Year 11: How to solve word problems on polynomials

In Year 11 mathematics, polynomials are used to model real situations: areas of shapes, volumes of boxes, profit of a business, and more. In this article, we will walk through a step-by-step method to solve a word problem that leads to a polynomial.

Key idea: A polynomial is an expression like

4 x^{3} - 100 x^{2} + 600 x

where the variable has whole-number powers and the terms are added or subtracted.

Step 1: Read the problem and identify what is being asked

Example problem: making an open-top box

A rectangular piece of cardboard is 30 cm long and 20 cm wide. Squares of side length $x$ centimetres are cut from each corner. The sides are then folded up to make an open-top box.

(a) Write a polynomial for the volume $V (x)$ of the box in terms of $x$ .
(b) For what values of $x$ will the volume of the box be zero?
(c) Which of these values make sense in the real situation?

First, we highlight the important information:

Original length: $30$ cm
Original width: $20$ cm
Cut-out square side length: $x$ cm
We want: a polynomial for $V (x)$ , the volume of the box

Also notice what the question is asking us to find:

A formula for $V (x)$
Values of $x$ that make $V (x) = 0$
Which solutions are realistic for the cardboard box

Step 2: Choose a variable and write expressions

The problem already uses $x$ for the length of each cut-out square. We stick with that:

Let $x$ be the side length of each small square (in cm).

Once we cut out squares and fold up the sides, the dimensions of the box will be:

Height of the box: $x$ (this comes from the side of the square we fold up)
New length of the base: $30 - 2 x$ (we remove $x$ from both ends)
New width of the base: $20 - 2 x$

We now have expressions for all three dimensions in terms of $x$ .

Step 3: Use the volume formula to build a polynomial

The volume of a rectangular prism is: $Volume = length \times width \times height$ .

Substitute in our expressions:

$V (x) = x \times (30 - 2 x) \times (20 - 2 x)$

At this point, we already have a polynomial written as a product of factors. Often, though, we expand it to standard polynomial form:

First multiply $(30 - 2 x)$ and $(20 - 2 x)$ :

$(30 - 2 x) (20 - 2 x)$
$= 30 \times 20 + 30 \times (- 2 x) + (- 2 x) \times 20 + (- 2 x) \times (- 2 x)$
$= 600 - 60 x - 40 x + 4 x^{2}$
$= 600 - 100 x + 4 x^{2}$

Now multiply by $x$ :

$V (x) = x \times (600 - 100 x + 4 x^{2})$
$= 600 x - 100 x^{2} + 4 x^{3}$

It is common to write polynomial terms in descending powers of $x$ :

Polynomial for the volume: $V (x) = 4 x^{3} - 100 x^{2} + 600 x$

Step 4: Factor the polynomial to solve the word problem

Part (b) asks: For what values of $x$ will the volume be zero? That means we solve:

$V (x) = 0$
$4 x^{3} - 100 x^{2} + 600 x = 0$

Step 4a: Factor out common factors

All terms share a factor of $4 x$ :

$4 x^{3} - 100 x^{2} + 600 x = 4 x (x^{2} - 25 x + 150)$

Step 4b: Factor the quadratic

Now we factor the quadratic expression $x^{2} - 25 x + 150$ . We look for two numbers that:

Multiply to $150$
Add to $- 25$

The pair $- 10$ and $- 15$ works because:

$(- 10) \times (- 15) = 150$
$(- 10) + (- 15) = - 25$

So we can write:

$x^{2} - 25 x + 150 = (x - 10) (x - 15)$

Therefore the fully factored form of the volume is:

$V (x) = 4 x (x - 10) (x - 15)$

Step 4c: Solve each factor equal to zero

To find when the volume is zero, set each factor to zero:

$4 x = 0$ → $x = 0$
$x - 10 = 0$ → $x = 10$
$x - 15 = 0$ → $x = 15$

So the polynomial has zeros at $x = 0$ , $10$ and $15$ .

Step 5: Interpret the solutions in context

Now we answer part (c): Which values make sense for the real cardboard box?

$x = 0$ : This would mean we do not cut out any squares. Then the “box” has no height. Volume is zero. This is mathematically correct but does not give a useful box.
$x = 10$ : Cutting out 10 cm squares makes the new width $20 - 2 \times 10 = 0$ . The base is a line segment, so the volume is zero.
$x = 15$ : Now the new length is $30 - 2 \times 15 = 0$ , so again the base collapses and the volume is zero.

Also, for the dimensions to be positive:

$30 - 2 x > 0$ → $x < 15$
$20 - 2 x > 0$ → $x < 10$
$x > 0$

Combining these, we get the realistic domain:

$0 < x < 10$

This means:

$x = 10$ and $x = 15$ are “edge” solutions where the box collapses.
For a real, three-dimensional box, we need $0 < x < 10$ centimetres.

General strategy for polynomial word problems

You can apply the same structure to many applications of polynomials in Year 11:

Read and highlight the important quantities and what is being asked.
Choose variables and write expressions for all lengths, areas, volumes, or other quantities.
Build the polynomial using known formulas (like area, volume, cost, or revenue).
Rewrite neatly in standard polynomial form, e.g. $a x^{3} + b x^{2} + c x + d$ .
Factor or solve the polynomial to answer the question.
Check the domain and interpret which solutions make sense in the real situation.

Practice problems

A rectangular garden is 8 m longer than it is wide. Its area (in m²) can be written as a polynomial: $A (w) = w (w + 8)$ , where $w$ is the width in metres.
(a) Expand the expression to write $A (w)$ in the form $a w^{2} + b w + c$ .
(b) If the area must be 240 m², solve the polynomial equation to find possible values of $w$ .
(c) Decide which value of $w$ is realistic.

The profit (in dollars) of selling $n$ items is modelled by the polynomial $P (n) = - n^{2} + 40 n - 96$ .
(a) Factor the polynomial.
(b) Find the values of $n$ for which the profit is zero.
(c) For which values of $n$ is the profit positive?

Try solving these using the same step-by-step “how to solve it” method. With practice, turning words into polynomials will become a lot more natural.