chapter builds on expanding and factorising polynomials. If you have not studied those + topics yet, use the callout below as a minimal toolkit. +
+ +Before you start
++ To simplify, factorise the numerator and denominator completely, then divide out any common + factors. +
+ +Rule
++ \[ \frac{ac}{bc} = \frac{a}{b} \quad (b \neq 0,\; c \neq 0) \] +
+Here \(c\) is a common factor of the top and bottom, not a term added to each.
+Example. Simplify \(\dfrac{x^2 - 1}{x^2 + x}\).
++ Factorise: \(x^2 - 1 = (x-1)(x+1)\) and \(x^2 + x = x(x+1)\). Cancel the common factor + \((x+1)\): + \[ \frac{(x-1)(x+1)}{x(x+1)} = \frac{x-1}{x}, \quad x \neq 0,\; x \neq -1. \] + The restrictions matter: at \(x = -1\) the original denominator is zero. +
+ +Example. Simplify \(\dfrac{2x^2 - 8}{x^2 - 4}\).
++ \[ \frac{2(x^2-4)}{x^2-4} = \frac{2(x-2)(x+2)}{(x-2)(x+2)} = 2, \quad x \neq 2,\; x \neq -2. \] +
+ ++ Use a common denominator (the LCM of the denominators), as with numeric + fractions. +
+ +Example. \(\dfrac{2}{x} + \dfrac{3}{x+1}\).
++ LCD is \(x(x+1)\): + \[ \frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1) + 3x}{x(x+1)} = \frac{5x+2}{x(x+1)}, \quad x \neq 0,\; x \neq -1. \] +
+ +Example. \(\dfrac{1}{x-2} - \dfrac{3}{x^2 - 4}\).
++ Note \(x^2 - 4 = (x-2)(x+2)\). With LCD \((x-2)(x+2)\): + \[ \frac{x+2}{(x-2)(x+2)} - \frac{3}{(x-2)(x+2)} = \frac{x-1}{(x-2)(x+2)}, \quad x \neq 2,\; x \neq -2. \] +
+ ++ Multiply numerators and denominators; factorise before cancelling. To divide, multiply by the + reciprocal of the divisor. +
+ +Divide
++ \[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} \] +
+Example. \(\dfrac{x^2 - 9}{2x} \times \dfrac{4}{x+3}\).
++ \[ \frac{(x-3)(x+3)}{2x} \times \frac{4}{x+3} = \frac{4(x-3)}{2x} = \frac{2(x-3)}{x}, \quad x \neq 0,\; x \neq -3. \] +
+ +Example. \(\dfrac{x^2 - 1}{x+1} \div \dfrac{x-1}{2}\).
++ \[ \frac{(x-1)(x+1)}{x+1} \times \frac{2}{x-1} = 2, \quad x \neq -1,\; x \neq 1. \] +
+ ++ These use the same rules; the algebra is longer. Work factorisation and the LCD carefully. +
+ +Example. \(\dfrac{1}{x} + \dfrac{1}{x+1}\) (already seen) gives + \(\dfrac{5x+2}{x(x+1)}\).
+ +Example. Simplify \(\dfrac{x}{x^2 - 5x + 6}\).
++ Factorise \(x^2 - 5x + 6 = (x-2)(x-3)\): + \[ \frac{x}{(x-2)(x-3)}, \quad x \neq 2,\; x \neq 3. \] +
+ ++ A formula links several quantities. Changing the subject means + rewriting the formula so that one chosen letter stands alone on one side — like solving an + equation for that letter. +
++ Undo operations in reverse order to how they build the right-hand side: if the + target variable is multiplied, divide; if it is inside a bracket, expand or divide through; + if it appears in a denominator, multiply both sides by that denominator. +
+ +Example. Make \(t\) the subject of \(v = \dfrac{d}{t}\) (\(t \neq 0\)).
++ Multiply both sides by \(t\): \(vt = d\). Divide by \(v\) (assuming \(v \neq 0\)): + \[ t = \frac{d}{v}. \] +
+ +Example. Make \(A\) the subject of \(P = \dfrac{F}{A}\) (\(A \neq 0\)).
++ Multiply by \(A\): \(PA = F\), so \(A = \dfrac{F}{P}\) (\(P \neq 0\)). +
+ +Example. Make \(R\) the subject of \(V = IR\).
++ Divide by \(I\) (assuming \(I \neq 0\)): \(R = \dfrac{V}{I}\). +
+ +| Formula | +Meaning (typical units) | +
|---|---|
| \(v = d/t\) | +speed = distance ÷ time | +
| \(P = F/A\) | +pressure = force ÷ area | +
| \(V = IR\) | +voltage = current × resistance | +
| \(\rho = m/V\) | +density = mass ÷ volume | +
+ For centuries, algebra was written in words. In the 16th and 17th centuries, mathematicians such + as François Viète began to use letters systematically for both unknowns and + known parameters. That made formulas — general relationships such as + \(v = d/t\) — easy to rearrange without repeating long verbal rules. +
++ Rational expressions (fractions built from polynomials) appear whenever you divide one polynomial + by another. Scientists from Galileo onward used ratios and proportionalities in + motion; later, Newton’s laws and electrical laws (e.g. \(V = IR\)) relied on + changing the subject of a formula to solve for the quantity you can measure. +
++ Today, algebraic fractions show up in simplifying models, in calculus limits, and whenever a + relationship is written as one quantity divided by another. +
+ ++ If \(c \neq 0\), then \(\dfrac{ac}{bc} = \dfrac{a}{b}\) because multiplying numerator and + denominator by the same non-zero number does not change the value of a numeric fraction; the + same idea holds when \(c\) is a polynomial factor shared by top and bottom. +
++ You cannot cancel in \(\dfrac{a+c}{b+c}\) in general — that would mean, for example, + \(\dfrac{1+2}{3+2} = \dfrac{1}{3}\), which is false. Only factors, not + terms, may be cancelled. +
+ ++ A formula like \(v = d/t\) builds \(v\) by first dividing \(d\) by \(t\). To recover \(t\), + undo division by multiplying: \(vt = d\), then undo multiplication by \(v\) (when \(v \neq 0\)) + to get \(t = d/v\). Each step reverses one operation that was applied to the subject you want. +
+ ++ Pause and reason before continuing. Discuss with a partner or write your work in a notebook. +
+ +Checkpoint
++ Simplify \(\dfrac{x^2 - 4}{x + 2}\). What value of \(x\) must be excluded? +
+Checkpoint
++ Can you cancel the \(x\) in \(\dfrac{x + 1}{x}\)? Explain briefly. +
+Checkpoint
++ Write \(\dfrac{3}{x} - \dfrac{1}{2x}\) as a single fraction in simplest form. +
+Checkpoint
++ Make \(d\) the subject of \(v = \dfrac{d}{t}\). State any conditions on \(t\) and \(v\). +
+Checkpoint
++ A wire obeys \(V = IR\). If voltage and resistance are known, which formula gives current? +
+Checkpoint
++ Simplify \(\dfrac{1}{x} + \dfrac{1}{x+1}\). What is the common denominator? +
++ For constant speed, distance \(d\), speed \(v\), and time \(t\) are related by \(v = d/t\). If a + train travels \(240\) km in \(3\) h, then \(v = 240/3 = 80\) km/h. If you know \(v\) and \(t\) + but need distance, rearrange to \(d = vt\). +
+ + + ++ Density \(\rho\) (rho) is mass per unit volume: \(\rho = m/V\). A block of mass \(2.7\) kg with + volume \(0.001\) m³ has \(\rho = 2.7/0.001 = 2700\) kg/m³. To find mass from \(\rho\) and \(V\), + use \(m = \rho V\). +
+ ++ Pressure \(P = F/A\) links force \(F\) and area \(A\). Hydraulic and weather models often solve + for \(F = PA\). In a resistor, \(V = IR\) gives voltage from current and resistance; rearrange + to \(I = V/R\) when you know voltage and resistance. +
+ + + ++ Work each question, then open the solution to check your reasoning. Difficulty increases + through the sets. +
+ +Answer: \(2\), with \(x \neq 0\).
+Steps: Cancel the common factor \(3x\).
+Answer: \(x\).
+Explanation: Cancel one factor \(x\) from numerator and denominator.
+Answer: \(d = vt\) (\(t \neq 0\) when rearranging from the original).
+Steps: Multiply both sides by \(t\).
+Answer: \(R = V/I\) (\(I \neq 0\)).
+Steps: Divide both sides by \(I\).
+Answer: \(x - 3\), with \(x \neq -3\).
+Steps: \(x^2 - 9 = (x-3)(x+3)\); cancel \((x+3)\).
+Answer: \(\dfrac{3}{x}\), \(x \neq 0\).
+Explanation: Same denominator already.
+Answer: \(\dfrac{x-1}{x}\), \(x \neq 0,\; x \neq -1\).
+Steps: Factorise; cancel \((x+1)\).
+Answer: \(m = \rho V\) (\(V \neq 0\) in the original).
+Steps: Multiply both sides by \(V\).
+Answer: \(\dfrac{3x+1}{x(x+1)}\), \(x \neq 0,\; x \neq -1\).
+Steps: LCD \(x(x+1)\); numerator \( (x+1) + 2x = 3x+1\).
+Answer: \(\dfrac{2}{x}\), \(x \neq 0\).
+Steps: Multiply; cancel a factor \(x\).
+Answer: \(\dfrac{x-1}{(x-2)(x+2)}\), \(x \neq 2,\; x \neq -2\).
+Steps: Write \(x^2-4 = (x-2)(x+2)\); common denominator \((x-2)(x+2)\).
+Answer: \(2\), with \(x \neq -1,\; x \neq 1\).
+Steps: Factorise \(x^2-1\); multiply by reciprocal; cancel \((x-1)\) and \((x+1)\).
+Answer: \(t = d/v = 2\) h.
+Steps: \(vt = d \Rightarrow t = d/v\); substitute values.
+Answer: \(2\), \(x \neq 2,\; x \neq -2\).
+Steps: Factorise numerator \(2(x^2-4)\) and denominator \((x-2)(x+2)\); cancel \((x-2)(x+2)\).
+Answer: \(P = 120/0.05 = 2400\) Pa; \(F = PA = 2400 \times 0.05 = 120\) N.
+Explanation: Rearranging checks that the formula is consistent.
++ Result: +
+ ++ Cancel only factors that appear in both numerator and denominator. +
++ Look around you. How many birds? How many cups? You can answer by + counting — pointing at one thing at a time until everything has been pointed at. + This chapter uses pictures of dots to stand for real things, like marbles or + fingers. +
+ ++ To count is to touch (or point at) each object once. Say a word for each touch: + “one, two, three…” The last word tells you how many there are. +
++ Try it on your fingers: one finger up for each thing you counted. If every object has a finger, + the number of fingers matches the number of objects. +
+ ++ When there is nothing to point at, we write 0 and say + zero. An empty plate, no toys in the box — zero dots: +
++ (none) +
+Zero comes before one when we count: first nothing, then one, then two, and so on.
+ +Remember
+Zero means “none here,” not “forgot to count.”
++ Each number has a digit (the symbol), a word, and a + picture of dots — one dot for each thing you counted. +
+| Digit | +Word | +Dots | +
|---|---|---|
| 1 | +one | ++ |
| 2 | +two | ++ |
| 3 | +three | ++ |
| 4 | +four | ++ |
| 5 | +five | ++ |
| 6 | +six | ++ |
| 7 | +seven | ++ |
| 8 | +eight | ++ |
| 9 | +nine | ++ |
| 10 | +ten | ++ |
Which group has more dots — the bigger picture wins:
++ A pattern is when something repeats or grows the same way each time. You can + often guess what comes next if you look at the pictures. +
+ +Each step adds exactly one dot:
++ Even counts split into pairs with nothing left over. We jump two dots at a time: +
++ Odd counts also grow by two dots each step, but there is always + one dot left alone — it does not fit in a pair. Look for the single dot in orange: +
+Patterns can use shapes, not only dots. Watch the pictures repeat:
++ Addition means putting two groups together. Three dots and four more dots make + seven dots in one big group. We write 3 + 4 = 7. +
+ +Adding
+Start with the first group, then place the second group beside it and count all the dots.
++ Multiplication is many equal groups. Three bags with two sweets each + is the same as three pairs of dots. We write 3 × 2 = 6 — “three groups of two.” +
++ You can lay the dots in a rectangle: 3 rows with 2 dots in each row fills a + block of 6 dots. +
+ ++ Division is about sharing equally or making equal groups. If + 12 biscuits are shared among 4 friends, each friend gets the same number. We write + 12 ÷ 4 = 3 — “twelve split into four equal piles gives three in each pile.” +
++ Division is the partner of multiplication. If 3 × 4 = 12, then + 12 ÷ 3 = 4 and 12 ÷ 4 = 3. One picture of dots can answer both + kinds of question. +
+ +Fair shares
++ When every pile has the same number of dots, the share is fair. If dots are + left over, the total does not split evenly — those extras are the remainder. +
++ You can count higher than ten on your fingers — people use place value. Ten + single dots can be thought of as one group of ten. Ten groups of ten make + one hundred. +
++ The number 47 means 4 tens and 7 ones — forty-seven dots, without drawing every + dot one by one. +
+ ++ The world is huge. Counting every grain of sand on a beach one by one would take forever, so we + use big number words to talk about amounts we cannot easily picture dot by dot. + They matter because governments, scientists, and shops all work with large totals. +
+ +| Word | +How we write it | +Why it matters | +
|---|---|---|
| hundred | +100 | +About how many people fit in a big sports field; pages in a long novel. | +
| thousand | +1,000 | +Steps you might walk in a long hike; students in a large school district. | +
| million | +1,000,000 | +People in a large city; dollars in a national budget line item. | +
| billion | +1,000,000,000 | +World population is several billion; stars in our galaxy are hundreds of billions. | +
+ Each big word is one thousand times the word before (ten hundreds make a + thousand; one thousand thousands make a million). We bundle in tens and thousands so we can + compare and add without losing track. +
+ ++ Use the tools below: count dots, add and multiply, share equally, and zoom from + one dot up to a billion to see how big numbers compare. +
++ Long before writing, people kept track of herds, harvests, and trade by making marks — tallies — + one mark for each item. Archaeologists have found tally bones tens of thousands of years old. +
++ Different cultures invented symbols for amounts. The idea of zero as a number took + time: Indian mathematicians such as Brahmagupta (7th century CE) wrote rules for + arithmetic that included zero. The Maya used a shell sign for zero in their calendar. Zero let + people write “none” clearly and build place-value systems later on. +
++ Multiplication grew from repeated addition in trade and land measurement — counting equal rows of + goods or fields without adding one row at a time forever. +
++ Division appears in sharing problems as old as fair trade: split grain, land, or money into equal + parts. Large-number names (million, billion) expanded as empires, astronomy, and finance needed to + record bigger totals than everyday counting. +
+ ++ Suppose the first group has \(a\) objects and the second has \(b\) objects, with no overlap. If + you put them in one pile and count from the start, you say \(a\) number words for the first + group, then continue with \(b\) more words. The last word is the total \(a + b\). That is why + \(3 + 4 = 7\): three objects, then four more, gives seven in all. +
+ ++ If there are \(r\) equal rows and each row has \(c\) dots, you can add + \(c + c + \cdots + c\) (\(r\) times) to get the total. Writing \(r \times c\) saves repeating + the same addition. For example, \(4 \times 3 = 3 + 3 + 3 + 3 = 12\). +
+ ++ If \(T\) dots are split into \(g\) equal piles with nothing left over, each pile has + \(T \div g\) dots. That is the same as asking: what number times \(g\) gives \(T\)? So + \(T \div g = q\) when \(g \times q = T\). Example: \(12 \div 4 = 3\) because \(4 \times 3 = 12\). +
+ ++ In base ten, each column is worth ten times the column to its right. So \(47 = 4 \times 10 + 7\). + Extending the pattern: \(1{,}000 = 10^3\), \(1{,}000{,}000 = 10^6\), and each “thousand step” multiplies + by \(1{,}000\) again. That is why we group digits in threes when writing millions and billions. +
+ +Pause and answer in your head or with a partner before moving on.
+ +Checkpoint
+Show the number \(7\) on your fingers. What digit symbol matches seven?
+Checkpoint
+Which is greater: \(5\) or \(9\)? How do you know?
+Checkpoint
+Is \(2 + 2 + 2\) the same total as \(3 \times 2\)? Explain with a picture of dots.
+Checkpoint
+What comes next in the pattern \(2, 4, 6, 8, \ldots\)?
+Checkpoint
+Share 15 dots into 3 equal piles. How many in each pile? What if you try 4 piles?
+Checkpoint
+Which is bigger: 1 million or 1 thousand? How do you know?
+Answer: \(5\).
+Answer: \(0\) (zero).
+Answer: \(7\).
+Steps: Two objects plus five more gives seven.
+Answer: \(10\).
+Answer: \(12\).
+Steps: \(4 + 4 + 4 = 12\), or four rows of three dots.
+Answer: \(8\) is larger.
+Answer: \(10\).
+Steps: \(5 + 3 = 8\), then \(8 + 2 = 10\).
+Answer: \(9\).
+Steps: \(3 \times 3 = 9\).
+Answer: \(10\).
+Explanation: Addition of whole numbers can be done in either order (commutative property).
+Answer: \(10\).
+Steps: \(2 + 2 + 2 + 2 + 2 = 10\); five rows of two dots (or two columns of five).
+Answer: \(3\) per pile.
+Steps: \(12 \div 4 = 3\) because \(4 \times 3 = 12\).
+Answer: Six zeros: 1,000,000.
++ Look around you. How many birds? How many cups? You can answer by + counting — pointing at one thing at a time until everything has been pointed at. + This chapter uses pictures of dots to stand for real things, like marbles or + fingers. +
+ ++ To count is to touch (or point at) each object once. Say a word for each touch: + “one, two, three…” The last word tells you how many there are. +
++ Try it on your fingers: one finger up for each thing you counted. If every object has a finger, + the number of fingers matches the number of objects. +
+ ++ When there is nothing to point at, we write 0 and say + zero. An empty plate, no toys in the box — zero dots: +
++ (none) +
+Zero comes before one when we count: first nothing, then one, then two, and so on.
+ +Remember
+Zero means “none here,” not “forgot to count.”
++ Each number has a digit (the symbol), a word, and a + picture of dots — one dot for each thing you counted. +
+| Digit | +Word | +Dots | +
|---|---|---|
| 1 | +one | ++ |
| 2 | +two | ++ |
| 3 | +three | ++ |
| 4 | +four | ++ |
| 5 | +five | ++ |
| 6 | +six | ++ |
| 7 | +seven | ++ |
| 8 | +eight | ++ |
| 9 | +nine | ++ |
| 10 | +ten | ++ |
Which group has more dots — the bigger picture wins:
++ A pattern is when something repeats or grows the same way each time. You can + often guess what comes next if you look at the pictures. +
+ +Each step adds exactly one dot:
++ Even counts split into pairs with nothing left over. We jump two dots at a time: +
++ Odd counts also grow by two dots each step, but there is always + one dot left alone — it does not fit in a pair. Look for the single dot in orange: +
+Patterns can use shapes, not only dots. Watch the pictures repeat:
++ Addition means putting two groups together. Three dots and four more dots make + seven dots in one big group. We write 3 + 4 = 7. +
+ +Adding
+Start with the first group, then place the second group beside it and count all the dots.
++ Multiplication is many equal groups. Three bags with two sweets each + is the same as three pairs of dots. We write 3 × 2 = 6 — “three groups of two.” +
++ You can lay the dots in a rectangle: 3 rows with 2 dots in each row fills a + block of 6 dots. +
+ ++ Division is about sharing equally or making equal groups. If + 12 biscuits are shared among 4 friends, each friend gets the same number. We write + 12 ÷ 4 = 3 — “twelve split into four equal piles gives three in each pile.” +
++ Division is the partner of multiplication. If 3 × 4 = 12, then + 12 ÷ 3 = 4 and 12 ÷ 4 = 3. One picture of dots can answer both + kinds of question. +
+ +Fair shares
++ When every pile has the same number of dots, the share is fair. If dots are + left over, the total does not split evenly — those extras are the remainder. +
++ You can count higher than ten on your fingers — people use place value. Ten + single dots can be thought of as one group of ten. Ten groups of ten make + one hundred. +
++ The number 47 means 4 tens and 7 ones — forty-seven dots, without drawing every + dot one by one. +
+ ++ The world is huge. Counting every grain of sand on a beach one by one would take forever, so we + use big number words to talk about amounts we cannot easily picture dot by dot. + They matter because governments, scientists, and shops all work with large totals. +
+ +| Word | +How we write it | +Why it matters | +
|---|---|---|
| hundred | +100 | +About how many people fit in a big sports field; pages in a long novel. | +
| thousand | +1,000 | +Steps you might walk in a long hike; students in a large school district. | +
| million | +1,000,000 | +People in a large city; dollars in a national budget line item. | +
| billion | +1,000,000,000 | +World population is several billion; stars in our galaxy are hundreds of billions. | +
+ Each big word is one thousand times the word before (ten hundreds make a + thousand; one thousand thousands make a million). We bundle in tens and thousands so we can + compare and add without losing track. +
+ ++ Use the tools below: count dots, add and multiply, share equally, and zoom from + one dot up to a billion to see how big numbers compare. +
++ Long before writing, people kept track of herds, harvests, and trade by making marks — tallies — + one mark for each item. Archaeologists have found tally bones tens of thousands of years old. +
++ Different cultures invented symbols for amounts. The idea of zero as a number took + time: Indian mathematicians such as Brahmagupta (7th century CE) wrote rules for + arithmetic that included zero. The Maya used a shell sign for zero in their calendar. Zero let + people write “none” clearly and build place-value systems later on. +
++ Multiplication grew from repeated addition in trade and land measurement — counting equal rows of + goods or fields without adding one row at a time forever. +
++ Division appears in sharing problems as old as fair trade: split grain, land, or money into equal + parts. Large-number names (million, billion) expanded as empires, astronomy, and finance needed to + record bigger totals than everyday counting. +
+ ++ Suppose the first group has \(a\) objects and the second has \(b\) objects, with no overlap. If + you put them in one pile and count from the start, you say \(a\) number words for the first + group, then continue with \(b\) more words. The last word is the total \(a + b\). That is why + \(3 + 4 = 7\): three objects, then four more, gives seven in all. +
+ ++ If there are \(r\) equal rows and each row has \(c\) dots, you can add + \(c + c + \cdots + c\) (\(r\) times) to get the total. Writing \(r \times c\) saves repeating + the same addition. For example, \(4 \times 3 = 3 + 3 + 3 + 3 = 12\). +
+ ++ If \(T\) dots are split into \(g\) equal piles with nothing left over, each pile has + \(T \div g\) dots. That is the same as asking: what number times \(g\) gives \(T\)? So + \(T \div g = q\) when \(g \times q = T\). Example: \(12 \div 4 = 3\) because \(4 \times 3 = 12\). +
+ ++ In base ten, each column is worth ten times the column to its right. So \(47 = 4 \times 10 + 7\). + Extending the pattern: \(1{,}000 = 10^3\), \(1{,}000{,}000 = 10^6\), and each “thousand step” multiplies + by \(1{,}000\) again. That is why we group digits in threes when writing millions and billions. +
+ +Pause and answer in your head or with a partner before moving on.
+ +Checkpoint
+Show the number \(7\) on your fingers. What digit symbol matches seven?
+Checkpoint
+Which is greater: \(5\) or \(9\)? How do you know?
+Checkpoint
+Is \(2 + 2 + 2\) the same total as \(3 \times 2\)? Explain with a picture of dots.
+Checkpoint
+What comes next in the pattern \(2, 4, 6, 8, \ldots\)?
+Checkpoint
+Share 15 dots into 3 equal piles. How many in each pile? What if you try 4 piles?
+Checkpoint
+Which is bigger: 1 million or 1 thousand? How do you know?
+Answer: \(5\).
+Answer: \(0\) (zero).
+Answer: \(7\).
+Steps: Two objects plus five more gives seven.
+Answer: \(10\).
+Answer: \(12\).
+Steps: \(4 + 4 + 4 = 12\), or four rows of three dots.
+Answer: \(8\) is larger.
+Answer: \(10\).
+Steps: \(5 + 3 = 8\), then \(8 + 2 = 10\).
+Answer: \(9\).
+Steps: \(3 \times 3 = 9\).
+Answer: \(10\).
+Explanation: Addition of whole numbers can be done in either order (commutative property).
+Answer: \(10\).
+Steps: \(2 + 2 + 2 + 2 + 2 = 10\); five rows of two dots (or two columns of five).
+Answer: \(3\) per pile.
+Steps: \(12 \div 4 = 3\) because \(4 \times 3 = 12\).
+Answer: Six zeros: 1,000,000.
++ 100 squares × 10,000 each = 1 million +
++ Each orange square is a whole million. This board shows 100 of them — imagine + 10 identical boards stacked (10 × 100 = 1,000 + millions) to reach one billion. +
+- Numbers on a line can lie to the right of zero (positive) or to the - left (negative). Temperatures below freezing, bank overdrafts, and positions - below sea level are everyday examples of directed numbers. + You already know how to count and work with whole numbers from 0 to + 10. On a number line, those values lie on or to the right of zero. + Numbers can also lie to the left of zero — we call them negative. + Temperatures below freezing, bank overdrafts, and positions below sea level are everyday examples + of directed numbers.
- Numbers on a line can lie to the right of zero (positive) or to the - left (negative). Temperatures below freezing, bank overdrafts, and positions - below sea level are everyday examples of directed numbers. + You already know how to count and work with whole numbers from 0 to + 10. On a number line, those values lie on or to the right of zero. + Numbers can also lie to the left of zero — we call them negative. + Temperatures below freezing, bank overdrafts, and positions below sea level are everyday examples + of directed numbers.
+ An algebraic expression combines numbers, variables, and operations. A + polynomial in one variable \(x\) is a sum of terms of the form + \(a x^n\) where \(a\) is a constant and \(n\) is a non-negative integer. Examples: + \(3x^2 - 5x + 1\), \(x^3 + 2x\), and \(7\). +
+ ++ In \(5x^2 - 3x + 4\), the terms are \(5x^2\), \(-3x\), and \(4\). The + coefficient of \(x^2\) is \(5\); the coefficient of \(x\) is \(-3\). The + degree of the polynomial is the highest power of \(x\) — here, \(2\). + Like terms have the same variable part (same power of \(x\)); only their + coefficients may differ. +
+ ++ Line up like terms and add or subtract their coefficients: + \[ (3x^2 + 2x - 1) + (x^2 - 5x + 4) = 4x^2 - 3x + 3. \] + When subtracting, distribute the minus sign to every term in the bracket: + \[ (2x^2 + x) - (x^2 - 3x + 2) = 2x^2 + x - x^2 + 3x - 2 = x^2 + 4x - 2. \] +
+ ++ Use the distributive law: each term in the first bracket multiplies each term in + the second. For two binomials, + \[ (a+b)(c+d) = ac + ad + bc + bd. \] + Example: + \[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6. \] +
++ A binomial times a trinomial works the same way — multiply each of the two terms by each of the + three: + \[ (x+1)(x^2 - x + 2) = x^3 - x^2 + 2x + x^2 - x + 2 = x^3 + x + 2. \] +
+ ++ After expanding, combine like terms so the answer is simplified. From above, + \(x^2 - 2x + 3x - 6\) becomes \(x^2 + x - 6\). Constants, \(x\)-terms, and \(x^2\)-terms are + collected separately. +
+ ++ Factorisation is the reverse of expansion: write an expression as a product of + simpler factors. Choose a method by inspecting the shape of the expression: +
+| Situation | +Method | +
|---|---|
| Every term shares a common factor | +Highest common factor (HCF) | +
| Four terms, pairs share factors | +Grouping | +
| Matches \(a^2-b^2\) or \((a\pm b)^2\) | +Standard identities | +
| Quadratic in \(x\): \(ax^2+bx+c\) | +Two-number search or ac-method | +
+ Factor out the largest factor common to every term: + \[ 6x^2 + 9x = 3x(2x + 3), \qquad 4x^3 - 8x^2 = 4x^2(x - 2). \] + Always check whether the bracket can be factorised further. +
+ ++ With four terms, split into two pairs and factor each pair: + \[ ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y). \] + Example: + \[ 2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3). \] +
+ +Identities
++ \[ a^2 - b^2 = (a+b)(a-b) \] + \[ (a+b)^2 = a^2 + 2ab + b^2 \] + \[ (a-b)^2 = a^2 - 2ab + b^2 \] +
++ Recognising these forms speeds up factorisation. For example, + \(x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)\) and + \(x^2 + 6x + 9 = (x+3)^2\). +
+ ++ To factorise a monic quadratic \(x^2 + bx + c\), find two numbers that + multiply to \(c\) and add to \(b\): + \[ x^2 + 5x + 6 = (x+2)(x+3) \quad\text{since } 2\times 3 = 6,\; 2+3 = 5. \] +
++ For a non-monic quadratic \(ax^2 + bx + c\) with \(a \neq 1\), use the + ac-method: find two numbers that multiply to \(ac\) and add to \(b\), split the + middle term, then factor by grouping: + \[ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3). \] +
++ This chapter stops at factorising expressions. Solving equations of the form \(ax^2+bx+c=0\) is + covered in the chapter on quadratic equations. +
+ ++ Use the area model to see how \((a+b)(c+d)\) splits into four partial products. Then practise + choosing a factorisation method and checking your factors in the explorer below. +
++ Long before modern symbols, people solved problems that required multiplying lengths and + collecting areas — the same ideas behind expanding and factorising. Babylonian scribes (around + 1800 BCE) recorded rectangle problems on clay tablets: given area and side relationships, + find unknown lengths. Their recipes are early cousins of “expand, then simplify.” +
+ +
+ + In the 9th century, al-Khwarizmi organised methods for equations and products + in Arabic texts that later spread to Europe. Renaissance mathematicians such as + Viète introduced letters for unknowns and constants, making it easier to see + patterns like \(a^2 - b^2\). By the 17th–18th centuries, polynomial algebra was central to + calculus and science; factorisation became a routine step before solving equations or + simplifying formulas. +
+ ++ Draw a rectangle with width \(a+b\) and height \(c+d\). The total area is \((a+b)(c+d)\). Split + the rectangle into four smaller rectangles with areas \(ac\), \(ad\), \(bc\), and \(bd\). The + sum of partial areas equals the whole: + \[ (a+b)(c+d) = ac + ad + bc + bd. \] + This is the distributive law in two dimensions — the same picture as in the area-model widget + above. +
+ ++ Set \(c = a\) and \(d = b\) in the area formula for a square of side \(a+b\): + \[ (a+b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. \] + For a square of side \(a-b\) (with \(a > b\)), remove a strip of width \(b\) from a square of + side \(a\); the remaining L-shape rearranges to show + \[ (a-b)^2 = a^2 - 2ab + b^2. \] + Difference of squares: a large square of side \(a\) minus a small square of side \(b\) leaves two + congruent rectangles, which rearrange to a rectangle of dimensions \((a+b)\) by \((a-b)\): + \[ a^2 - b^2 = (a+b)(a-b). \] +
+ ++ Factorise \(3x^2 + 12x - 2x - 8\). +
++ Pause and reason before continuing. Discuss with a partner or write your work in a notebook. +
+ +Checkpoint
++ In \(4x^3 - 2x^2 + 5\), what is the degree? Which terms are like terms with \(4x^3\)? +
+Checkpoint
++ Expand \((2x-1)(x+4)\) without skipping steps. How many terms do you get before collecting? +
+Checkpoint
++ How would you factorise \(x^2 - 16\)? Which identity applies? +
+Checkpoint
++ Expand \(x(x+1) - (x+1)\). What common factor appears? What is the factorised form? +
++ A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Its area is + \[ (x+3)(2x-1) = 2x^2 + 5x - 3 \quad\text{cm}^2. \] + The perimeter is \(2\bigl((x+3)+(2x-1)\bigr) = 2(3x+2) = 6x+4\) cm. Expansion links algebraic + descriptions of sides to measurable quantities. +
+ ++ If a shop sells \(n\) items at price \((50 - 0.5n)\) dollars each, revenue is + \(n(50 - 0.5n) = 50n - 0.5n^2\). Expanding a product of two linear factors often appears before + finding maximum revenue (a quadratic in \(n\)). +
+ ++ Notice that \(11^2 = 121\), \(111^2 = 12321\). In general, expanding \((10^k + 10^{k-1} + + \cdots + 1)^2\) explains palindromic squares — a playful link between identities and arithmetic + patterns. +
+ ++ Work each question, then open the solution to check your reasoning. Difficulty increases + through the sets. +
+ +Answer: \(3x + 6\).
+Steps: \(3 \cdot x + 3 \cdot 2 = 3x + 6\).
+Answer: \(7x - 1\).
+Explanation: Combine like terms \(2x + 5x = 7x\).
+Answer: \(3(2x + 3)\).
+Steps: HCF is \(3\); \(6x \div 3 = 2x\), \(9 \div 3 = 3\).
+Answer: \(x^2 + 2x - 8\).
+Steps: \(x^2 - 2x + 4x - 8 = x^2 + 2x - 8\).
+Answer: \((x+3)(x-3)\).
+Explanation: Difference of squares: \(x^2 - 3^2\).
+Answer: \((x+3)^2\).
+Explanation: Perfect square: \(3 \times 3 = 9\), \(3 + 3 = 6\).
+Answer: \((2x-1)(x+3)\).
+Steps: \(2x(x+3) - 1(x+3) = (2x-1)(x+3)\).
+Answer: \((2x+1)(x+3)\).
+Steps: \(ac = 6\); split \(7x = 6x + x\); \(2x(x+3) + 1(x+3) = (2x+1)(x+3)\).
+Answer: \((x+2)(x-1)\) after factorising (expansion gives \(x^2 + x - 2\)).
+Steps: Factor common \((x+2)\): \((x+2)(x-1)\). Check: \((x+2)(x-1) = x^2 + x - 2\).
+Answer: \((2x^2 + 5x - 3)\) cm².
+Calculation: \((x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\).
++ An algebraic expression combines numbers, variables, and operations. A + polynomial in one variable \(x\) is a sum of terms of the form + \(a x^n\) where \(a\) is a constant and \(n\) is a non-negative integer. Examples: + \(3x^2 - 5x + 1\), \(x^3 + 2x\), and \(7\). +
+ ++ In \(5x^2 - 3x + 4\), the terms are \(5x^2\), \(-3x\), and \(4\). The + coefficient of \(x^2\) is \(5\); the coefficient of \(x\) is \(-3\). The + degree of the polynomial is the highest power of \(x\) — here, \(2\). + Like terms have the same variable part (same power of \(x\)); only their + coefficients may differ. +
+ ++ Line up like terms and add or subtract their coefficients: + \[ (3x^2 + 2x - 1) + (x^2 - 5x + 4) = 4x^2 - 3x + 3. \] + When subtracting, distribute the minus sign to every term in the bracket: + \[ (2x^2 + x) - (x^2 - 3x + 2) = 2x^2 + x - x^2 + 3x - 2 = x^2 + 4x - 2. \] +
+ ++ Use the distributive law: each term in the first bracket multiplies each term in + the second. For two binomials, + \[ (a+b)(c+d) = ac + ad + bc + bd. \] + Example: + \[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6. \] +
++ A binomial times a trinomial works the same way — multiply each of the two terms by each of the + three: + \[ (x+1)(x^2 - x + 2) = x^3 - x^2 + 2x + x^2 - x + 2 = x^3 + x + 2. \] +
+ ++ After expanding, combine like terms so the answer is simplified. From above, + \(x^2 - 2x + 3x - 6\) becomes \(x^2 + x - 6\). Constants, \(x\)-terms, and \(x^2\)-terms are + collected separately. +
+ ++ Factorisation is the reverse of expansion: write an expression as a product of + simpler factors. Choose a method by inspecting the shape of the expression: +
+| Situation | +Method | +
|---|---|
| Every term shares a common factor | +Highest common factor (HCF) | +
| Four terms, pairs share factors | +Grouping | +
| Matches \(a^2-b^2\) or \((a\pm b)^2\) | +Standard identities | +
| Quadratic in \(x\): \(ax^2+bx+c\) | +Two-number search or ac-method | +
+ Factor out the largest factor common to every term: + \[ 6x^2 + 9x = 3x(2x + 3), \qquad 4x^3 - 8x^2 = 4x^2(x - 2). \] + Always check whether the bracket can be factorised further. +
+ ++ With four terms, split into two pairs and factor each pair: + \[ ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y). \] + Example: + \[ 2x^2 + 6x - x - 3 = 2x(x+3) - 1(x+3) = (2x-1)(x+3). \] +
+ +Identities
++ \[ a^2 - b^2 = (a+b)(a-b) \] + \[ (a+b)^2 = a^2 + 2ab + b^2 \] + \[ (a-b)^2 = a^2 - 2ab + b^2 \] +
++ Recognising these forms speeds up factorisation. For example, + \(x^2 - 9 = x^2 - 3^2 = (x+3)(x-3)\) and + \(x^2 + 6x + 9 = (x+3)^2\). +
+ ++ To factorise a monic quadratic \(x^2 + bx + c\), find two numbers that + multiply to \(c\) and add to \(b\): + \[ x^2 + 5x + 6 = (x+2)(x+3) \quad\text{since } 2\times 3 = 6,\; 2+3 = 5. \] +
++ For a non-monic quadratic \(ax^2 + bx + c\) with \(a \neq 1\), use the + ac-method: find two numbers that multiply to \(ac\) and add to \(b\), split the + middle term, then factor by grouping: + \[ 2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3). \] +
++ This chapter stops at factorising expressions. Solving equations of the form \(ax^2+bx+c=0\) is + covered in the chapter on quadratic equations. +
+ ++ Use the area model to see how \((a+b)(c+d)\) splits into four partial products. Then practise + choosing a factorisation method and checking your factors in the explorer below. +
++ Long before modern symbols, people solved problems that required multiplying lengths and + collecting areas — the same ideas behind expanding and factorising. Babylonian scribes (around + 1800 BCE) recorded rectangle problems on clay tablets: given area and side relationships, + find unknown lengths. Their recipes are early cousins of “expand, then simplify.” +
+ +
+ + In the 9th century, al-Khwarizmi organised methods for equations and products + in Arabic texts that later spread to Europe. Renaissance mathematicians such as + Viète introduced letters for unknowns and constants, making it easier to see + patterns like \(a^2 - b^2\). By the 17th–18th centuries, polynomial algebra was central to + calculus and science; factorisation became a routine step before solving equations or + simplifying formulas. +
+ ++ Draw a rectangle with width \(a+b\) and height \(c+d\). The total area is \((a+b)(c+d)\). Split + the rectangle into four smaller rectangles with areas \(ac\), \(ad\), \(bc\), and \(bd\). The + sum of partial areas equals the whole: + \[ (a+b)(c+d) = ac + ad + bc + bd. \] + This is the distributive law in two dimensions — the same picture as in the area-model widget + above. +
+ ++ Set \(c = a\) and \(d = b\) in the area formula for a square of side \(a+b\): + \[ (a+b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2. \] + For a square of side \(a-b\) (with \(a > b\)), remove a strip of width \(b\) from a square of + side \(a\); the remaining L-shape rearranges to show + \[ (a-b)^2 = a^2 - 2ab + b^2. \] + Difference of squares: a large square of side \(a\) minus a small square of side \(b\) leaves two + congruent rectangles, which rearrange to a rectangle of dimensions \((a+b)\) by \((a-b)\): + \[ a^2 - b^2 = (a+b)(a-b). \] +
+ ++ Factorise \(3x^2 + 12x - 2x - 8\). +
++ Pause and reason before continuing. Discuss with a partner or write your work in a notebook. +
+ +Checkpoint
++ In \(4x^3 - 2x^2 + 5\), what is the degree? Which terms are like terms with \(4x^3\)? +
+Checkpoint
++ Expand \((2x-1)(x+4)\) without skipping steps. How many terms do you get before collecting? +
+Checkpoint
++ How would you factorise \(x^2 - 16\)? Which identity applies? +
+Checkpoint
++ Expand \(x(x+1) - (x+1)\). What common factor appears? What is the factorised form? +
++ A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Its area is + \[ (x+3)(2x-1) = 2x^2 + 5x - 3 \quad\text{cm}^2. \] + The perimeter is \(2\bigl((x+3)+(2x-1)\bigr) = 2(3x+2) = 6x+4\) cm. Expansion links algebraic + descriptions of sides to measurable quantities. +
+ ++ If a shop sells \(n\) items at price \((50 - 0.5n)\) dollars each, revenue is + \(n(50 - 0.5n) = 50n - 0.5n^2\). Expanding a product of two linear factors often appears before + finding maximum revenue (a quadratic in \(n\)). +
+ ++ Notice that \(11^2 = 121\), \(111^2 = 12321\). In general, expanding \((10^k + 10^{k-1} + + \cdots + 1)^2\) explains palindromic squares — a playful link between identities and arithmetic + patterns. +
+ ++ Work each question, then open the solution to check your reasoning. Difficulty increases + through the sets. +
+ +Answer: \(3x + 6\).
+Steps: \(3 \cdot x + 3 \cdot 2 = 3x + 6\).
+Answer: \(7x - 1\).
+Explanation: Combine like terms \(2x + 5x = 7x\).
+Answer: \(3(2x + 3)\).
+Steps: HCF is \(3\); \(6x \div 3 = 2x\), \(9 \div 3 = 3\).
+Answer: \(x^2 + 2x - 8\).
+Steps: \(x^2 - 2x + 4x - 8 = x^2 + 2x - 8\).
+Answer: \((x+3)(x-3)\).
+Explanation: Difference of squares: \(x^2 - 3^2\).
+Answer: \((x+3)^2\).
+Explanation: Perfect square: \(3 \times 3 = 9\), \(3 + 3 = 6\).
+Answer: \((2x-1)(x+3)\).
+Steps: \(2x(x+3) - 1(x+3) = (2x-1)(x+3)\).
+Answer: \((2x+1)(x+3)\).
+Steps: \(ac = 6\); split \(7x = 6x + x\); \(2x(x+3) + 1(x+3) = (2x+1)(x+3)\).
+Answer: \((x+2)(x-1)\) after factorising (expansion gives \(x^2 + x - 2\)).
+Steps: Factor common \((x+2)\): \((x+2)(x-1)\). Check: \((x+2)(x-1) = x^2 + x - 2\).
+Answer: \((2x^2 + 5x - 3)\) cm².
+Calculation: \((x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3\).
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