Polynomials • Topic 2 of 2

Relationship Between Zeros and Coefficients of Quadratic Polynomials

What is the relationship between zeros and coefficients? For a quadratic polynomial p(x) = \(ax^{2}+bx+c\) (a ≠ 0), if α (alpha) and β (beta) are its two zeros, then:

Sum of zeros: α + β = −(coefficient of x) / (coefficient of \(x^{2}\)) = \(-\tfrac{b}{a}\).

Product of zeros: α × β = (constant term) / (coefficient of \(x^{2}\)) = \(\tfrac{c}{a}\).

Where do these come from? If α and β are the zeroes, then \(ax^{2}+bx+c=a(x-\alpha)(x-\beta)\). Expanding the right side gives \(a\,x^{2}-a(\alpha+\beta)x+a\alpha\beta\). Comparing coefficients with \(ax^{2}+bx+c\): \(-a(\alpha+\beta)=b\) so \(\alpha+\beta=-\tfrac{b}{a}\), and \(a\alpha\beta=c\) so \(\alpha\beta=\tfrac{c}{a}\).

Why does this matter? These formulas let us:

  • Find the sum and product of zeros without solving the equation.
  • Construct a quadratic polynomial when we know its zeros.
  • Verify whether given values could be the zeros of a polynomial.
  • Evaluate symmetric expressions in the zeroes (such as \(\alpha^{2}+\beta^{2}\) or \(\tfrac{1}{\alpha}+\tfrac{1}{\beta}\)) quickly.

Real-life example: Imagine a rectangular garden whose length (α) and width (β) are the zeroes of a polynomial. Knowing the sum (half the perimeter) and the product (the area) lets you reconstruct the quadratic that describes the dimensions.

Forming a quadratic from given zeroes

If α and β are the zeroes, a quadratic with those zeroes is

\(p(x)=k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right]\), where k is any non-zero constant.

In words: \(x^{2}-(\text{sum of zeroes})x+(\text{product of zeroes})\). Taking k = 1 gives the simplest such polynomial; any non-zero k gives a valid answer because multiplying by a constant does not change the zeroes. If the sum or product is a fraction, choose k to clear denominators and get integer coefficients.

Useful symmetric identities

Many problems ask for an expression in α and β without solving for them. Express everything in terms of (α + β) and αβ:

  • \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\)
  • \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}\)
  • \((\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta\)
  • \(\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)\)

Extension: zeroes and coefficients of a cubic polynomial

For a cubic \(p(x)=ax^{3}+bx^{2}+cx+d\) (a ≠ 0) with zeroes α, β and γ:

  • Sum of zeroes: \(\alpha+\beta+\gamma=-\dfrac{b}{a}\)
  • Sum of products taken two at a time: \(\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}\)
  • Product of zeroes: \(\alpha\beta\gamma=-\dfrac{d}{a}\)

These follow from \(a(x-\alpha)(x-\beta)(x-\gamma)\) expanded and compared with \(ax^{3}+bx^{2}+cx+d\). Notice the alternating signs: minus, plus, minus.

Common mistakes to avoid:

  • Forgetting the leading coefficient a — the formulas are −b/a and c/a, not −b and c.
  • Sign error in the sum: it is −b/a, so a negative b makes the sum positive.
  • When forming a polynomial, dropping the minus sign on the middle term — it must be \(x^{2}-(\text{sum})x+(\text{product})\).
  • For a cubic, mixing up the middle two formulas, or missing the minus sign on the product (−d/a).
  • Trying to compute \(\alpha^{2}+\beta^{2}\) as \((\alpha+\beta)^{2}\) — you must subtract \(2\alpha\beta\).
Parabola p(x) = ax squared + bx + c with zeros alpha and beta, sum of zeros equals minus b over a, product equals c over a Zeros α, β of p(x) = ax^2 + bx + c x y y = p(x) α β the two x-intercepts are the zeros α and β Sum of zeros equals minus b over a and product of zeros equals c over a, verified with p(x) = 2x squared minus 5x + 3 Zeros ↔ coefficients Sum of zeros α + β = −b / a Product of zeros α × β = c / a Check: p(x) = 2x^2 − 5x + 3 → zeros 3/2 and 1 Sum = 3/2 + 1 = 5/2 = −b/a = −(−5)/2 Product = 3/2 × 1 = 3/2 = c/a = 3/2 Building a quadratic from its zeros: x squared minus (sum) x plus (product) Building the polynomial from its zeros Given zeros α = 3, β = −2 Sum = 1 Product = −6 p(x) = x^2 − (sum)x + product = x^2 − (1)x + (−6) = x^2 − x − 6 check: p(3)=0, p(−2)=0
Solved Examples
1
Worked Example
Find the sum and product of the zeros of the polynomial p(x) = \(3x^{2}-7x+4\).
Solution
  1. Step 1: Identify coefficients: a = 3, b = −7, c = 4.
  2. Step 2: Sum of zeros = −b/a = −(−7)/3 = \(\tfrac{7}{3}\).
  3. Step 3: Product of zeros = c/a = \(\tfrac{4}{3}\).

Answer: Sum = \(\tfrac{7}{3}\), Product = \(\tfrac{4}{3}\).

2
Worked Example
Find a quadratic polynomial whose zeros are 4 and −5.
Solution
  1. Step 1: Let α = 4, β = −5.
  2. Step 2: Sum = α + β = 4 + (−5) = −1.
  3. Step 3: Product = α × β = 4 × (−5) = −20.
  4. Step 4: Polynomial = \(x^{2}-(\text{sum})x+\text{product}=x^{2}-(-1)x+(-20)=x^{2}+x-20\).

Answer: \(x^{2}+x-20\) (or any non-zero multiple).

3
Worked Example
Verify the relationship between the zeroes and coefficients of \(x^{2}-2x-8\).
Solution
  1. Step 1: Factorise: \(x^{2}-2x-8=(x-4)(x+2)\), so the zeroes are 4 and −2.
  2. Step 2: Here a = 1, b = −2, c = −8.
  3. Step 3: Sum of zeroes = 4 + (−2) = 2; formula −b/a = −(−2)/1 = 2. Matches.
  4. Step 4: Product of zeroes = 4 × (−2) = −8; formula c/a = −8/1 = −8. Matches.

Answer: Sum 2 = −b/a and product −8 = c/a, verified.

4
Worked Example
Find a quadratic polynomial whose sum of zeroes is \(\tfrac{1}{4}\) and product of zeroes is −1.
Solution
  1. Step 1: Use \(x^{2}-(\text{sum})x+\text{product}=x^{2}-\tfrac{1}{4}x-1\).
  2. Step 2: Multiply by k = 4 to clear the fraction: \(4x^{2}-x-4\).

Answer: \(4x^{2}-x-4\).

5
Worked Example
Find a quadratic polynomial whose zeroes are \(2+\sqrt{3}\) and \(2-\sqrt{3}\).
Solution
  1. Step 1: Sum = \((2+\sqrt{3})+(2-\sqrt{3})=4\).
  2. Step 2: Product = \((2+\sqrt{3})(2-\sqrt{3})=2^{2}-(\sqrt{3})^{2}=4-3=1\).
  3. Step 3: Polynomial = \(x^{2}-(\text{sum})x+\text{product}=x^{2}-4x+1\).

Answer: \(x^{2}-4x+1\).

6
Worked Example
If one zero of \(x^{2}-5x+k\) is 2, find k and the other zero.
Solution
  1. Step 1: a = 1, b = −5, c = k. Sum of zeroes = −b/a = 5.
  2. Step 2: One zero is 2, so the other = 5 − 2 = 3.
  3. Step 3: Product of zeroes = c/a = k, and product = 2 × 3 = 6, so k = 6.

Answer: k = 6 and the other zero is 3.

7
Worked Example
If α and β are the zeroes of \(x^{2}-6x+k\) and α − β = 2, find k.
Solution
  1. Step 1: a = 1, b = −6, c = k, so α + β = −b/a = 6.
  2. Step 2: Given α − β = 2. Add the two equations: 2α = 8, so α = 4.
  3. Step 3: Then β = 6 − 4 = 2.
  4. Step 4: Product αβ = c/a = k, so k = 4 × 2 = 8.

Answer: k = 8.

8
Worked Example
If α and β are the zeroes of \(x^{2}-7x+10\), find \(\alpha^{2}+\beta^{2}\).
Solution
  1. Step 1: α + β = −b/a = 7 and αβ = c/a = 10.
  2. Step 2: Use the identity \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\).
  3. Step 3: Substitute: \((7)^{2}-2(10)=49-20=29\).

Answer: \(\alpha^{2}+\beta^{2}=29\).

9
Worked Example
If α and β are the zeroes of \(x^{2}-3x-2\), find \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\).
Solution
  1. Step 1: α + β = 3 and αβ = −2.
  2. Step 2: \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}\).
  3. Step 3: Substitute: \(\dfrac{3}{-2}=-\tfrac{3}{2}\).

Answer: \(-\tfrac{3}{2}\).

10
Worked Example
The zeroes of \(x^{2}-px+40\) are positive and differ by 6. Find p.
Solution
  1. Step 1: Let the zeroes be α and β with product αβ = c/a = 40 and difference α − β = 6.
  2. Step 2: Use the identity \((\alpha+\beta)^{2}=(\alpha-\beta)^{2}+4\alpha\beta\).
  3. Step 3: Substitute: \((\alpha+\beta)^{2}=6^{2}+4(40)=36+160=196\), so α + β = 14 (positive zeroes).
  4. Step 4: The zeroes are then 10 and 4 (sum 14, product 40, difference 6). Since α + β = −b/a = p, we get p = 14.

Answer: p = 14 (zeroes 10 and 4).

11
Worked Example
Verify the relationship between the zeroes and coefficients of the cubic \(x^{3}-x^{2}-2x\), whose zeroes are 0, 2 and −1.
Solution
  1. Step 1: Here a = 1, b = −1, c = −2, d = 0; zeroes α = 0, β = 2, γ = −1.
  2. Step 2: Sum α + β + γ = 0 + 2 + (−1) = 1; formula −b/a = −(−1)/1 = 1. Matches.
  3. Step 3: Sum of pairwise products = (0)(2) + (2)(−1) + (−1)(0) = −2; formula c/a = −2/1 = −2. Matches.
  4. Step 4: Product αβγ = 0 × 2 × (−1) = 0; formula −d/a = −0/1 = 0. Matches.

Answer: All three relationships are verified.

12
Worked Example
Find a cubic polynomial whose zeroes are 2, −3 and 4.
Solution
  1. Step 1: Sum α + β + γ = 2 + (−3) + 4 = 3.
  2. Step 2: Sum of pairwise products = (2)(−3) + (−3)(4) + (4)(2) = −6 − 12 + 8 = −10.
  3. Step 3: Product αβγ = 2 × (−3) × 4 = −24.
  4. Step 4: A cubic is \(x^{3}-(\text{sum})x^{2}+(\text{pair sum})x-(\text{product})=x^{3}-3x^{2}-10x+24\).

Answer: \(x^{3}-3x^{2}-10x+24\).

Key Points

  • For quadratic \(ax^{2}+bx+c\) with zeros α, β: α + β = −b/a, αβ = c/a
  • These relationships work without solving the polynomial
  • To form a polynomial from zeros: \(k\left[x^{2}-(\text{sum})x+(\text{product})\right]\)
  • If sum and product are fractions, multiply by a suitable k to clear denominators
  • When one zero is 0, the product is 0 and so c = 0
  • If zeros are equal (α = β), then the discriminant \(b^{2}-4ac=0\)
  • Use \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\) and \(\tfrac{1}{\alpha}+\tfrac{1}{\beta}=\tfrac{\alpha+\beta}{\alpha\beta}\) for symmetric expressions
  • For a cubic \(ax^{3}+bx^{2}+cx+d\): sum = −b/a, pairwise-product sum = c/a, product = −d/a
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.For $ax^2+bx+c$, the sum of the zeros is:
Explanation: Sum $=-b/a$.
Q2.The product of the zeros of $ax^2+bx+c$ is:
Explanation: Product $=c/a$.
Q3.For $x^2-5x+6$, the sum of zeros is:
Explanation: $-(-5)/1=5$.
Q4.A quadratic with zeros $2$ and $3$ is:
Explanation: Sum $5$, product $6$.
Practice more MCQs → 📝 Assignment · 35 marks