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Module 2 · L14 of 15 ~35 min ⚡ +95 XP available

Applications of Polynomials

A cardboard box. A company's profit curve. A projectile's path. Behind every real-world shape and trend lies a polynomial equation waiting to be set up and solved. In this lesson you'll go beyond pure algebra and apply polynomial techniques to physical problems — and discover why checking solutions against context is just as important as solving the equation.

Today's hook — You cut squares of side $x$ from the corners of a 20 cm square sheet, fold up the flaps, and make a box. If the volume must be 500 cm³, you get a cubic equation with three roots — but only some of them make physical sense. By the end of this lesson you'll know exactly how to set up any applied polynomial and which solutions to keep.

0/5QUESTS

Recall — your gut answer first

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A box is made from a square sheet of side 20 cm by cutting equal squares from each corner and folding up the sides. Without calculating — if the volume is 500 cm³, what equation gives the height $x$ of the box? What type of equation is it? Write your prediction before reading on.

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The two core moves

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Every applied polynomial problem has exactly two phases — and most marks are lost by skipping phase two.

Phase 1 — Set up. Translate the physical scenario into a polynomial equation. Define your variable clearly, express all dimensions and quantities in terms of it, and write the governing equation. This usually produces a cubic or quartic.

Phase 2 — Filter. Solve the equation algebraically, then check every solution against the physical constraints of the problem. Length must be positive. A side of a box cannot exceed half the sheet size. Time must be non-negative. Reject any solution that violates these constraints — they are extraneous solutions.

Always check solutions against physical constraints

Define the variable first

Always write "Let $x$ = …" explicitly. This forces you to think about what it represents — and its physical domain.

State the constraint early

Write down the physical domain of $x$ (e.g., $0 < x < 15$) before solving. Then check each solution against it.

Interpret the answer

Always state what the valid solution(s) mean in context: "The height of the box is $x = 10$ cm." A bare number scores only partial marks.

What you'll master

Know

Key facts

A mathematical model is a polynomial equation representing a real-world scenario
An extraneous solution satisfies the equation but not the physical constraints
Physical constraints define the valid domain of the variable

Understand

Concepts

How dimensions of geometric objects translate into polynomial expressions
Why some algebraic roots must be rejected in an applied context
The modelling cycle: define → set up → solve → interpret → validate

Can do

Skills

Set up polynomial equations from volume, area, and geometric constraint problems
Solve using factorisation techniques from earlier in the module
Identify and reject extraneous solutions; interpret valid solutions in context

Key terms

Mathematical modelAn equation or system of equations that represents a real-world situation in mathematical terms.

Extraneous solutionA value that satisfies the mathematical equation but violates the physical constraints of the problem (e.g., a negative length).

Physical constraintA condition imposed by the real-world context — for example, dimensions must be positive, or a cut-out cannot exceed half the sheet.

Modelling cycleThe process: define variable → translate to equation → solve → check constraints → interpret in context.

Box volume: the classic application

core concept

The box-cutting problem is the most common applied polynomial in HSC Extension 1. Here is the full worked solution for a 30 cm sheet.

Problem: A square sheet of side 30 cm has squares of side $x$ cut from each corner. The sides are folded up to make an open box. Find $x$ for a volume of 1000 cm³.

Setting up the model:

Let $x$ = side length of cut-out square (cm).
Physical constraint: $0 < x < 15$ (since $30 - 2x > 0 \Rightarrow x < 15$).
Base side after cutting: $30 - 2x$; height: $x$.
Volume: $V = x(30-2x)^2 = 1000$.

Solving:

$x(900 - 120x + 4x^2) = 1000$

$4x^3 - 120x^2 + 900x - 1000 = 0$

$x^3 - 30x^2 + 225x - 250 = 0$ (divide by 4)

Test $x = 10$: $1000 - 3000 + 2250 - 250 = 0$ ✓

Factorise: $(x - 10)(x^2 - 20x + 25) = 0$

$x = 10$ or $x = \dfrac{20 \pm \sqrt{400 - 100}}{2} = 10 \pm 5\sqrt{3}$

Checking constraints ($0 < x < 15$):

$x = 10$: valid ✓
$x = 10 + 5\sqrt{3} \approx 18.7$: exceeds 15 → reject ✗
$x = 10 - 5\sqrt{3} \approx 1.34$: valid ✓

$$x = 10 \text{ cm} \quad \text{or} \quad x = (10 - 5\sqrt{3}) \text{ cm} \approx 1.34 \text{ cm}$$

Two valid solutions — which to report? Both $x=10$ and $x \approx 1.34$ satisfy the volume requirement. Report both and let the context decide if one is preferred. If the problem says "find the height" (not "find all possible heights"), list both values and note they both satisfy the equation and the physical constraint.

Box cutting formula: V = x(L - 2x)^2 where L is the sheet side and x is the cut-out size; Always state the constraint 0 < x < L/2 before solving

Pause — copy the box-cutting formula into your book: $V = x(L-2x)^2$ where $L$ is the sheet side and $x$ is the cut-out size; always state the domain constraint $0 < x < L/2$ before solving.

Quick check: A 20 cm square sheet has $x$ cm cut from each corner. What is the valid domain for $x$?

Polynomial models in context

core concept

We just saw the box-cutting problem: $V = x(L-2x)^2$ with constraint $0 < x < L/2$. That raises a question: in any polynomial modelling problem, what general steps do you follow from the problem description to the final answer? This card answers it → (1) define the variable; (2) write the polynomial; (3) solve; (4) check constraints; (5) interpret the result in context.

Polynomials appear in many real-world contexts beyond geometry. When a polynomial models a quantity (profit, height, population), the roots correspond to key moments — zero crossings — in that quantity's behaviour.

Example: The polynomial $P(x) = x^3 - 6x^2 + 11x - 6$ models the profit (in thousands of dollars) of a company $x$ years after startup. When was the profit zero?

Solution:

$P(1) = 1 - 6 + 11 - 6 = 0$ → $(x-1)$ is a factor.

Divide: $P(x) = (x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)$

Roots: $x = 1, 2, 3$ years.

Interpretation: The profit was zero at 1, 2, and 3 years after startup.

Context matters. In the profit model, $x$ represents time since startup — so $x < 0$ is meaningless. Similarly, a root at $x = 0.5$ would mean 6 months — potentially valid. Always ask: does this root make sense given what $x$ represents?

Modelling steps: define variable → write polynomial equation → solve → check constraints → interpret; Roots of the model equation correspond to zeros of the modelled quantity

Pause — copy the polynomial modelling steps into your book: (1) define variable; (2) write polynomial equation; (3) solve; (4) check constraints (domain); (5) interpret — roots correspond to zeros of the modelled quantity.

Did you get this? True or false: all roots of a polynomial model equation are valid answers to the applied problem.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BOX VOLUME

A square sheet of side 30 cm has $x$ cm cut from each corner. The volume of the resulting box is 1000 cm³. Show that $x^3 - 30x^2 + 225x - 250 = 0$ and find all valid values of $x$.

$V = x(30-2x)^2 = 1000 \Rightarrow x(900-120x+4x^2) = 1000 \Rightarrow 4x^3-120x^2+900x-1000=0$

Expand and collect. Divide through by 4 to simplify: $x^3 - 30x^2 + 225x - 250 = 0$.

PROBLEM 2 · PROFIT MODEL

The polynomial $P(x) = x^3 - 6x^2 + 11x - 6$ models profit (thousands of dollars) $x$ years after startup. Find when the profit was zero and identify any extraneous solutions.

$P(1) = 1-6+11-6 = 0$ ✓. Divide: $P(x) = (x-1)(x^2-5x+6) = (x-1)(x-2)(x-3)$.

Test small positive integers first as potential roots. Fully factorise once one root is found.

PROBLEM 3 · PROJECTILE PATH

The path of a projectile is modelled by $h = -2x^2 + 8x + 10$ where $h$ is height (m) and $x$ is horizontal distance (m). Find where the projectile hits the ground. Reject any extraneous solution.

Set $h = 0$: $-2x^2 + 8x + 10 = 0 \Rightarrow x^2 - 4x - 5 = 0 \Rightarrow (x-5)(x+1) = 0$.

Divide by $-2$ to simplify. Factorise the resulting quadratic.

Fill the gap: In the box problem with sheet side $L$, the height of the box equals the size $x$, and the base side equals . The constraint is $0 < x < $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Accepting all algebraic roots

Students solve the equation correctly but report all roots — including negative lengths or dimensions larger than the sheet. Every applied problem requires you to filter roots through physical constraints. Not doing so loses a mark every time.

Trap 02

Not defining the variable

Writing the equation without stating "Let $x$ = …" means you can't identify what the physical constraint should be. Always define the variable and its units before writing a single equation.

Trap 03

Forgetting to interpret the answer

Stopping at "$x = 10$" without saying "so the height is 10 cm" loses the final mark in almost every exam question. Always close with a sentence that connects the number to the real-world quantity.

Did you get this? True or false: for the projectile model $h = -2x^2 + 8x + 10$, the root $x = -1$ represents where the projectile was launched from.

Odd one out: Three of the following represent steps in the applied polynomial modelling process. Which one is the odd one out?

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A rectangle has length 4 cm more than its width. If the area is 60 cm², write and solve an equation to find the dimensions.

A box with a square base has height equal to twice the side of the base. If the volume is 250 cm³, find the dimensions. (Constraint: side must be positive.)

For the projectile $h = -2x^2 + 8x + 10$: (a) find the launch height (at $x=0$); (b) find where the projectile lands; (c) explain why one root is rejected.

Squares of side $x$ cm are cut from the corners of a 24 cm square sheet to form a box. Find $x$ if the volume is 864 cm³. State the constraint and check each root.

Explain in your own words why checking solutions against physical constraints is just as important as solving the polynomial equation correctly.

Revisit your thinking

Earlier you were asked: For a 20 cm square sheet with $x$ cm cut from each corner, what equation gives the height if the volume is 500 cm³?

Height $= x$, base side $= 20-2x$, so $V = x(20-2x)^2 = 500$. This expands to $4x^3 - 80x^2 + 400x - 500 = 0$, or $x^3 - 20x^2 + 100x - 125 = 0$. This is a cubic equation. The constraint is $0 < x < 10$. Testing $x=5$: $125 - 500 + 500 - 125 = 0$ ✓ — so $x=5$ cm is one valid solution.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. A rectangle has length 4 cm more than its width. If the area is 60 cm², find the dimensions. State any rejected solutions. (2 marks)

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ApplyBand 43 marks

Q2. A box with square base has height equal to twice the side of the base. If the volume is 250 cm³, find the dimensions. Show full working. (3 marks)

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AnalyseBand 52 marks

Q3. The path of a projectile is $h = -2x^2 + 8x + 10$ where $h$ is height (m) and $x$ is horizontal distance (m). Find where it hits the ground and explain why one root is rejected. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $w(w+4)=60 \Rightarrow w^2+4w-60=0 \Rightarrow (w+10)(w-6)=0$; $w=6$ (reject $w=-10$, negative); dimensions $6 \times 10$ cm. 2. $s^2 \cdot 2s = 250 \Rightarrow 2s^3=250 \Rightarrow s^3=125 \Rightarrow s=5$; base $5\times5$, height $10$ cm. 3. (a) $h(0)=10$ m; (b) $-2x^2+8x+10=0 \Rightarrow x^2-4x-5=0 \Rightarrow (x-5)(x+1)=0$; $x=5$ m; (c) reject $x=-1$ (negative distance). 4. $x(24-2x)^2=864 \Rightarrow x^3-24x^2+144x-216=0=(x-6)(x^2-18x+36)=0$; $x=6$ or $x=9\pm3\sqrt{5}$; check: $x=9+3\sqrt{5}\approx15.7>12$ rejected; $x=9-3\sqrt{5}\approx2.3$ valid; both $x=6$ and $x\approx2.3$ valid. 5. The algebraic equation may have roots outside the physically possible range; accepting all roots would mean reporting impossible dimensions or negative times.

Q1 (2 marks): Let $w$ = width; $(w)(w+4) = 60 \Rightarrow w^2+4w-60=0 \Rightarrow (w-6)(w+10)=0$ [1]; $w=-10$ rejected (negative length); $w=6$, length $=10$ cm [1].

Q2 (3 marks): Let $s$ = base side; height $= 2s$; $V = s^2 \cdot 2s = 2s^3 = 250$ [1]; $s^3 = 125$, so $s = 5$ cm [1]; base $5 \times 5$ cm, height $10$ cm [1].

Q3 (2 marks): Set $h=0$: $x^2-4x-5=0$; $(x-5)(x+1)=0$ [0.5]; $x=5$ or $x=-1$; reject $x=-1$ (horizontal distance cannot be negative) [0.5]; the projectile hits the ground at $x = 5$ m [1].

Boss battle · The Model Builder

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Key takeaways

Translate word problems into polynomial equations: define variable → express dimensions → set up equation.
Solve using factorisation or other techniques from this module.
Always check solutions against physical constraints — reject extraneous solutions.
Interpret valid solutions in context, with units.
Next lesson: Module Synthesis & Exam Technique — all Module 2 topics integrated for the exam.

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