Polynomials

Chapter 2 · Mathematics · Class 10 30 min read

Why This Matters

Before we start, a few words from earlier classes turn up again and again here — variable, coefficient, constant term. A 30-second refresher so they’re fresh:

A polynomial is an expression made from a variable (like x) using only three things: adding, subtracting, and whole-number powers. For example, x² − 3x − 4 is a polynomial. Polynomials show up everywhere in maths and science. The path of a ball you throw is a polynomial. So is the area of a square garden as it grows bigger. So is the profit of a shop as it changes its price. They are one of the most useful tools you will learn.

The biggest question we ask about any polynomial is simple: for what value of x does it become 0? The x-values that make this happen are called its zeroes. Zeroes are hiding inside many real problems. The moment a ball hits the ground is a zero. The price at which a business neither gains nor loses money is a zero. Solving a real problem very often means finding a zero.

In this chapter you’ll see two lovely ideas. First, zeroes have a picture: they are the exact spots where the polynomial’s graph crosses the x-axis. Second, the zeroes are secretly connected to the coefficients (the numbers in front of x). Without solving anything, those numbers already tell you the sum and product of the zeroes. This hidden link will help you again and again.

The Big Idea

A zero of a polynomial p(x) is a value of x that makes p(x) equal to 0. On a graph, the zeroes are the x-intercepts — the points where the curve y = p(x) touches or crosses the x-axis. The zeroes are also linked to the coefficients. For a quadratic ax² + bx + c, the sum of the zeroes is −b/a and the product of the zeroes is c/a.

Let’s Break It Down

Polynomials, degree and zeroes

The degree of a polynomial is the highest power of the variable in it. For example, in x² − 3x − 4 the highest power is 2, so its degree is 2. We give polynomials names based on their degree:

Naming polynomials by degree
DegreeNameGeneral formExample
1Linearax + b2x − 3
2Quadraticax² + bx + cx² − 3x − 4
3Cubicax³ + bx² + cx + d2x³ − 5x² − 14x + 8

(In every case a ≠ 0. If a were 0, the highest term would disappear and the degree would drop.) Expressions like 1/x or √x + 2 are not polynomials. This is because the powers in a polynomial must be whole numbers. Here 1/x means x to the power −1, and √x means x to the power 1/2 — neither is a whole number.

The value of p(x) at x = k is written p(k). You get it by putting k in place of x everywhere in the polynomial. We say a number k is a zero of p(x) when p(k) = 0. For example, take p(x) = x² − 3x − 4:

  • p(−1) = (−1)² − 3(−1) − 4 = 1 + 3 − 4 = 0, and
  • p(4) = 16 − 12 − 4 = 0,

Both inputs give an output of 0, so −1 and 4 are the zeroes of x² − 3x − 4.

For a linear polynomial ax + b, we find the zero by setting ax + b = 0 and solving for x. This gives one zero: x = −b/a. Notice that the zero is built from the coefficients a and b. That is an early hint that zeroes and coefficients are linked.

The geometrical meaning of a zero

To picture zeroes on a graph, you first need to be comfortable reading points off the x and y axes. Here’s a quick reminder of how that works:

Here is the picture that makes everything clear. Suppose you draw the graph of y = p(x). The zeroes are the x-values where the graph crosses or touches the x-axis. Why? On the x-axis, y is always 0. So at those points y = p(x) = 0, which is exactly what a zero means.

A quadratic always makes a curve called a parabola — a smooth U-shape. It opens upward when a > 0 and downward when a < 0. Figure 2.1 below shows how the zeroes of x² − 3x − 4 (which we found to be −1 and 4) show up as the two points where the curve meets the x-axis:

The upward parabola y = x² − 3x − 4 crossing the x-axis at x = −1 and x = 4, with those two intercepts marked as the zeroes.
Figure 2.1 — The upward (U-shaped) parabola y = x² − 3x − 4 drawn on the coordinate axes (x-axis left-right, y-axis up-down). The red curve dips below the axis and crosses it at two points, marked with red dots at (−1, 0) and (4, 0). Its lowest point, the vertex, sits between them. Those two crossing points are exactly the zeroes −1 and 4 we found by substitution: a zero of the polynomial is the same thing as an x-intercept of its graph.

A parabola can sit against the x-axis in only three ways. So a quadratic can have either two, one, or no real zeroes. Figure 2.2 below lays out all three side by side:

Three upward parabolas: one crossing the x-axis at two points (two distinct zeroes), one touching at a single point (one repeated zero), and one sitting entirely above the axis (no real zeroes).
Figure 2.2 — Three upward parabolas showing the only three ways a quadratic graph can meet the x-axis. (a) Two distinct zeroes: the curve dips below the axis and cuts through it at two separate points (two red dots). (b) One zero, repeated: the curve just touches the axis at a single point and turns back up (one red dot). (c) No real zeroes: the whole curve stays above the axis and never touches it. A downward parabola (a < 0) gives the same three cases, flipped upside-down.

This picture explains a key fact:

A polynomial of degree n has at most n zeroes. So a quadratic (degree 2) has at most 2 zeroes, and a cubic (degree 3) has at most 3.

The graph shows this is believable — a parabola can only meet the axis up to twice. But why is the limit exactly the degree, and never more? Here is the real reason, and it doesn’t need a graph at all.

Each different zero quietly “uses up” one factor. We’ll prove this properly in a moment (it’s called the factor theorem), but the idea is short: if α is a zero, then (x − α) is a factor of the polynomial. Two different zeroes mean two different factors, like (x − α)(x − β). And factors multiply together, so their degrees add up — two linear factors make degree 2, three make degree 3, and so on.

Now watch what goes wrong if there were too many zeroes. Suppose a quadratic — degree 2 — had 3 different zeroes. Then it would have 3 factors, (x − α)(x − β)(x − γ), which is degree 3. But the polynomial is only degree 2. A polynomial can’t be degree 2 and degree 3 at the same time. So this is impossible. The most zeroes a degree-2 polynomial can have is 2. The same reasoning caps a cubic at 3, and a degree-n polynomial at n. Figure 2.3 below walks through this argument:

Each distinct zero adds one factor of the form (x minus zero). Two zeroes give (x minus alpha)(x minus beta), degree 2. Three zeroes give three factors, degree 3. If a degree-2 quadratic had 3 different zeroes it would be degree 3, which contradicts degree 2, so that is impossible. Hence a quadratic has at most 2 zeroes, a cubic at most 3, and degree n at most n.
Figure 2.3 — Why a degree-n polynomial has at most n zeroes. The two blue rows show the counting rule: 2 zeroes force the two factors (x − α)(x − β), which is degree 2; 3 zeroes force three factors, which is degree 3 — each distinct zero adds one more factor and factors' degrees add up. The red box shows the contradiction: if a quadratic (degree 2) had 3 zeroes it would be degree 3, but degree 3 ≠ degree 2, which is impossible. The green box states the conclusion: a quadratic has at most 2 zeroes, a cubic at most 3, and a degree-n polynomial at most n.

Let’s put that “at most n” rule to work on a cubic whose graph cuts the axis three times:

Reading zeroes off a graph

A graph of y = p(x) crosses the x-axis at exactly 3 points and p(x) is a cubic. How many zeroes does it have, and is that allowed?

Zeroes are tied to the coefficients

Now for the surprise. Take any quadratic and factorise it. For example, p(x) = 2x² − 8x + 6 = 2(x − 1)(x − 3), so its zeroes are 1 and 3. Now look at what the coefficients already tell us:

  • Sum of zeroes = 1 + 3 = 4. And −b/a = −(−8)/2 = 4. Same answer.
  • Product of zeroes = 1 × 3 = 3. And c/a = 6/2 = 3. Same answer.

This is not luck. It is true for every quadratic, and we can prove it.

The proof leans on one quiet fact we keep using: if α is a zero of a polynomial, then (x − α) is a factor of it. This is called the factor theorem. It sounds like a big claim, so let’s see exactly why it must be true — it comes straight from ordinary division.

Dividing p(x) by (x minus alpha) gives p(x) equals (x minus alpha) times q(x) plus a number r. Putting x equals alpha makes the (x minus alpha) part zero, so p(alpha) equals r. If alpha is a zero then p(alpha) is zero, so r is zero and (x minus alpha) divides p(x) exactly — it is a factor.
Figure 2.4 — The factor theorem proved in four steps. Top blue box: dividing p(x) by (x − α) always gives p(x) = (x − α)·q(x) + r, where the remainder r is just a number. Middle lines: put x = α, so the (x − α) part becomes 0·q(α) and vanishes, leaving p(α) = r. Orange box: so the remainder r equals p(α). Bottom green box: if α is a zero then p(α) = 0, so r = 0 and the division is exact — p(x) = (x − α)·q(x), meaning (x − α) is a factor.

With the factor theorem in hand, the proof of the sum and product rules falls out — its very first step is just to write the zeroes as factors:

Proof: sum and product of the zeroes of ax² + bx + c

If α and β are the zeroes of p(x) = ax² + bx + c (a ≠ 0), show that α + β = −b/a and αβ = c/a.

That whole proof is just one simple idea. We wrote the same quadratic in two ways, placed them one above the other, and matched the matching terms. This is much easier to see in a picture than to read in words. Figure 2.5 below lines the two forms up column by column:

A two-row line-up: the expanded factor form a x squared minus a(alpha plus beta) x plus a alpha beta sits above the standard form a x squared plus b x plus c, with matching columns linked by dashes. The x-terms give b equals minus a times (alpha plus beta), so alpha plus beta equals minus b over a; the constant terms give c equals a alpha beta, so alpha beta equals c over a.
Figure 2.5 — Matching the quadratic written two ways. Top row (factor form): ax² − a(α + β)x + a·αβ, the result of building the quadratic from its zeroes. Bottom row (standard form): ax² + bx + c. The three columns — the x² terms, the x terms, and the constants — are lined up, with dashed leader lines joining the matching pieces. Comparing the x terms gives b = −a(α + β), so α + β = −b/a (the orange box, sum of zeroes). Comparing the constants gives c = a·αβ, so αβ = c/a (the green box, product of zeroes).

An easy way to remember it: sum = −(coefficient of x)/(coefficient of x²) and product = (constant)/(coefficient of x²).

There is also a nice way to see both rules right on the graph. A parabola is symmetric — its left and right halves are mirror images. So its two zeroes sit at equal distances from the centre line, called the axis of symmetry. That centre line is at x = −b/2a, which is the average (the middle) of the two zeroes. If you double the average, you get the sum, which is −b/a. Also, the curve crosses the y-axis at the value c. Since c = a·αβ, dividing by a gives the product c/a. Figure 2.6 below shows both rules on the actual graph:

The parabola y = x squared minus 3x minus 4 with zeroes alpha = minus 1 and beta = 4 marked on the x-axis. A dashed vertical axis of symmetry at x = 1.5 sits exactly halfway between them, each zero being 2.5 away, so the zeroes add to minus b over a = 3. The graph crosses the y-axis at minus 4, which equals a times the product of the zeroes, so the product is c over a = minus 4.
Figure 2.6 — Reading the sum and product of the zeroes straight off the graph of y = x² − 3x − 4. The two zeroes are marked on the x-axis as red dots: α = −1 on the left, β = 4 on the right. The green dashed vertical line is the axis of symmetry at x = 1.5, sitting exactly halfway between them — each zero is the same distance 2.5 from it (shown by the two green segments). Because they balance around x = (α + β)/2 = −b/2a, the zeroes add to −b/a = 3. The blue dot where the curve crosses the y-axis is the y-intercept (0, c) = (0, −4); since c = a·αβ, the product is αβ = c/a = −4.

Let’s test these two rules on a fresh quadratic — find its zeroes, then check the sum and product against the coefficients:

Find the zeroes and verify the relationship

Find the zeroes of x² + 7x + 10 and verify the sum and product against the coefficients.

This link works both ways. If someone gives you the sum and product of the zeroes, you can build the quadratic back. A polynomial whose zeroes add up to S and multiply to P is x² − Sx + P (here we take a = 1).

Let’s use that to build a quadratic when we’re only told the sum and product of its zeroes:

Build a quadratic from its zeroes

Find a quadratic polynomial whose zeroes have sum −3 and product 2.

The same idea for cubics

A cubic ax³ + bx² + cx + d with zeroes α, β, γ follows three matching rules. We prove them the same way as before — by expanding a(x − α)(x − β)(x − γ) and matching terms:

  • α + β + γ = −b/a (sum of zeroes)
  • αβ + βγ + γα = c/a (sum of products, two at a time)
  • αβγ = −d/a (product of zeroes)

Notice the pattern in the signs: −b/a, then +c/a, then −d/a. The sign keeps flipping (minus, plus, minus) as you move down the coefficients.

Figure 2.7 below shows where all three cubic rules come from — it’s the very same matching trick, just one degree higher:

The factor form of a cubic, a times (x minus alpha)(x minus beta)(x minus gamma), expanded and lined up under a x cubed plus b x squared plus c x plus d. Matching terms gives three relations: alpha plus beta plus gamma equals minus b over a (the sum), alpha beta plus beta gamma plus gamma alpha equals c over a (the pairs), and alpha beta gamma equals minus d over a (the product). The signs alternate minus, plus, minus down the coefficients.
Figure 2.7 — The same matching trick for a cubic. The top line is the expanded factor form ax³ − a(α+β+γ)x² + a(αβ+βγ+γα)x − a·αβγ; the line below it is the standard form ax³ + bx² + cx + d. Matching the columns gives three relations, shown in colour-coded boxes: the x² terms give α + β + γ = −b/a (the sum), the x terms give αβ + βγ + γα = c/a (the pairs), and the constants give αβγ = −d/a (the product). Notice the sign pattern down the coefficients: −b/a, then +c/a, then −d/a (minus, plus, minus). A quadratic is just the first and last of these three.

Common Mistakes

⚠️ Common mistake
What students think

The 'zero' of a polynomial means the value of the polynomial at x = 0, that is p(0).

Why it seems right

The word 'zero' makes you think of x = 0. And p(0) is easy to work out. So the two ideas get mixed up.

What actually happens

A zero is an INPUT x that makes the OUTPUT p(x) = 0. It is a solution of p(x) = 0. But p(0) is just the value when x = 0 (it is the constant term, where the graph meets the y-axis). p(0) is usually NOT zero.

⚠️ Common mistake
What students think

For ax² + bx + c, the sum of the zeroes is b/a and the product is −c/a.

Why it seems right

Both formulas use b/a and c/a. So it is easy to forget which one has the minus sign.

What actually happens

The correct rule is sum = −b/a (the minus goes with the SUM) and product = +c/a. Quick check: x² − 5x + 6 has zeroes 2 and 3. Sum 5 = −(−5)/1 ✓, product 6 = 6/1 ✓.

⚠️ Common mistake
What students think

Every quadratic has two real zeroes (the graph always cuts the x-axis twice).

Why it seems right

Most quadratics in textbooks are picked so they factorise nicely into two real zeroes. So having two zeroes starts to feel normal.

What actually happens

A parabola can also just TOUCH the x-axis (one repeated zero), or MISS it completely (no real zeroes). For example, x² + 1 is never 0 for any real x. So a quadratic has at most two real zeroes, but it may have one or even none.

⚠️ Common mistake
What students think

x² + 1 has the zeroes 1 and −1.

Why it seems right

x² − 1 factorises to (x−1)(x+1) with zeroes 1 and −1. And x² + 1 looks almost the same, so the sign gets copied by mistake.

What actually happens

To find the zeroes of x² + 1, set x² + 1 = 0, which means x² = −1. No real number squared gives a negative answer. So x² + 1 has NO real zeroes (its parabola stays fully above the x-axis). It is x² − 1 that has zeroes 1 and −1.

Quick Check

Which of these is NOT a polynomial?

What are the zeroes of the polynomial whose graph is shown crossing the x-axis at x = 2 and x = −5?

For the quadratic 3x² + 5x − 2, what is the sum of its zeroes?

At most how many zeroes can a cubic polynomial have?

Practice Problems

Easy

easy

Find the zeroes of 4u² + 8u and verify the relationship with the coefficients.

easy

Find a quadratic polynomial whose zeroes have sum 0 and product −15.

Medium

These next ones don’t factorise at a glance, so we’ll lean on “splitting the middle term” from Class 9. Here’s how that method works:

medium

Find the zeroes of 6x² − 7x − 3 and verify the sum and product against the coefficients.

The next problem has no middle term to split, but a square-minus-a-square identity unlocks it. A quick reminder of that identity:

medium

Find the zeroes of x² − 3 and verify the relationship between zeroes and coefficients.

Challenge

challenge

If α and β are the zeroes of x² − 5x + 6, find the value of α² + β² without finding α and β individually.

Summary

You should now be able to explain:

  • A polynomial’s degree is its highest power. Degrees 1, 2, 3 are called linear, quadratic, cubic. A quadratic has the form ax² + bx + c with a ≠ 0.
  • A zero is an input k that makes p(k) = 0. It is not the same as p(0).
  • On a graph, the zeroes of p(x) are the x-values where y = p(x) meets the x-axis.
  • A quadratic’s graph is a parabola (opens up if a > 0, down if a < 0). It has two, one or no real zeroes. In general, a polynomial of degree n has at most n zeroes.
  • For a quadratic ax² + bx + c: sum of zeroes = −b/a and product = c/a. (You can prove both by expanding a(x − α)(x − β).)
  • The other way round: a quadratic with sum S and product P is x² − Sx + P.
  • For a cubic ax³ + bx² + cx + d: α + β + γ = −b/a, αβ + βγ + γα = c/a, αβγ = −d/a.

What’s Next

You now know that zeroes are where one graph meets the x-axis. Next, in Pair of Linear Equations in Two Variables, we draw two straight-line graphs on the same axes. Then we ask where they meet each other. That meeting point is the one solution that fits both equations at the same time. It is the same idea — picture meeting algebra — taken one step further.

Frequently Asked Questions

What is a zero of a polynomial and how do you find it?

A zero of a polynomial p(x) is a value of x that makes p(x) equal to 0. To find it, set the polynomial equal to zero and solve for x. For example, for p(x) = x² − 3x − 4, solving x² − 3x − 4 = 0 gives x = 4 and x = −1, so 4 and −1 are its zeroes.

What does the geometrical meaning of zeroes of a polynomial mean?

When you draw the graph of y = p(x), the zeroes are the x-coordinates of the points where the graph meets the x-axis. A linear polynomial's graph is a straight line crossing the x-axis at one point. A quadratic's graph is a U-shaped curve (parabola) that can cross the x-axis at two points, one point, or not at all.

What is the relationship between zeroes and coefficients of a quadratic polynomial?

For a quadratic polynomial ax² + bx + c with zeroes α and β, the sum of zeroes α + β = −b/a and the product of zeroes αβ = c/a. These formulas let you find the sum and product of zeroes directly from the coefficients, without solving the equation first.

How many zeroes can a polynomial of degree n have?

A polynomial of degree n has at most n zeroes. So a linear polynomial (degree 1) has exactly 1 zero, a quadratic (degree 2) has at most 2 zeroes, and a cubic (degree 3) has at most 3 zeroes. The graph of a degree-n polynomial can cross the x-axis at most n times.

How do you form a quadratic polynomial if you know the sum and product of its zeroes?

Use the formula x² − (sum of zeroes)x + (product of zeroes). For example, if the sum of zeroes is 3 and the product is −4, the quadratic is x² − 3x − 4. This works because expanding (x − α)(x − β) always gives x² − (α + β)x + αβ.