Polynomials
Why This Matters
A polynomial is just an expression built from a variable using adding, subtracting and whole-number powers — things like x² − 3x − 4. They’re the workhorses of all of maths and science: the path of a thrown ball, the area of a widening square garden, the profit of a shop as it changes its price — all are described by polynomials.
The single most important question you can ask about a polynomial is: for what value of x does it become 0? Those special inputs are called its zeroes, and they’re everywhere in disguise — the moment a ball hits the ground, the break-even price of a business, the points where a bridge cable touches its supports. Solving almost any real problem eventually comes down to finding a zero.
In this chapter you’ll see two beautiful things. First, zeroes have a picture: they are exactly the spots where the polynomial’s graph crosses the x-axis. Second, the zeroes are secretly tied to the coefficients — without solving anything, the numbers in front of x already tell you the sum and product of the zeroes. That hidden link will save you again and again.
The Big Idea
A zero of a polynomial p(x) is a value of x that makes p(x) = 0. Pictured on a graph, the zeroes are precisely the x-intercepts — where y = p(x) meets the x-axis. And the zeroes aren’t independent of the coefficients: for a quadratic ax² + bx + c, the sum of the zeroes is −b/a and the product is c/a.
Let’s Break It Down
Polynomials, degree and zeroes
The degree of a polynomial is its highest power of the variable. By degree we name the common types:
| Degree | Name | General form | Example |
|---|---|---|---|
| 1 | Linear | ax + b | 2x − 3 |
| 2 | Quadratic | ax² + bx + c | x² − 3x − 4 |
| 3 | Cubic | ax³ + bx² + cx + d | 2x³ − 5x² − 14x + 8 |
(In every case a ≠ 0 — otherwise the leading term vanishes and the degree drops.) Expressions like 1/x or √x + 2 are not polynomials: powers must be whole numbers.
The value of p(x) at x = k, written p(k), is what you get by substituting x = k. A real number k is a zero of p(x) if p(k) = 0. For example, with p(x) = x² − 3x − 4:
- p(−1) = (−1)² − 3(−1) − 4 = 1 + 3 − 4 = 0, and
- p(4) = 16 − 12 − 4 = 0,
so −1 and 4 are the zeroes of x² − 3x − 4.
For a linear polynomial ax + b, setting ax + b = 0 gives the single zero x = −b/a — already a hint that zeroes and coefficients are linked.
The geometrical meaning of a zero
Here’s the picture that makes everything click. If you plot y = p(x), the zeroes are exactly the x-coordinates where the graph crosses (or touches) the x-axis — because crossing the axis means y = 0, which means p(x) = 0.
A quadratic always graphs as a parabola — a smooth U-shape that opens upward when a > 0 and downward when a < 0. Watch how the zeroes of x² − 3x − 4 (which we found to be −1 and 4) appear as its two x-intercepts:
Because a parabola can sit relative to the x-axis in only three ways, a quadratic has either two, one, or no real zeroes:
This is the geometry behind a key fact:
A polynomial of degree n has at most n zeroes — because its graph can cross the x-axis at most n times. So a quadratic has at most 2 zeroes, a cubic at most 3.
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 the x-intercepts, and the graph meets the x-axis at 3 points, so p(x) has 3 zeroes.
- A cubic has degree 3, and a degree-n polynomial has at most n zeroes — so at most 3.
- 3 zeroes is exactly the maximum allowed for a cubic, so yes, it’s allowed (this cubic has the most zeroes it possibly can).
Zeroes are tied to the coefficients
Now the surprise. Take any quadratic and factorise it — say p(x) = 2x² − 8x + 6 = 2(x − 1)(x − 3), so its zeroes are 1 and 3. Look at what the coefficients “know”:
- Sum of zeroes = 1 + 3 = 4 = −(−8)/2 = −b/a
- Product of zeroes = 1 × 3 = 3 = 6/2 = c/a
This is not a coincidence — it’s true for every quadratic, and we can prove it.
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.
- If α and β are the zeroes, then (x − α) and (x − β) are factors, so ax² + bx + c = a(x − α)(x − β) for the leading coefficient a.
- Expand the right side: a(x − α)(x − β) = a[x² − (α + β)x + αβ] = ax² − a(α + β)x + aαβ.
- Match this term-by-term with ax² + bx + c: the x-coefficients give b = −a(α + β), and the constants give c = aαβ.
- Divide through by a: α + β = −b/a and αβ = c/a. ∎ (Sum = −b/a, Product = c/a — worth memorising.)
A handy mnemonic: sum = −(coefficient of x)/(coefficient of x²) and product = (constant)/(coefficient of x²).
Find the zeroes of x² + 7x + 10 and verify the sum and product against the coefficients.
- Factorise: x² + 7x + 10 = (x + 2)(x + 5). So it’s zero when x = −2 or x = −5 — the zeroes are −2 and −5.
- Here a = 1, b = 7, c = 10. Sum of zeroes = −2 + (−5) = −7, and −b/a = −7/1 = −7. ✓
- Product of zeroes = (−2)(−5) = 10, and c/a = 10/1 = 10. ✓ The relationship holds.
The link runs both ways — if you’re given the sum and product, you can build the quadratic, because a polynomial with zeroes summing to S and multiplying to P is x² − Sx + P (taking a = 1).
Find a quadratic polynomial whose zeroes have sum −3 and product 2.
- Use x² − (sum)x + (product). Here sum = −3 and product = 2.
- So the polynomial is x² − (−3)x + 2 = x² + 3x + 2.
- Check by factorising: x² + 3x + 2 = (x + 1)(x + 2), zeroes −1 and −2 — sum −3 ✓, product 2 ✓. (Any multiple k(x² + 3x + 2) also works.)
The same idea for cubics
A cubic ax³ + bx² + cx + d with zeroes α, β, γ obeys three matching relations (proved the same way, by expanding a(x − α)(x − β)(x − γ)):
- 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 with the signs: −b/a, then +c/a, then −d/a, alternating as you go down the coefficients.
Common Mistakes
The 'zero' of a polynomial means the value of the polynomial at x = 0, i.e. p(0).
The word 'zero' naturally makes you think of x = 0, and p(0) is an easy number to compute, so the two get mixed up.
A zero is an INPUT x that makes the OUTPUT p(x) = 0 — a solution of p(x) = 0. p(0) is just the value at x = 0 (the constant term, the y-intercept) and is usually NOT zero.
For ax² + bx + c, the sum of the zeroes is b/a and the product is −c/a.
The formulas involve b/a and c/a, so it's easy to misremember which one carries the minus sign.
It's sum = −b/a (the minus is on the SUM) and product = +c/a. A quick check: x² − 5x + 6 has zeroes 2 and 3 — sum 5 = −(−5)/1 ✓, product 6 = 6/1 ✓.
Every quadratic has two real zeroes (the graph always cuts the x-axis twice).
Most textbook quadratics are chosen to factorise nicely into two real zeroes, so it feels like the norm.
A parabola can also just TOUCH the axis (one repeated zero) or MISS it entirely (no real zeroes) — like x² + 1, which is never 0 for any real x. A quadratic has at most two real zeroes, but it can have one or none.
x² + 1 has the zeroes 1 and −1.
x² − 1 factorises to (x−1)(x+1) with zeroes ±1, and x² + 1 looks almost the same, so the sign slips.
x² + 1 = 0 needs x² = −1, which no real number satisfies — it has NO real zeroes (its parabola sits entirely above the axis). It's x² − 1 that has zeroes ±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
Find the zeroes of 4u² + 8u and verify the relationship with the coefficients.
Factorise: 4u² + 8u = 4u(u + 2). It is zero when u = 0 or u + 2 = 0, so the zeroes are 0 and −2.
Here a = 4, b = 8, c = 0. Sum = 0 + (−2) = −2, and −b/a = −8/4 = −2 ✓.
Product = 0 × (−2) = 0, and c/a = 0/4 = 0 ✓.
Find a quadratic polynomial whose zeroes have sum 0 and product −15.
Use x² − (sum)x + (product) = x² − (0)x + (−15) = x² − 15.
Check: x² − 15 = 0 gives x = ±√15, which sum to 0 ✓ and multiply to −15 ✓.
Medium
Find the zeroes of 6x² − 7x − 3 and verify the sum and product against the coefficients.
Split the middle term: 6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3).
Zeroes: 3x + 1 = 0 → x = −1/3; and 2x − 3 = 0 → x = 3/2. So the zeroes are −1/3 and 3/2.
Here a = 6, b = −7, c = −3. Sum = −1/3 + 3/2 = −2/6 + 9/6 = 7/6, and −b/a = 7/6 ✓.
Product = (−1/3)(3/2) = −3/6 = −1/2, and c/a = −3/6 = −1/2 ✓.
Find the zeroes of x² − 3 and verify the relationship between zeroes and coefficients.
Use the identity a² − b² = (a − b)(a + b): x² − 3 = (x − √3)(x + √3).
So it is zero when x = √3 or x = −√3 — the zeroes are √3 and −√3.
Here a = 1, b = 0, c = −3. Sum = √3 + (−√3) = 0, and −b/a = 0 ✓.
Product = (√3)(−√3) = −3, and c/a = −3/1 = −3 ✓.
Challenge
If α and β are the zeroes of x² − 5x + 6, find the value of α² + β² without finding α and β individually.
From the coefficients: α + β = −(−5)/1 = 5, and αβ = 6/1 = 6.
Use the identity α² + β² = (α + β)² − 2αβ. This lets us answer using only the sum and product.
α² + β² = (5)² − 2(6) = 25 − 12 = 13.
(Check: the zeroes are 2 and 3, and 2² + 3² = 4 + 9 = 13 ✓. The point is we never needed to solve for them.)
Summary
You should now be able to explain:
- A polynomial’s degree is its highest power; degrees 1, 2, 3 are linear, quadratic, cubic (general form ax² + bx + c for a quadratic, a ≠ 0).
- A zero is an input k with p(k) = 0 — not the same as p(0).
- Geometrically, the zeroes of p(x) are the x-coordinates where y = p(x) meets the x-axis.
- A quadratic graphs as a parabola (up if a > 0, down if a < 0) and has two, one or no real zeroes; a polynomial of degree n has at most n zeroes.
- For a quadratic ax² + bx + c: sum of zeroes = −b/a, product = c/a (both provable by expanding a(x − α)(x − β)).
- Reverse it: a quadratic with given 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 a graph meets the x-axis. Next, in Pair of Linear Equations in Two Variables, we put two straight-line graphs on the same axes and ask where they meet each other — the point that satisfies both equations at once. The same picture-meets-algebra idea, one dimension up.