Chapter 2 · Polynomials

2.1 Polynomials and the Geometrical Meaning of Zeroes

A zero of a polynomial is an input that makes its value $0$. On a graph, the zeroes are exactly the places where the curve meets the $x$-axis.

At a school science fair, a water rocket is launched straight up from the roof of a building $15$ m high. Its height above the ground after $t$ seconds is about $h(t)=15+10t-5t^2$ metres. After $1$ second, $h(1)=15+10-5=20$ m. After $2$ seconds, $h(2)=15+20-20=15$ m. After $3$ seconds, $h(3)=15+30-45=0$. A height of $0$ means the rocket has reached the ground. So the question "when does it land?" is really the question "for which $t$ is $h(t)=0$?"

The expression $15+10t-5t^2$ is a polynomial, and a value of $t$ that makes it equal to $0$ is called a zero of the polynomial. In this section we name the different kinds of polynomials, practise finding their values and zeroes, and then see the picture behind it all: on a graph, the zeroes are exactly the points where the curve meets the $x$-axis. That picture also tells us how many zeroes a polynomial can have.

Polynomials, degree and value

You met polynomials in Class 9. Let us fix the words carefully, because the rest of the chapter depends on them.

📘 Definition 2.1 — Polynomial and its degree

A polynomial in $x$ is an expression of the form $$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0 ,$$ where $a_n, a_{n-1}, \dots, a_1, a_0$ are real numbers (the coefficients) and every power of $x$ is a whole number $0, 1, 2, 3, \dots$
If $a_n\ne0$, the highest power $n$ is the degree of the polynomial. A non-zero constant, such as $7$, has degree $0$. (For the zero polynomial $0$ the degree is not defined.)

So $4x^3-2x+9$ is a polynomial of degree $3$, but $\sqrt x+1$ (power $\tfrac12$) and $x+\tfrac1x$ (power $-1$) are not polynomials. The coefficients may be any real numbers, even $\sqrt2$ or $-\tfrac35$; it is only the powers that must be whole numbers.

📘 Definition 2.2 — Linear, quadratic and cubic polynomials

A polynomial of degree $1$ is called linear: $\;ax+b$, with $a\ne0$.
A polynomial of degree $2$ is called quadratic: $\;ax^2+bx+c$, with $a\ne0$.
A polynomial of degree $3$ is called cubic: $\;ax^3+bx^2+cx+d$, with $a\ne0$.
The condition $a\ne0$ matters: if $a=0$, the highest power disappears and the degree drops.

✏️ Example 1 — Polynomial or not? Degree and type

Which of these are polynomials in one variable? For each polynomial, give its degree and its type.
(a) $3x^2-5x+\sqrt2$ (b) $2x+\dfrac3x$ (c) $\sqrt x+7$ (d) $7-4y^3+y$ (e) $x(x-1)(x+2)$ (f) $9$

💡 Show solution▶

(a) The powers are $2, 1, 0$, all whole numbers. The coefficient $\sqrt2$ is allowed. Polynomial, degree $2$, quadratic.

(b) $\dfrac3x=3x^{-1}$, and $-1$ is not a whole number. Not a polynomial.

(d) The variable is $y$, and the highest power is $3$. Order does not matter: $7-4y^3+y=-4y^3+y+7$. Polynomial, degree $3$, cubic.

(e) Multiply out first: $x(x-1)(x+2)=x(x^2+x-2)=x^3+x^2-2x$. Polynomial, degree $3$, cubic.

(f) $9=9x^0$. A constant polynomial, degree $0$ (it is not linear, quadratic or cubic).

📘 Definition 2.3 — Value and zero of a polynomial

If $p(x)$ is a polynomial and $k$ is a real number, the number obtained by putting $x=k$ in $p(x)$ is the value of $p(x)$ at $x=k$, written $p(k)$.
A real number $k$ is called a zero of the polynomial $p(x)$ if $p(k)=0$.

✏️ Example 2 — Values and zeroes

Let $p(x)=x^2-3x-4$. Find $p(0)$, $p(2)$, $p(-2)$, $p(-1)$ and $p(4)$. Which of the numbers $0, 2, -2, -1, 4$ are zeroes of $p(x)$?

💡 Show solution▶

Substitute $p(0)=0-0-4=-4$.
$p(2)=4-6-4=-6$.
$p(-2)=(-2)^2-3(-2)-4=4+6-4=6$.

Careful with signs $p(-1)=(-1)^2-3(-1)-4=1+3-4=0$. Write the brackets: a common slip is $-1^2=-1$, which gives the wrong value $-6$.

Last one $p(4)=16-12-4=0$.

Answer $p(-1)=0$ and $p(4)=0$, so $-1$ and $4$ are zeroes of $p(x)$. The numbers $0$, $2$ and $-2$ are not.

Check $(x+1)(x-4)=x^2-4x+x-4=x^2-3x-4$. ✓ Each bracket is $0$ at one of the zeroes.

The zero of a linear polynomial. For $p(x)=ax+b$ with $a\ne0$, solve $ax+b=0$: $\;x=-\dfrac ba$. So a linear polynomial has exactly one zero, $$\text{zero of }ax+b=-\frac ba=-\frac{\text{constant term}}{\text{coefficient of }x}.$$

✏️ Example 3 — Zeroes of linear polynomials

(a) Find the zero of $2x-5$. (b) Find the zero of $3x+12$.
(c) A water tank holds $1200$ L. When the outlet is opened, the water left after $t$ minutes is $V(t)=1200-40t$ litres. Find the zero of $V(t)$ and explain what it means.

💡 Show solution▶

(a) $2x-5=0\Rightarrow x=\dfrac52$. Check: $2\times\tfrac52-5=0$. ✓

(b) $3x+12=0\Rightarrow x=-\dfrac{12}{3}=-4$. Check: $3(-4)+12=0$. ✓

The geometrical meaning of zeroes

Draw the graph of $y=p(x)$ by plotting the points $(x, p(x))$. If $k$ is a zero, then $p(k)=0$, so the point $(k, 0)$ lies on the graph — and every point with $y$-coordinate $0$ lies on the $x$-axis. Turning this round:

🧭 The key picture

The zeroes of $p(x)$ are exactly the $x$-coordinates of the points where the graph of $y=p(x)$ meets the $x$-axis. So to count the zeroes, count the meeting points; to read a zero, read the $x$-coordinate of a meeting point.

Linear polynomials. The graph of $y=ax+b$ ($a\ne0$) is a straight line that is not parallel to the $x$-axis, so it meets the $x$-axis at exactly one point, $\left(-\tfrac ba, 0\right)$. For example, $y=2x+3$ meets the axis at $\left(-\tfrac32, 0\right)$, and $-\tfrac32$ is the zero of $2x+3$.

Quadratic polynomials. The graph of $y=ax^2+bx+c$ ($a\ne0$) is a curve called a parabola. It opens upwards like ∪ when $a\gt0$ and downwards like ∩ when $a\lt0$. A parabola can meet the $x$-axis in three ways:

at two distinct points — the quadratic has two different zeroes;
at exactly one point, just touching the axis — the two zeroes are equal (one zero, counted twice);
at no point — the quadratic has no real zero.

Use Interactive 1 to see all three cases. Changing $c$ slides the parabola up and down without changing its shape.

✏️ Example 4 — Zeroes from a table and a graph

Make a table of values of $p(x)=x^2-2x-3$ for $x=-2, -1, 0, 1, 2, 3, 4$. Describe its graph and find the zeroes of $p(x)$.

💡 Show solution▶

Table For example $p(-2)=4+4-3=5$ and $p(1)=1-2-3=-4$.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y=p(x)$	$5$	$0$	$-3$	$-4$	$-3$	$0$	$5$

Graph Plot the points $(-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), (4, 5)$ and join them with a smooth curve. Since $a=1\gt0$ it is a ∪-shaped parabola. Its lowest point is $(1, -4)$, and the values are symmetric about $x=1$.

Read the zeroes The graph meets the $x$-axis at $(-1, 0)$ and $(3, 0)$. So the zeroes are $-1$ and $3$.

Check $p(-1)=1+2-3=0$ ✓ and $p(3)=9-6-3=0$ ✓. Also $(x+1)(x-3)=x^2-2x-3$. ✓

Cubic polynomials. The graph of a cubic can turn at most twice, so it can cross the $x$-axis once, twice or three times — but never more. Its two ends point in opposite directions (one up, one down), because for large $x$ the $ax^3$ term is much bigger than the others and changes sign with $x$. So an unbroken cubic graph must cross the axis at least once.

📐 Theorem 2.1 — How many zeroes?

A polynomial of degree $n$ (with $n\ge1$) has at most $n$ zeroes. In particular:

a linear polynomial $ax+b$ ($a\ne0$) has exactly one zero, $-\tfrac ba$;
a quadratic polynomial has at most $2$ zeroes (its graph meets the $x$-axis at $0$, $1$ or $2$ points);
a cubic polynomial has at most $3$ zeroes (its graph meets the $x$-axis at $1$, $2$ or $3$ points).

You do not need to prove this theorem. The pictures give a convincing reason: a straight line crosses the axis once, a parabola turns once and so can cross at most twice, and a cubic turns at most twice and so can cross at most three times. Interactive 2 lets you test it.

✏️ Example 5 — Counting zeroes from a description of the graph

How many zeroes does $p(x)$ have if the graph of $y=p(x)$ is
(a) a parabola opening downwards whose highest point is $(1, 4)$;
(b) a parabola opening upwards whose lowest point is $(2, 3)$;
(c) a parabola whose lowest point is $(-3, 0)$;
(d) a cubic curve that meets the $x$-axis at $(-2, 0)$, $(0, 0)$ and $(2, 0)$;
(e) the horizontal straight line $y=4$?

💡 Show solution▶

(a) The highest point is above the axis and the curve goes down on both sides, so it crosses the axis twice. 2 zeroes.

(b) The lowest point is already above the axis, so the whole curve stays above it. No zero.

(d) Three meeting points: 3 zeroes, $-2$, $0$ and $2$ — the most a cubic can have.

(e) The line $y=4$ is parallel to the $x$-axis and never meets it. No zero. (This is the constant polynomial $p(x)=4$, not a linear one.)

✏️ Example 6 — A cubic with three zeroes

Make a table of values of $p(x)=x^3-4x$ for $x=-3, -2, \dots, 3$. Find the zeroes of $p(x)$ and check them by factorising.

💡 Show solution▶

Table For example $p(-3)=-27+12=-15$ and $p(-1)=-1+4=3$.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y=p(x)$	$-15$	$0$	$3$	$0$	$-3$	$0$	$15$

Read the graph The curve comes up from the bottom left, crosses the axis at $x=-2$, rises to a hump, comes down through $x=0$, dips, and rises again through $x=2$. The zeroes are $-2$, $0$ and $2$.

Check $x^3-4x=x(x^2-4)=x(x-2)(x+2)$, which is $0$ exactly when $x=0$, $x=2$ or $x=-2$. ✓ Three zeroes — the maximum for a cubic, as Theorem 2.1 says.

⚠️ Common mistakes with zeroes and graphs

1. Reading the wrong axis. The zeroes are where the graph meets the $x$-axis. The point where it meets the $y$-axis is $(0, p(0))$ — that gives the constant term, not a zero.

2. Writing the point instead of the number. If the graph meets the axis at $(3, 0)$, the zero is the number $3$, not "$(3, 0)$".

3. Thinking "opens downwards" means "no zeroes". Whether a parabola meets the axis depends on where its turning point is, not only on which way it opens (Example 5).

4. Sign slips with negative inputs. $(-2)^2=4$ but $-2^2=-4$. Always put a negative input in brackets before squaring or cubing.

Exercises

Is each expression a polynomial in one variable? If it is, give its degree and say whether it is linear, quadratic, cubic or none of these.

1 $4x^2-3x+7$
2 $2-y^2-y^3+2y^8$
3 $3\sqrt t+t\sqrt2$
4 $y+\dfrac2y$
5 $x^{10}+y^3+t^{50}$
6 $\sqrt3\,x^2-5$
7 $5$
8 $(2x-1)(x^2+3)$
9 $x(x+1)-x^2$

💡 Show solutions▶

1 Polynomial, degree $2$, quadratic.

2 Polynomial in $y$, highest power $8$: degree $8$, none of the three types.

3 $\sqrt t=t^{1/2}$ has a power that is not a whole number. Not a polynomial. (The term $t\sqrt2$ is fine — $\sqrt2$ is only a coefficient.)

4 $\dfrac2y=2y^{-1}$. Not a polynomial.

5 It contains three different variables $x$, $y$, $t$, so it is not a polynomial in one variable.

6 Polynomial, degree $2$, quadratic (coefficient $a=\sqrt3$).

7 Constant polynomial, degree $0$.

8 $(2x-1)(x^2+3)=2x^3-x^2+6x-3$: degree $3$, cubic.

9 $x(x+1)-x^2=x^2+x-x^2=x$: degree $1$, linear. (Simplify before deciding — it only looks quadratic.)

Values and zeroes.

1 For $p(x)=5x-4x^2+3$, find $p(0)$, $p(-1)$ and $p(2)$.
2 For $p(y)=y^2-y+1$, find $p(0)$, $p(1)$ and $p(2)$.
3 Verify that $-\tfrac13$ is a zero of $3x+1$.
4 Check whether $2$ and $3$ are zeroes of $x^2-5x+6$.
5 Check whether $1$ and $-1$ are zeroes of $x^3-x^2-x+1$.
6 Find the zero of $2x+5$.
7 Find the zero of $3x-2$.
8 Find the zero of $cx+d$, where $c\ne0$.
9 Find $k$ if $2$ is a zero of $x^2+kx-10$.
10 Find $k$ if $-1$ is a zero of $2x^2-3x+k$.
11 Phone battery. A phone at $80\%$ charge loses $4\%$ of charge every hour while streaming, so its charge after $t$ hours is $C(t)=80-4t$ (in $\%$). Find the zero of $C(t)$ and say what it means.

💡 Show solutions▶

1 $p(0)=3$; $\;p(-1)=-5-4+3=-6$; $\;p(2)=10-16+3=-3$.

2 $p(0)=1$; $\;p(1)=1-1+1=1$; $\;p(2)=4-2+1=3$.

3 $3\left(-\tfrac13\right)+1=-1+1=0$. ✓ So $-\tfrac13$ is a zero.

4 $p(2)=4-10+6=0$ and $p(3)=9-15+6=0$. Both are zeroes.

5 $p(1)=1-1-1+1=0$ and $p(-1)=-1-1+1+1=0$. Both are zeroes.

6 $2x+5=0\Rightarrow x=-\tfrac52$.

7 $3x-2=0\Rightarrow x=\tfrac23$.

8 $cx+d=0\Rightarrow x=-\tfrac dc$.

9 $p(2)=4+2k-10=0\Rightarrow 2k=6\Rightarrow k=3$.

10 $p(-1)=2(1)-3(-1)+k=2+3+k=0\Rightarrow k=-5$.

11 $80-4t=0\Rightarrow t=20$. The battery runs out after $20$ hours of streaming.

III

Zeroes and graphs.

1 The graph of $y=p(x)$ meets the $x$-axis only at $(-3, 0)$ and $(5, 0)$. How many zeroes does $p(x)$ have, and what are they?
2 The graph of a quadratic polynomial touches the $x$-axis only at $(4, 0)$. What are its zeroes?
3 The graph of a cubic polynomial meets the $x$-axis at $(-1, 0)$, $(0, 0)$ and $(2, 0)$. Write its zeroes. Can it have a fourth zero?
4 A parabola opens upwards and its lowest point is $(1, 2)$. How many zeroes does the quadratic have?
5 A parabola opens downwards and its highest point is $(-2, 5)$. How many zeroes does the quadratic have?
6 Make a table of $y=x^2+x-2$ for $x=-3, -2, \dots, 2$ and read off the zeroes.
7 Where does the graph of $y=3x-6$ meet the $x$-axis? Where does it meet the $y$-axis?
8 Can a quadratic polynomial have three zeroes? Give a reason.
9 Can the graph of a cubic polynomial miss the $x$-axis completely? Give a reason.
10 Show that the graph of $y=x^2-4x+4$ touches the $x$-axis at just one point, and find it.

💡 Show solutions▶

1 Two meeting points, so $2$ zeroes: $-3$ and $5$.

2 It touches at one point, so the two zeroes are equal: $4$ and $4$ (one zero, $4$).

3 Zeroes $-1$, $0$, $2$. No fourth zero: a cubic has at most $3$ zeroes (Theorem 2.1).

4 The lowest point is above the axis, so the parabola never meets it: no zero.

5 The highest point is above the axis and the curve falls on both sides: $2$ zeroes.

6 Values: $x=-3\colon 4$; $\;-2\colon 0$; $\;-1\colon -2$; $\;0\colon -2$; $\;1\colon 0$; $\;2\colon 4$. Zeroes $-2$ and $1$. (Check: $(x+2)(x-1)=x^2+x-2$.)

7 $x$-axis: $3x-6=0\Rightarrow x=2$, the point $(2, 0)$. $y$-axis: put $x=0$, $y=-6$, the point $(0, -6)$.

8 No. A polynomial of degree $2$ has at most $2$ zeroes; a parabola can meet the $x$-axis at most twice.

9 No. For large positive $x$ and large negative $x$ the term $ax^3$ is the biggest, and it has opposite signs at the two ends. So one end of the graph is above the axis and the other is below, and the unbroken curve must cross the axis at least once.

10 $x^2-4x+4=(x-2)^2$. A square is never negative, and it is $0$ only when $x=2$. So the graph never goes below the axis and meets it only at $(2, 0)$: one zero, $2$ (two equal zeroes).

Zeroes in real life.

1 Water rocket. For $h(t)=15+10t-5t^2$ (metres, $t$ in seconds), find $h(0)$, $h(1)$, $h(2)$ and $h(3)$. Which value of $t$ is a zero, and what does it mean?
2 Show that $t=-1$ is also a zero of $h(t)$ in part 1. Why do we ignore it?
3 Tea stall. A stall's daily profit (in ₹) when it charges $x$ rupees per cup is $P(x)=-10x^2+300x-2000$. Find $P(10)$, $P(15)$ and $P(20)$. Which of these prices are zeroes of $P$, and what do they mean for the owner?
4 Open box. Squares of side $x$ cm are cut from the corners of a $12$ cm × $12$ cm card and the sides folded up. The volume is $V(x)=x(12-2x)^2$ cm³. Write $V(x)$ as a polynomial and give its degree. Find $V(2)$, and find the zeroes of $V$. Why does no box exist at the zeroes?
5 Arch gateway. The height (in m) of an arch at a horizontal distance $x$ m from its centre line is $y=10-0.1x^2$. Find the zeroes of this polynomial and the width of the arch at ground level.

💡 Show solutions▶

1 $h(0)=15$, $h(1)=20$, $h(2)=15$, $h(3)=15+30-45=0$ (all in m). $t=3$ is a zero: the rocket lands $3$ s after launch.

2 $h(-1)=15-10-5=0$. ✓ But $t=-1$ means one second before the launch, when the rocket was not flying, so it has no meaning here.

3 $P(10)=-1000+3000-2000=0$; $\;P(15)=-2250+4500-2000=250$; $\;P(20)=-4000+6000-2000=0$. So ₹$10$ and ₹$20$ are zeroes: at those prices the stall just breaks even (no profit, no loss). At ₹$15$ per cup it makes a profit of ₹$250$ a day.

4 $V(x)=x(144-48x+4x^2)=4x^3-48x^2+144x$, degree $3$ (cubic). $V(2)=2\times8^2=128$ cm³. Zeroes: $x=0$ or $12-2x=0$, so $0$ and $6$. At $x=0$ nothing is folded up (no height); at $x=6$ the corner squares use up the whole card (no base). Either way the volume is $0$.

5 $10-0.1x^2=0\Rightarrow x^2=100\Rightarrow x=10$ or $x=-10$. The arch meets the ground $10$ m on each side of the centre line, so its width at ground level is $20$ m.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: board papers show a graph and ask for the number of zeroes, ask you to check or use a zero ($p(k)=0$), and set short case studies where a parabola models an arch or a path. Always say why: "the graph meets the $x$-axis at … points".

E1. Board · 1 mark The graph of $y=p(x)$ is a parabola that opens downwards, and its highest point is $(2, -3)$. The number of zeroes of $p(x)$ is (A) $0$ (B) $1$ (C) $2$ (D) $3$

💡 Show solution▶

Idea The highest point $(2, -3)$ is below the $x$-axis, and every other point of a ∩-shaped parabola is lower still. So the graph never reaches the axis.

Trap (C) comes from "opens downwards, so it crosses twice" — that is only true when the highest point is above the axis.

Answer (A) $0$.

E2. Board · 1 mark If $-2$ is a zero of $p(x)=x^2+3x+k$, then $k$ equals (A) $2$ (B) $-2$ (C) $10$ (D) $-10$

💡 Show solution▶

Idea $p(-2)=0$: $\;(-2)^2+3(-2)+k=4-6+k=0$, so $k=2$.

Trap (D) comes from writing $3(-2)=+6$: then $4+6+k=0$ gives $k=-10$.

Answer (A) $2$. Check: $x^2+3x+2=(x+1)(x+2)$ has $-2$ as a zero. ✓

E3. Board · 1 mark Assertion (A): The polynomial $x^3+x$ has exactly one zero. Reason (R): A cubic polynomial has at most three zeroes.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check A $x^3+x=x(x^2+1)$. Since $x^2+1\ge1$ for every real $x$, the product is $0$ only when $x=0$. So A is true.

Check R True (Theorem 2.1). But "at most three" does not explain why there is exactly one; the factorisation does.

Answer (B). Option (A) is the tempting choice if you only check that both statements are true.

E4. Board · 2 marks Check whether $2$ and $-3$ are zeroes of $p(x)=x^2+x-6$. Hence write down the points where the graph of $y=p(x)$ meets the $x$-axis.

💡 Show solution▶

Substitute $p(2)=4+2-6=0$ and $p(-3)=9-3-6=0$, so both are zeroes. (1 mark)

Graph A quadratic has at most two zeroes, so these are all of them. The graph meets the $x$-axis at $(2, 0)$ and $(-3, 0)$. (1 mark)

E5. Board · 3 marks Complete a table of values of $p(x)=x^2-x-6$ for $x=-3, -2, \dots, 4$. Use it to describe the graph of $y=p(x)$, find the zeroes of $p(x)$, and state the point where the graph cuts the $y$-axis.

💡 Show solution▶

Table (1 mark)

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$p(x)$	$6$	$0$	$-4$	$-6$	$-6$	$-4$	$0$	$6$

Graph $a=1\gt0$, so it is a ∪-shaped parabola, symmetric about $x=\tfrac12$ (the values repeat on both sides). (1 mark)

Answer It meets the $x$-axis at $(-2, 0)$ and $(3, 0)$, so the zeroes are $-2$ and $3$; it cuts the $y$-axis at $(0, -6)$. Check: $(x+2)(x-3)=x^2-x-6$. ✓ (1 mark)

E6. Board · Case study Footbridge arch. The arch of a park footbridge is a parabola. Measuring $x$ metres from the left foot of the arch along the ground, the height of the arch is $h(x)=-0.05x^2+x$ metres.
(i) Find the zeroes of $h(x)$ and hence the span of the arch (the distance between its two feet). (2 marks)
(ii) Find the height of the arch at the middle of the span. (1 mark)
(iii) A lamp post is to stand under the arch at $x=4$ m. What is the greatest height it can have? (1 mark)

💡 Show solution▶

(i) $h(x)=-0.05x(x-20)$. It is $0$ when $x=0$ or $x=20$, so the zeroes are $0$ and $20$, and the span is $20-0=20$ m.

(ii) The middle of the span is $x=10$: $\;h(10)=-0.05(100)+10=-5+10=5$ m.

(iii) $h(4)=-0.05(16)+4=-0.8+4=3.2$ m. The lamp post can be at most $3.2$ m tall.

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. Can a quadratic polynomial predict where a jet of water from a garden hose lands?

Why it matters. Engineers model fountains, irrigation sprinklers, the path of a cricket ball and the cables of suspension bridges with parabolas. Finding where the curve meets the ground — a zero of the polynomial — tells a designer how far a sprinkler reaches or how wide a fountain basin must be.

What to do.

Hold a garden hose horizontally at a fixed height (for example $0.8$ m, resting on a chair) and keep the tap steady. Ask a friend to photograph the jet against a wall or a fence with a tape measure along the ground.
From the photo, read the height $y$ of the jet above the ground at horizontal distances $x=0, 0.5, 1.0, 1.5, \dots$ m.
Type the pairs $(x, y)$ into a spreadsheet and draw a scatter chart. Add a polynomial trendline of order 2 (Google Sheets and Excel both offer this) and display its equation $y=ax^2+bx+c$.
Use the equation: substitute values of $x$ until you find where $y$ is $0$ (or as close to $0$ as you can). That zero is your predicted landing distance.
Compare the prediction with the landing point you actually measured. Repeat with the tap opened further.

Tools: Garden hose, tape measure, phone camera, spreadsheet (Google Sheets / Excel)

Hand in: Your table of measurements, the chart with its trendline equation, the predicted and measured landing distances, and two sentences on why they differ.

Going further. Is the coefficient $a$ in your equation positive or negative? Explain how you could have known before doing any calculation.

Teacher/parent note: for a horizontal jet the fitted $a$ is negative (∩ shape) and $b$ is close to $0$; the predicted zero usually agrees with the measured landing point to within about $10\%$. Air resistance and reading errors from the photo explain most of the difference. Water play outdoors — choose a place where getting wet is fine.

🧪 Self-check quiz

0/5 correct

Q1. The degree of $5x^3-2x^7+x-1$ is

Q2. Which of these is not a polynomial?

Q3. The zero of $4x-3$ is

Q4. The graph of a quadratic polynomial just touches the $x$-axis at one point. Then the polynomial has

Q5. The greatest number of zeroes a cubic polynomial can have is

← Chapter 2 — Polynomials 2.2 Relationship between Zeroes and Coefficients →