Polynomials • Topic 1 of 2

Zeros of a Polynomial

What is a zero of a polynomial? A zero of a polynomial p(x) is a value of x for which p(x) = 0. In other words, it is the value that makes the entire polynomial equal to zero. Zeros are also called roots of the polynomial.

Example: For p(x) = x − 3, when x = 3, p(3) = 3 − 3 = 0. So 3 is a zero of p(x).

Real-life analogy: Think of a polynomial as a machine. You put a number (x) into the machine, and it gives an output. A zero is that special input number that makes the machine output exactly zero — like finding the exact temperature where water freezes (0°C) on a thermometer scale.

Key properties:

  • A linear polynomial (degree 1, like ax + b) has exactly 1 zero, namely \(x=-\tfrac{b}{a}\).
  • A quadratic polynomial (degree 2, like \(ax^{2}+bx+c\)) has at most 2 zeros.
  • A cubic polynomial (degree 3) has at most 3 zeros.
  • In general a polynomial of degree n has at most n zeros.
  • Graphically, the zeros of a polynomial are the x‑intercepts (where the graph crosses or touches the x‑axis).
  • Finding zeros means solving the equation p(x) = 0.

Geometrical meaning of the zeroes of a polynomial

If we draw the graph of y = p(x), the zeroes are exactly the x-coordinates of the points where the graph meets the x-axis. This is because on the x-axis the y-value is 0, and p(x) = 0 is precisely what defines a zero. So you can literally read the zeroes off a graph by spotting where the curve cuts (or touches) the x-axis.

Linear polynomial p(x) = ax + b: its graph is a straight line, and a straight line that is not horizontal cuts the x-axis at exactly one point. Hence a linear polynomial has exactly one zero.

Quadratic polynomial \(p(x)=ax^{2}+bx+c\): its graph is a parabola (opening upwards when a > 0, downwards when a < 0). Reading the graph, three situations arise:

  • The parabola cuts the x-axis at two distinct points → two distinct zeroes (happens when \(b^{2}-4ac>0\)).
  • The parabola just touches the x-axis at one point → two equal zeroes / one repeated zero (when \(b^{2}-4ac=0\)).
  • The parabola lies entirely above or below the x-axis → no real zero (when \(b^{2}-4ac<0\)).

Cubic polynomial \(p(x)=ax^{3}+bx^{2}+cx+d\): its graph can cross the x-axis at one, two or three points, so a cubic has at most three zeroes. Counting the x-axis crossings on the graph tells you the number of real zeroes directly.

How to find zeroes algebraically: Set p(x) = 0 and solve. For a quadratic you can usually factorise by splitting the middle term: rewrite \(ax^{2}+bx+c\) by splitting b into two parts whose product is a×c. For example \(x^{2}-5x+6\) splits as \(x^{2}-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)\), giving zeroes 2 and 3.

Division Algorithm for polynomials

Just as with whole numbers (Dividend = Divisor × Quotient + Remainder), polynomials obey the Division Algorithm: if p(x) and g(x) are polynomials with g(x) ≠ 0, then there exist unique polynomials q(x) (quotient) and r(x) (remainder) such that

\(p(x)=g(x)\cdot q(x)+r(x)\), where either r(x) = 0 or degree of r(x) < degree of g(x).

We carry this out by long division: divide the highest-degree term of the dividend by the highest-degree term of the divisor, multiply back, subtract, bring down the next term, and repeat until the remainder has degree less than the divisor.

Using division to find the remaining zeroes: if you already know one or more zeroes of a polynomial, each known zero c gives a factor (x − c). Divide p(x) by that known factor; the quotient is a lower-degree polynomial whose zeroes are the remaining zeroes. For a cubic with one known zero you divide out (x − c) and are left with a quadratic you can factorise.

Common mistakes to avoid:

  • Confusing the value p(c) with a zero — a zero is a value of x, not the output.
  • Forgetting that “touching” the x-axis still counts as two equal zeroes, not one.
  • While doing long division, not writing missing-power terms as 0 (e.g. write \(x^{3}+0\cdot x^{2}+0\cdot x-8\) so columns line up).
  • Sign slips when subtracting during long division — subtracting a negative adds.
  • Stopping factorisation too early: a quadratic factor like \(x^{2}-4\) still splits into (x − 2)(x + 2).
Graph of the linear polynomial p(x) = 2x − 4 cutting the x-axis once at x = 2 A line has exactly one zero (one x-intercept) x y p(x) = 2x − 4 x = 2 zero = x-coordinate where the graph meets the x-axis Parabola of p(x) = x squared minus 4 cutting the x-axis at x = −2 and x = 2, its two zeros A parabola can cut the x-axis at two points x y p(x) = x^2 − 4 x = −2 x = 2 2 x-axis crossings → 2 zeros (here −2 and 2) Three cases for a quadratic graph: two zeros, two equal zeros, and no real zero Number of real zeros = number of x-axis crossings Two zeros b^2 − 4ac > 0 Two equal zeros b^2 − 4ac = 0 No real zero b^2 − 4ac < 0
Solved Examples
1
Worked Example
Check whether x = 2 is a zero of the polynomial p(x) = \(x^{2}-4x+4\).
Solution
  1. Step 1: Substitute x = 2 into p(x): \(p(2)=(2)^{2}-4(2)+4\).
  2. Step 2: Calculate: 4 − 8 + 4 = 0.
  3. Step 3: Since p(2) = 0, x = 2 is a zero.

Answer: Yes, x = 2 is a zero.

2
Worked Example
Find the zero of the linear polynomial p(x) = 5x − 15.
Solution
  1. Step 1: Set p(x) = 0, so 5x − 15 = 0.
  2. Step 2: Add 15 to both sides: 5x = 15.
  3. Step 3: Divide both sides by 5: x = 3.
  4. Step 4: Verify: 5(3) − 15 = 15 − 15 = 0.

Answer: The zero is x = 3.

3
Worked Example
Find the zeroes of the quadratic polynomial \(x^{2}-5x+6\) by splitting the middle term.
Solution
  1. Step 1: We need two numbers whose product is a×c = 1×6 = 6 and whose sum is b = −5. These are −2 and −3.
  2. Step 2: Split the middle term: \(x^{2}-2x-3x+6\).
  3. Step 3: Group: \(x(x-2)-3(x-2)=(x-2)(x-3)\).
  4. Step 4: Set each factor to zero: x − 2 = 0 gives x = 2; x − 3 = 0 gives x = 3.

Answer: The zeroes are 2 and 3.

4
Worked Example
Find the zeroes of \(x^{2}-3x-10\).
Solution
  1. Step 1: Product a×c = 1×(−10) = −10; required sum b = −3. The numbers are −5 and +2.
  2. Step 2: \(x^{2}-5x+2x-10=x(x-5)+2(x-5)\).
  3. Step 3: Factorise: (x − 5)(x + 2) = 0.
  4. Step 4: x = 5 or x = −2.

Answer: The zeroes are 5 and −2.

5
Worked Example
Find the zeroes of \(6x^{2}-7x-3\) by splitting the middle term.
Solution
  1. Step 1: a×c = 6×(−3) = −18; required sum = −7. The numbers are −9 and +2.
  2. Step 2: \(6x^{2}-9x+2x-3\).
  3. Step 3: Group: \(3x(2x-3)+1(2x-3)=(2x-3)(3x+1)\).
  4. Step 4: 2x − 3 = 0 gives x = \(\tfrac{3}{2}\); 3x + 1 = 0 gives x = \(-\tfrac{1}{3}\).

Answer: The zeroes are \(\tfrac{3}{2}\) and \(-\tfrac{1}{3}\).

6
Worked Example
Find the zeroes of \(t^{2}-15\).
Solution
  1. Step 1: Set \(t^{2}-15=0\), so \(t^{2}=15\).
  2. Step 2: Take square roots on both sides: \(t=\pm\sqrt{15}\).

Answer: The zeroes are \(\sqrt{15}\) and \(-\sqrt{15}\).

7
Worked Example
Looking at a graph, a parabola y = p(x) touches the x-axis at exactly one point, x = 1, and lies above the x-axis everywhere else. How many zeroes does p(x) have and what are they?
Solution
  1. Step 1: The graph meets the x-axis only at x = 1, so 1 is a zero.
  2. Step 2: A quadratic that merely touches (does not cross) the x-axis has two equal zeroes, i.e. a repeated zero.
  3. Step 3: Therefore both zeroes are 1 (here p(x) is of the form \(k(x-1)^{2}\)).

Answer: Two equal zeroes, both x = 1.

8
Worked Example
The graph of a cubic polynomial cuts the x-axis at x = −1, x = 0 and x = 2. How many real zeroes does it have, and write a possible cubic.
Solution
  1. Step 1: Each x-axis crossing is a zero, so the zeroes are −1, 0 and 2 — three real zeroes.
  2. Step 2: Each zero c gives a factor (x − c): factors are (x + 1), x and (x − 2).
  3. Step 3: A possible cubic is \(p(x)=x(x+1)(x-2)\).
  4. Step 4: Expand: \(x(x^{2}-x-2)=x^{3}-x^{2}-2x\).

Answer: Three real zeroes (−1, 0, 2); a possible cubic is \(x^{3}-x^{2}-2x\).

9
Worked Example
Divide \(p(x)=2x^{2}+3x+1\) by g(x) = x + 2 and write the quotient and remainder.
Solution
  1. Step 1: Divide leading terms: \(2x^{2}\div x = 2x\). Multiply: 2x(x + 2) = \(2x^{2}+4x\).
  2. Step 2: Subtract: \((2x^{2}+3x+1)-(2x^{2}+4x)=-x+1\).
  3. Step 3: Divide leading terms: \(-x\div x=-1\). Multiply: −1(x + 2) = −x − 2.
  4. Step 4: Subtract: \((-x+1)-(-x-2)=3\). Degree of 3 is 0, less than degree of g(x), so stop.
  5. Step 5: Quotient q(x) = 2x − 1, remainder r(x) = 3. Check: (x + 2)(2x − 1) + 3 = \(2x^{2}+3x-2+3=2x^{2}+3x+1\).

Answer: Quotient = 2x − 1, Remainder = 3.

10
Worked Example
Divide \(p(x)=x^{3}-3x^{2}+5x-3\) by \(g(x)=x^{2}-2\) using the division algorithm.
Solution
  1. Step 1: \(x^{3}\div x^{2}=x\). Multiply: \(x(x^{2}-2)=x^{3}-2x\). Subtract: \((x^{3}-3x^{2}+5x-3)-(x^{3}-2x)=-3x^{2}+7x-3\).
  2. Step 2: \(-3x^{2}\div x^{2}=-3\). Multiply: \(-3(x^{2}-2)=-3x^{2}+6\). Subtract: \((-3x^{2}+7x-3)-(-3x^{2}+6)=7x-9\).
  3. Step 3: Degree of 7x − 9 is 1, less than degree 2 of g(x), so we stop.
  4. Step 4: Quotient q(x) = x − 3, remainder r(x) = 7x − 9.

Answer: Quotient = x − 3, Remainder = 7x − 9.

11
Worked Example
Find all zeroes of \(p(x)=x^{3}-4x^{2}+x+6\), given that x = 2 is a zero.
Solution
  1. Step 1: Since x = 2 is a zero, (x − 2) is a factor.
  2. Step 2: Divide p(x) by (x − 2). Long division gives quotient \(x^{2}-2x-3\) with remainder 0.
  3. Step 3: Factorise the quotient: \(x^{2}-2x-3=(x-3)(x+1)\).
  4. Step 4: Set each factor to zero: x = 2, x = 3, x = −1.

Answer: The zeroes are 2, 3 and −1.

12
Worked Example
Two zeroes of \(p(x)=2x^{4}-3x^{3}-3x^{2}+6x-2\) are \(\sqrt{2}\) and \(-\sqrt{2}\). Find the other two zeroes.
Solution
  1. Step 1: If \(\pm\sqrt{2}\) are zeroes then \((x-\sqrt{2})(x+\sqrt{2})=x^{2}-2\) is a factor of p(x).
  2. Step 2: Divide p(x) by \(x^{2}-2\). The division gives quotient \(2x^{2}-3x+1\) with remainder 0.
  3. Step 3: Factorise the quotient by splitting the middle term: \(2x^{2}-3x+1=2x^{2}-2x-x+1=2x(x-1)-1(x-1)=(x-1)(2x-1)\).
  4. Step 4: Set to zero: x − 1 = 0 gives x = 1; 2x − 1 = 0 gives x = \(\tfrac{1}{2}\).

Answer: The other two zeroes are 1 and \(\tfrac{1}{2}\).

Key Points

  • A zero of a polynomial p(x) is a value c such that p(c) = 0
  • Zeros are also called roots of the polynomial
  • A polynomial of degree n has at most n zeros
  • Graphically, zeros are the x‑intercepts — where the graph cuts or touches the x-axis
  • A parabola cutting the x-axis twice has two zeroes; touching it gives two equal zeroes; missing it gives no real zero
  • To find zeros, set the polynomial equal to zero and solve (factorise quadratics by splitting the middle term)
  • Linear polynomials (ax + b) have exactly one zero: x = −b/a
  • Division Algorithm: \(p(x)=g(x)\,q(x)+r(x)\) with r(x) = 0 or deg r < deg g
  • If a zero c is known, divide out the factor (x − c) to find the remaining zeroes
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.A zero of $p(x)$ is a value of $x$ for which:
Explanation: Where the polynomial vanishes.
Q2.The degree of a quadratic polynomial is:
Explanation: Highest power $2$.
Q3.A quadratic polynomial has at most:
Explanation: Degree $2\Rightarrow$ at most $2$ zeros.
Q4.The zero of $x-3$ is:
Explanation: $x-3=0\Rightarrow x=3$.
Practice more MCQs → 📝 Assignment · 35 marks