Relationship between zeroes and coefficients

Every quadratic $ax^2+bx+c$ hides two numbers — a sum and a product — that you can read straight off its coefficients without ever solving the equation. Factorise $2x^2-8x+6$ and you get zeroes $1$ and $3$ ; add them and you get $4$ , multiply them and you get $3$ . Notice anything? $4 = -(-8)/2$ and $3 = 6/2$ . That is not a coincidence — it is a fixed law linking every polynomial's zeroes to its coefficients, and it works in reverse too: give a target sum and product, and the law hands you back a polynomial. This article builds that law for quadratics from the factor theorem, states its cubic cousin, and shows both directions of the shortcut in action — reading zeroes' sum/product from coefficients, and building a polynomial from a given sum and product.

The core insight

For $p(x) = ax^2+bx+c$ with zeroes $\alpha,\beta$ , the sum $\alpha+\beta = -b/a$ and the product $\alpha\beta = c/a$ follow directly from the factor theorem, and the same idea extends to a cubic's three zeroes.

The core idea

Quadratic case. For $p(x) = ax^2 + bx + c$ (with $a \neq 0$ ) having zeroes $\alpha$ and $\beta$ :

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}

where: $a$ is the coefficient of $x^2$ , $b$ the coefficient of $x$ , $c$ the constant term, and $\alpha,\beta$ the two zeroes. In words, the sum of the zeroes is minus the coefficient of $x$ over the coefficient of $x^2$ , and the product of the zeroes is the constant term over the coefficient of $x^2$ .

Cubic case. For $p(x) = ax^3+bx^2+cx+d$ (with $a\neq 0$ ) having zeroes $\alpha,\beta,\gamma$ , the same pattern extends one degree further — these are stated and verified at Class 10 level, not derived by polynomial division:

\alpha+\beta+\gamma = -\frac{b}{a}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}

where: $a,b,c,d$ are the coefficients of $x^3, x^2, x, 1$ respectively. Notice the alternating sign pattern: sum is $-b/a$ , sum of pairwise products is $+c/a$ , full product is $-d/a$ .

Running it backwards. If a quadratic's zeroes are known only through their sum $S$ and product $P$ , the polynomial can be rebuilt without ever finding $\alpha$ and $\beta$ individually:

p(x) = k\left(x^2 - Sx + P\right)

where: $S = \alpha+\beta$ , $P = \alpha\beta$ , and $k$ is any nonzero constant (usually $k=1$ , or chosen to clear fractions). This works because matching $x^2-Sx+P$ against $ax^2+bx+c$ with $a=1$ gives exactly $-b/a=S$ and $c/a=P$ . Why does this shortcut matter at all? Without it, "find the sum of the zeroes of $5x^2-13x+2$ " would demand solving the quadratic in full — applying the quadratic formula, simplifying a surd, adding two messy roots — just to throw away the individual values and keep only their sum. The coefficient relationships skip that entire detour: read $a=5,\ b=-13$ straight off the equation and the sum is $-(-13)/5 = 13/5$ immediately, no factorisation or formula needed. The same shortcut runs the other way for construction problems: rather than guessing two numbers that add and multiply to given targets, the formula $x^2-Sx+P$ manufactures them automatically. Both directions rest on exactly the same algebraic fact, proved once below and reused everywhere.

Figure

Parabola for 2x squared minus 8x plus 6 crossing the x axis at 1 and 3 with the vertex marked

Derivation ASum and product of a quadratic's zeroes from the factor theorem

The goal is to prove, using nothing beyond the factor theorem and comparing coefficients, that for $p(x)=ax^2+bx+c$ with zeroes $\alpha$ and $\beta$ , the sum $\alpha+\beta$ must equal $-b/a$ and the product $\alpha\beta$ must equal $c/a$ , so that these two numbers can always be read off the coefficients directly instead of being computed after solving for $\alpha$ and $\beta$ separately.

Step 1. Since $\alpha$ and $\beta$ are the zeroes of $p(x)$ , the factor theorem says $(x-\alpha)$ and $(x-\beta)$ are both factors, so $p(x)$ can be written as $p(x) = k(x-\alpha)(x-\beta)$ for some nonzero constant $k$ .

Step 2. Expand the right-hand side: $k(x-\alpha)(x-\beta) = k\left[x^2 - (\alpha+\beta)x + \alpha\beta\right] = kx^2 - k(\alpha+\beta)x + k\alpha\beta$ .

Step 3. Compare this with $p(x) = ax^2+bx+c$ term by term. Matching the $x^2$ coefficients gives $k=a$ . Matching the $x$ coefficients gives $-k(\alpha+\beta) = b$ , so $-a(\alpha+\beta)=b$ , hence $\alpha+\beta = -b/a$ . Matching the constant terms gives $k\alpha\beta = c$ , so $a\alpha\beta = c$ , hence $\alpha\beta = c/a$ .

Step 4. Both formulae fall out of a single comparison, and the same three-step pattern (factor theorem, expand, compare coefficients) is exactly what is used to state — though not derive at this level — the three cubic relationships for $\alpha+\beta+\gamma$ , $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ .

Worked example

If the zeroes of $p(x) = 2x^2 - 8x + 6$ are $\alpha$ and $\beta$ , verify the relationship between the zeroes and the coefficients, then find a quadratic whose sum and product of zeroes are $\dfrac{1}{4}$ and $-1$ .

Step 1. Factorise: $2x^2-8x+6 = 2(x^2-4x+3) = 2(x-1)(x-3)$ , so the zeroes are $\alpha=1$ and $\beta=3$ .

Step 2. Compute directly: $\alpha+\beta = 1+3 = 4$ and $\alpha\beta = 1\times3 = 3$ .

Step 3. Compare with the formulae using $a=2,\ b=-8,\ c=6$ : $-\dfrac{b}{a} = -\dfrac{-8}{2} = 4 = \alpha+\beta$ and $\dfrac{c}{a} = \dfrac{6}{2} = 3 = \alpha\beta$ . Both relationships hold.

Step 4. Now build a new quadratic from $S=\dfrac14$ and $P=-1$ : $p(x) = k\left(x^2-\dfrac14x-1\right)$ . Taking $k=4$ to clear the fraction gives $p(x) = 4x^2 - x - 4$ . Check: $a=4,\ b=-1,\ c=-4$ , so $-b/a = 1/4 = S$ and $c/a = -1 = P$ , matching the given values.

Final answer: the zeroes of $2x^2-8x+6$ are $1$ and $3$ , verifying $\alpha+\beta=4$ and $\alpha\beta=3$ ; and $4x^2-x-4$ is a quadratic with sum $\dfrac14$ and product $-1$ .

Competitive stretch

If $\alpha$ and $\beta$ are the zeroes of $p(x) = x^2-5x+6$ , find the value of $\alpha^2+\beta^2$ without finding $\alpha$ and $\beta$ individually.

Step 1. Read off the coefficients: $a=1,\ b=-5,\ c=6$ , so $\alpha+\beta = -b/a = 5$ and $\alpha\beta = c/a = 6$ .

Step 2. Use the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ , which follows from expanding $(\alpha+\beta)^2 = \alpha^2+2\alpha\beta+\beta^2$ and rearranging.

Step 3. Substitute: $\alpha^2+\beta^2 = 5^2 - 2(6) = 25-12 = 13$ .

Final answer: $\alpha^2+\beta^2 = 13$ , obtained entirely from the coefficients — a pattern worth remembering, since combinations like $\alpha^2+\beta^2$ , $\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta}$ , or $(\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha\beta$ can always be rewritten in terms of the sum and product alone.

Common traps

Dropping the negative sign in the sum formula. The sum of zeroes is $-b/a$ , not $b/a$ — easy to lose under exam pressure.
Reading coefficients from a non-standard arrangement. Always rearrange to standard form $ax^2+bx+c$ before naming $a$ , $b$ , $c$ .
Assuming the reconstructed polynomial is unique. Any nonzero multiple $k(x^2-Sx+P)$ has the same sum and product of zeroes, so infinitely many polynomials share a given $S$ and $P$ .
Trying to solve the cubic relationships for $\alpha,\beta,\gamma$ individually. The three cubic equations let you verify given zeroes or combine them, not extract each zero separately at this level.
Forgetting that combinations like $\alpha^2+\beta^2$ or $1/\alpha+1/\beta$ can be built from $S$ and $P$ alone. Students often reach for the quadratic formula to find $\alpha$ and $\beta$ first, when rewriting the target expression in terms of the sum and product is faster and avoids surds entirely.

Ask the tutor: if you want another example verifying the cubic relationships, or help setting up $x^2-Sx+P$ from a word problem, ask the AI tutor to work through it with you step by step.

Why is there a minus sign in the sum formula but not in the product formula?

It comes straight out of expanding $(x-\alpha)(x-\beta) = x^2-(\alpha+\beta)x+\alpha\beta$ . The middle term picks up a minus sign because each bracket contributes $-\alpha$ and $-\beta$ , and when you add those two you get $-(\alpha+\beta)$ . The constant term is a plain product $\alpha\beta$ with no sign flip, since it comes from multiplying $(-\alpha)$ and $(-\beta)$ , which gives $+\alpha\beta$ . So the asymmetry is built into the algebra of expanding two brackets, not a rule to memorise separately.

Can I use these formulae to find the individual zeroes of a cubic?

Not directly at this level. The three cubic relationships give you $\alpha+\beta+\gamma$ , $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ , which is three equations in three unknowns — but solving that system for $\alpha,\beta,\gamma$ individually needs methods beyond the Class 10 syllabus. What you can do is verify zeroes that are already given (substitute and check the three sums/products match), or use the relationships to find combinations like $\alpha+\beta+\gamma$ without factorising the cubic at all.

If I'm only given the sum and product of zeroes, is x squared minus Sx plus P the only possible answer?

No — it is the simplest one, but not the only one. Any nonzero constant multiple $k(x^2-Sx+P)$ has exactly the same zeroes, and therefore the same sum $S$ and product $P$ , because multiplying a polynomial by a constant does not change where it equals zero. Exam questions usually expect the simplest form, taking $k=1$ , or a value of $k$ that clears any fractions to give whole-number coefficients — but technically every constant multiple is a correct answer.

Relationship between zeroes and coefficients

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads