viewof fixedCost = Inputs.range([0, 5000], {
value: 2000,
step: 100,
label: "Fixed cost F (β¬)"
})
viewof variableCost = Inputs.range([0, 100], {
value: 3,
step: 0.5,
label: "Variable cost per unit v (β¬)"
})
viewof price = Inputs.range([0, 200], {
value: 8,
step: 0.5,
label: "Selling price per unit p (β¬)"
})
NoteLearning objectives
- Simplify and rewrite an algebraic expression.
- Reveal a structure (common factor, notable identity).
- Solve first-degree equations in context.
1 Diagnostic Activity
Answer the following questions without a calculator. They check the prerequisite knowledge for this module.
- Factor \(x^2 + 4x\).
- Expand \(2(3x - 5)\).
- Solve \(5x - 8 = 12\).
NoteCheck your answers
- \(x^2 + 4x = x(x + 4)\)
- \(2(3x - 5) = 6x - 10\)
- \(5x - 8 = 12 \Rightarrow 5x = 20 \Rightarrow x = 4\)
Use your results to identify topics needing attention.
2 Opening Problem
A small cafΓ© has fixed monthly costs of β¬2,000 and variable costs of β¬3 per unit sold. Each unit sells at β¬8. How many units must the cafΓ© sell each month to cover all its costs?
Write down your estimate before continuing.
3 Core Concepts
Algebraic Transformations
Algebraic transformations are the set of rules and techniques used to change the form of an algebraic expression without changing its value.
These operations let us simplify, expand, factor, or rewrite expressions in order to:
- make them easier to manipulate;
- reveal a structure (such as a notable identity or a common factor); or
- make it easier to solve equations.
Factoring
Factoring consists in rewriting an expression β for example \(C(x) = x^{2} - 3x\) β in another form that contains a product. On the right-hand side, the variable \(x\) appears in both \(x^{2}\) and \(3x\), so we can use \(x\) as a common factor. The factored form becomes
\[ C(x) = x(x - 3). \]
Factoring makes the behaviour of the function easier to read: we see at once that the function is zero when \(x = 0\) or \(x = 3\).
TipSteps of factoring
- Identify the greatest common factor.
- Divide each term by that factor.
- Rewrite the expression as a product.
Exercises β factor
CautionBasic common factor (click for solutions)
- \(4x + 8\) β GCF \(4\): \(\;4(x + 2)\)
- \(-21x - 14\) β GCF \(-7\): \(\;-7(3x + 2)\)
- \(35x + 49\) β GCF \(7\): \(\;7(5x + 7)\)
- \(144 + 108x\) β GCF \(36\): \(\;36(4 + 3x)\)
- \(x^{2} + 4x\) β GCF \(x\): \(\;x(x + 4)\)
- \(8x^{2} + 4x\) β GCF \(4x\): \(\;4x(2x + 1)\)
- \(x^{3} + x^{2}\) β GCF \(x^{2}\): \(\;x^{2}(x + 1)\)
- \(60x^{3} + 90x^{2}\) β GCF \(30x^{2}\): \(\;30x^{2}(2x + 3)\)
- \(20xy + 30x\) β GCF \(10x\): \(\;10x(2y + 3)\)
- \(20x^{2}y + 30x\) β GCF \(10x\): \(\;10x(2xy + 3)\)
- \(45x^{2}y - 15xy\) β GCF \(15xy\): \(\;15xy(3x - 1)\)
- \(7x^{2}y + 21xy^{2}\) β GCF \(7xy\): \(\;7xy(x + 3y)\)
CautionCommon factor that is a bracket (click for solutions)
- \(3(x + 2) + 5(x + 2)\) β GCF \((x + 2)\): \(\;8(x + 2)\)
- \(2x(3x - 5) + 4(3x - 5)\) β GCF \((3x - 5)\): \(\;2(3x - 5)(x + 2)\)
- \(6(x - 3) - 5(-x + 3)\) β GCF \((x - 3)\): \(\;11(x - 3)\)
- \((3x + 2)(2x - 5) + (4x - 1)(2x - 5)\) β GCF \((2x - 5)\): \(\;(2x - 5)(7x + 1)\)
CautionGrouping β the GCF is \((x + 3)\) (click for solutions)
- \(2x^{2} + 6x + x + 3 = 2x(x + 3) + (x + 3)\) β \((x + 3)(2x + 1)\)
- \(3x^{2} + 9x + 2x + 6 = 3x(x + 3) + 2(x + 3)\) β \((x + 3)(3x + 2)\)
Expanding
Expanding is the inverse operation of factoring: it turns an expression written as a product into a sum of terms, generally in polynomial form.1 It relies on distributivity, as in the rule \(a(b + c) = ab + ac\). For example, expanding \(3(x + 4)\) gives \(3x + 12\), while \((x + 2)(x - 5)\) expands to \(x^{2} - 3x - 10\).
Expanding lets us simplify or compare expressions, prepare an equation for solving, or reveal the standard form of a polynomial.
Worked example β a cost function
\[ C(x) = 5x(x + 20)(y + 3) \]
- Simple distributivity: \(5x(x + 20) = 5x^{2} + 100x\).
- Double distributivity: \((5x^{2} + 100x)(y + 3) = 5x^{2}y + 15x^{2} + 100xy + 300x\).
We obtain \(C(x) = 5x^{2}y + 100xy + 15x^{2} + 300x\), made up of increasing variable costs (\(5x^{2}y + 15x^{2}\)) and proportional variable costs (\(100xy + 300x\)).
CautionExercises β expand (click for solutions)
- \(3(x + 4)\) β \(3x + 12\)
- \(-2(x - 7)\) β \(-2x + 14\)
- \(-4(x^{2} - 3x + 7)\) β \(-4x^{2} + 12x - 28\)
- \(-4x(x^{2} - 3x + 7)\) β \(-4x^{3} + 12x^{2} - 28x\)
- \((x + 2)(x + 3)\) β \(x^{2} + 5x + 6\)
- \((2x - 5)(x^{2} + x + 1)\) β \(2x^{3} - 3x^{2} - 3x - 5\)
- \((x + 1)(x - 2)(x + 3)\) β \(x^{3} + 2x^{2} - 5x - 6\)
Notable Identities
Notable identities are fundamental algebraic equalities that let us expand or factor certain expressions quickly, without a term-by-term calculation. The most common are:
\[ (a + b)^{2} = a^{2} + 2ab + b^{2}, \quad (a - b)^{2} = a^{2} - 2ab + b^{2}, \quad (a - b)(a + b) = a^{2} - b^{2}. \]
For instance, \(x^{2} + 6x + 9\) is immediately recognised as \((x + 3)^{2}\).
CautionDerivation β the square of a sum (click to expand)
Starting from the definition of a square and using distributivity,
\[ (a + b)^{2} = (a + b)(a + b) = a\,a + a\,b + b\,a + b\,b = a^{2} + ab + ab + b^{2} = a^{2} + 2ab + b^{2}. \]
The middle term \(2ab\) comes from the two identical cross-products \(ab\) and \(ba\). Replacing \(b\) by \(-b\) gives \((a - b)^{2} = a^{2} - 2ab + b^{2}\), and keeping only the cross-products that survive gives \((a - b)(a + b) = a^{2} - b^{2}\).
CautionExpand with notable identities (click for solutions)
- \((x + 5)^{2}\) β \(x^{2} + 10x + 25\)
- \((2x - 5)^{2}\) β \(4x^{2} - 20x + 25\)
- \((-x + 5)^{2}\) β \(x^{2} - 10x + 25\)
- \((2x - 5)(2x + 5)\) β \(4x^{2} - 25\)
CautionFactor with notable identities (click for solutions)
- \(x^{2} + 6x + 9\) β \((x + 3)^{2}\)
- \(9x^{2} - 24x + 16\) β \((3x - 4)^{2}\)
- \(-4x^{2} + 12x - 9\) β \(-(2x - 3)^{2}\)
- \(16x^{2} - 25\) β \((4x + 5)(4x - 5)\)
- \(49x^{2} - 1\) β \((7x + 1)(7x - 1)\)
- \(49x^{4} - x^{2}\) β \(x^{2}(7x + 1)(7x - 1)\)
- \((x + 2)^{2} - (x - 2)^{2}\) β \(8x\)
Solving First-Degree Equations
TipGeneral method β first-degree equation \(ax + b = 0\)
- Expand / remove parentheses if needed.
- Group the terms in \(x\) on one side and the constants on the other.
- Isolate \(x\): \(\;x = -\dfrac{b}{a}\) if \(a \neq 0\).
ImportantExam tip
If \(a = 0\): the equation is either impossible (no solution) when \(b \neq 0\), or an identity (infinitely many solutions) when \(b = 0\). Always check the coefficient of \(x\) before dividing.
Worked examples
- \(3x - 7 = 11 \Rightarrow 3x = 18 \Rightarrow x = 6\).
- \(2(x - 3) = 5 - x \Rightarrow 2x - 6 = 5 - x \Rightarrow 3x = 11 \Rightarrow x = \dfrac{11}{3}\).
- \(\dfrac{x - 1}{3} + \dfrac{2x + 5}{6} = 2\): multiply by \(6\) to get \(2(x - 1) + (2x + 5) = 12\), i.e. \(4x + 3 = 12\), so \(x = \dfrac{9}{4}\).
Break-Even Formula
TipBreak-even quantity
\[ Q^* = \frac{F}{p - v} \]
where \(F\) = fixed cost, \(p\) = selling price, and \(v\) = variable cost per unit. The break-even quantity is the number of units where revenue exactly covers total cost.
CautionMixed exercises (click for solutions)
- Simplify \(\dfrac{6x - 9}{3}\) β \(2x - 3\)
- Factor \(4x^{2} - 12x\) β \(4x(x - 3)\)
- Expand \((3x - 2)(x + 5)\) β \(3x^{2} + 13x - 10\)
- Solve \(5x + 7 = 2x - 5\) β \(x = -4\)
4 Worked Example
Worked example β break-even for a cafΓ©
Using the Opening Problem values: \(F = 2{,}000\), \(v = 3\), \(p = 8\).
Revenue function:
\[R(q) = 8q\]
Total cost function:
\[C(q) = 2{,}000 + 3q\]
Break-even: set \(R = C\):
\[ 8q = 2{,}000 + 3q \Rightarrow 5q = 2{,}000 \Rightarrow q^* = 400 \text{ units.} \]
Verification: \(R(400) = 8 \times 400 = 3{,}200\) and \(C(400) = 2{,}000 + 3 \times 400 = 3{,}200\). β
5 Guided Practice
A freelance designer charges β¬50 per hour. Her monthly fixed costs (software, office) total β¬600. How many hours must she bill to break even?
TipHint
Set up \(R(h) = 50h\) and \(C(h) = 600\). Then solve \(R(h) = C(h)\).
CautionShow the solution
Revenue function: \(R(h) = 50h\)
Total cost function: \(C(h) = 600\)
Break-even: \(50h = 600 \Rightarrow h = 12\) hours.
Verification: \(R(12) = C(12) = 600\). β
6 Interactive Exploration
Use the sliders below to explore how fixed cost, variable cost, and selling price interact to determine the break-even quantity.
Investigation
- Set the selling price equal to the variable cost. What happens to the break-even quantity? Why?
- Double the fixed cost. How does \(Q^*\) change? Is the relationship linear?
- A competitor offers the same product at β¬6. What break-even quantity would your cafΓ© need at this price?
7 Applied Problem
A technology startup developing a mobile application estimates: fixed monthly costs of β¬8,500 (salaries, hosting, licences); variable cost of β¬0.50 per active user per month; subscription price of β¬4.99 per user per month.
- Write the revenue function \(R(u)\) and the total cost function \(C(u)\), where \(u\) is the number of active users.
- Calculate the break-even number of users.
- If the startup currently has 1,800 users, is it profitable? By how much?
- What selling price would reduce the break-even point to 2,000 users, keeping all other parameters fixed?
8 Common Errors
WarningCommon error β distributing a factor incorrectly
A student writes \(3(x + 4) = 3x + 4\).
The correct calculation is:
\[3(x + 4) = 3x + 12.\]
The factor 3 must multiply every term inside the parentheses. Omitting to multiply 4 by 3 is the most frequent expansion error.
9 Check Your Understanding
- Factor \(6x^2 - 9x\).
- Why does the break-even formula require \(p > v\)? What does it mean economically if \(p \leq v\)?
- A restaurant has \(F =\) β¬3,000/month, \(v =\) β¬4/cover, \(p =\) β¬18/cover. Calculate the break-even number of covers.
- A student solves \(2(x - 3) = x + 4\) and gets \(x = -2\). Is this correct? Identify any error.
- Expand \((2x - 5)^2\) using a notable identity.
10 Summary
- Algebraic transformations (factoring, expanding) change the form of an expression without changing its value β they reveal structure that makes equations easier to solve.
- The break-even quantity \(Q^* = F / (p - v)\) is the point where revenue equals total cost; it only exists when the selling price exceeds the variable cost per unit.
- Solving a first-degree equation means isolating the unknown by performing identical operations on both sides β always verify the solution by substituting back.
11 Independent Work
- Complete factoring exercises 1β15 in the exercise bank.
- Complete expanding exercises 1β10.
- Solve break-even problems AβD.
- Record any errors in your error log.
- Review the glossary entries for: factoring, expanding, break-even point, contribution margin.
12 Continue
Footnotes
An expression formed only from products and sums of constants and variables (usually written \(x\), \(y\), \(z\), β¦).β©οΈ