Relationship with coefficients

Once you begin studying algebra, one of the primary things that you will encounter is polynomials. Previously, we looked at determining the zeroes of polynomials, which is the process of determining the values of the variable x that make the entire polynomial equal to zero. Is there a short cut in determining how the zeroes of a polynomial will behave (i.e., the sum of the zeroes, their product and their combinations) without calculating or factoring the zeroes of the polynomial?

This is where the relationship between the zeroes and coefficients of a polynomial matters. This relationship is called Vieta's Formulas, and it serves as a mathematical bridge connecting the structural parts of the polynomial (the coefficients) with the behavior of the polynomial (the zeroes).

In this article, we will explore the connection between the zeroes and coefficients of quadratic, cubic, and polynomials of higher degree, provide an explanation as to why this relationship exists, and provide multiple worked out examples.

1.0Understanding the Terms: Zeroes vs. Coefficients

Before we look at the mathematical shortcuts, let us ensure we have a crystal-clear understanding of our two main players: coefficients and zeroes.

Consider a standard polynomial expression:

$$P(x) = ax^2 + bx + c$$

• The Coefficients: These are the constant real numbers multiplied by the variables, alongside the standalone constant term. In the quadratic expression above, $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term.
• The Zeroes (or Roots): These are the specific values of $x$, traditionally designated by the Greek letters $\alpha$ (alpha) and $\beta$ (beta), that satisfy the equation $P(x) = 0$.

Normally, finding $\alpha$ and $\beta$ requires factoring or using the quadratic formula. However, the core secret of polynomial algebra is that the coefficients $a$, $b$, and $c$ already contain the exact DNA of $\alpha$ and $\beta$.

2.0The Relationship in Quadratic Polynomials (Degree 2)

Let us begin with the most common polynomial you will meet in school algebra: the quadratic polynomial. Its general form is written as:

$$P(x) = ax^2 + bx + c \quad (\text{where } a \neq 0)$$

If we assume that this polynomial has two real or complex zeroes, $\alpha$ and $\beta$, the relationship dictates two fundamental rules:

1. Sum of Zeroes

The sum of the zeroes is equal to the negative ratio of the coefficient of $x$ to the coefficient of $x^2$.

$$\alpha + \beta = -\frac{b}{a}$$

2. Product of Zeroes

The product of the zeroes is equal to the ratio of the constant term to the coefficient of $x^2$.

$$\alpha \cdot \beta = \frac{c}{a}$$

Why Does This Work? (The Derivation)

If $\alpha$ and $\beta$ are the zeroes of $P(x) = ax^2 + bx + c$, then according to the Factor Theorem, $(x - \alpha)$ and $(x - \beta)$ must be factors of the polynomial. Therefore, we can rewrite the polynomial as a product of its factors multiplied by a constant scaling factor $k$:

$$ax^2 + bx + c = k(x - \alpha)(x - \beta)$$

Let us expand the right-hand side of the equation:

$$ax^2 + bx + c = k[x^2 - \beta x - \alpha x + \alpha\beta]$$

$$ax^2 + bx + c = k[x^2 - (\alpha + \beta)x + \alpha\beta]$$

$$ax^2 + bx + c = kx^2 - k(\alpha + \beta)x + k\alpha\beta$$

Now, we compare the coefficients of the like terms on both sides:

• Comparing the coefficients of $x^2$: $a = k$
• Comparing the coefficients of $x$: $b = -k(\alpha + \beta)$
• Comparing the constant terms: $c = k\alpha\beta$$

Since $k = a$, we can substitute $a$ into the other equations:

$$b = -a(\alpha + \beta) \implies \alpha + \beta = -\frac{b}{a}$$

$$c = a(\alpha\beta) \implies \alpha\beta = \frac{c}{a}$$

This proof demonstrates how the roots and coefficients are permanently bound together.

3.0The Relationship in Cubic Polynomials (Degree 3)

As the degree of a polynomial grows, the relationship expands elegantly. A cubic polynomial has a degree of 3, meaning it can have up to three zeroes: $\alpha$, $\beta$, and $\gamma$ (gamma).

The general standard form of a cubic polynomial is:

$$P(x) = ax^3 + bx^2 + cx + d \quad (\text{where } a \neq 0)$$

For a cubic polynomial, we observe three distinct relationships:

1. Sum of All Zeroes

The sum of the individual roots follows the same starting pattern as the quadratic format:

$$\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{Coefficient of } x^2}{\text{Coefficient of } x^3}$$

2. Sum of the Products of Zeroes Taken Two at a Time

This relationship accounts for all possible paired combinations of roots:

$$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{Coefficient of } x}{\text{Coefficient of } x^3}$$

3. Product of All Zeroes

The product of all three roots maps directly to the constant term:

$$\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{Constant term}}{\text{Coefficient of } x^3}$$

Notice the Sign Pattern: Look closely at the signs as the degree progresses: it starts with a negative sign ($-\frac{b}{a}$), switches to a positive sign ($\frac{c}{a}$), and then returns to a negative sign ($-\frac{d}{a}$). This alternating sign pattern continues indefinitely for polynomials of any degree!

4.0Higher-Degree Polynomials: Vieta's General Formulas

In high-level competitive exams there are many situations where you could be playing with different types of polynomials, especially biquadratic polynomials (fourth degree), or general n-th (n being an arbitrary integer) degree expressions. The alternating rule that we've just discovered provides us with a way to apply Vieta's formulas, irrespective of the degree of the polynomial.

$$P(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \dots + a_1 x + a_0$$

If this polynomial has roots $x_1, x_2, \dots, x_n$:

• Sum of roots: $\sum x_i = -\frac{a_{n-1}}{a_n}$
• Sum of product of roots two at a time: $\sum x_i x_j = \frac{a_{n-2}}{a_n}$
• Product of all roots: $x_1 x_2 \dots x_n = (-1)^n \frac{a_0}{a_n}$

5.0Forming a Polynomial from Given Zeroes

This algebraic link also operates in reverse. If a problem gives you the values of the zeroes, you can construct the original polynomial instantly without long binomial multiplication.

For a Quadratic Polynomial:

If you are given the sum of zeroes ($S$) and the product of zeroes ($P$), the quadratic polynomial is constructed as:

$$P(x) = x^2 - (\text{Sum of Zeroes})x + (\text{Product of Zeroes})$$

$$P(x) = x^2 - S x + P$$

For a Cubic Polynomial:

If you know the individual roots or their combined sums, the cubic polynomial can be built using:

$$P(x) = x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - (\alpha\beta\gamma)$$

6.0Solved Examples

Example 1: Find the sum and product of the zeroes for the quadratic polynomial $P(x) = 3x^2 - 9x + 12$.

Solution:

Identify the coefficients by comparing the expression to the standard form $ax^2 + bx + c$:

• $a = 3$
• $b = -9$
• $c = 12$

Now, apply the formulas directly:

• Sum of zeroes ($\alpha + \beta$):
• $$\alpha + \beta = -\frac{b}{a} = -\frac{-9}{3} = \frac{9}{3} = 3$$
• Product of zeroes ($\alpha\beta$):
• $$\alpha\beta = \frac{c}{a} = \frac{12}{3} = 4$$
• Answer: The sum of the zeroes is 3, and the product of the zeroes is 4.

Example 2: Find the zeroes of the quadratic polynomial $P(x) = x^2 + 7x + 10$, and verify the relationship between the zeroes and the coefficients.

Solution:

Step 1: Find the actual roots using factorization.

Set $x^2 + 7x + 10 = 0$. We split the middle term into $5x$ and $2x$:

$$x^2 + 5x + 2x + 10 = 0$$

$$x(x + 5) + 2(x + 5) = 0$$

$$(x + 2)(x + 5) = 0$$

Equating each factor to zero gives us our roots:

• $\alpha = -2$
• $\beta = -5$

Step 2: Verify using coefficients.

From the polynomial, we have $a = 1$, $b = 7$, and $c = 10$.

• Verification for Sum:
• • Actual sum of calculated roots: $\alpha + \beta = (-2) + (-5) = -7$
  • Formula prediction: $-\frac{b}{a} = -\frac{7}{1} = -7$
  • Both values match perfectly.
• Verification for Product:
• • Actual product of calculated roots: $\alpha \cdot \beta = (-2) \cdot (-5) = 10$
  • Formula prediction: $\frac{c}{a} = \frac{10}{1} = 10$
  • Both values match perfectly.

Example 3: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $P(x) = 2x^2 - 5x + 7$, find the value of $\alpha^2 + \beta^2$.

Solution:

This is a favorite question type in competitive and foundation exams. Notice that you do not need to calculate the actual roots, which are complex numbers in this case. Instead, we can use algebraic identities.

First, write down the known relationships from the coefficients ($a = 2, b = -5, c = 7$):

$$\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}$$

$$\alpha\beta = \frac{7}{2}$$

We know the algebraic identity for the sum of squares:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

Substitute our known values into the identity:

$$\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2\left(\frac{7}{2}\right)$$

$$\alpha^2 + \beta^2 = \frac{25}{4} - 7$$

$$\alpha^2 + \beta^2 = \frac{25 - 28}{4} = -\frac{3}{4}$$

• Answer: The value of $\alpha^2 + \beta^2$ is $-\frac{3}{4}$.

Example 4: Find a quadratic polynomial whose zeroes are $3 + \sqrt{2}$ and $3 - \sqrt{2}$.

Solution:

Let $\alpha = 3 + \sqrt{2}$ and $\beta = 3 - \sqrt{2}$.

Step 1: Calculate the Sum ($S$) of the zeroes.

$$S = \alpha + \beta = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 3 + 3 = 6$$

Step 2: Calculate the Product ($P$) of the zeroes.

Using the difference of squares identity, $(a+b)(a-b) = a^2 - b^2$:

$$P = \alpha \cdot \beta = (3 + \sqrt{2})(3 - \sqrt{2}) = (3)^2 - (\sqrt{2})^2$$

$$P = 9 - 2 = 7$$

Step 3: Construct the polynomial.

$$P(x) = x^2 - S x + P$$

$$P(x) = x^2 - 6x + 7$$

• Answer: The required polynomial is $x^2 - 6x + 7$.

Example 5: For the cubic polynomial $P(x) = x^3 - 6x^2 + 11x - 6$, verify the product of all its zeroes if the roots are known to be 1, 2, and 3.

Solution:

Let the roots be $\alpha = 1$, $\beta = 2$, and $\gamma = 3$.

From the polynomial, the coefficients are $a = 1$, $b = -6$, $c = 11$, and $d = -6$.

• Method 1 (Direct multiplication):
• $$\alpha\beta\gamma = 1 \cdot 2 \cdot 3 = 6$$
• Method 2 (Using Vieta's Formula):
• $$\alpha\beta\gamma = -\frac{d}{a} = -\frac{-6}{1} = 6$$

Both methods yield the exact same answer, confirming the validity of the cubic coefficient formula.

7.0Common Mistakes Students Make

To avoid losing unnecessary points on exams, keep an eye out for these frequent pitfalls:

• Dropping the Negative Sign in $-\frac{b}{a}$: This is the single most common error. If $b$ is already negative in the polynomial (e.g., $-5x$), then $-\frac{b}{a}$ becomes $-\frac{-5}{a} = +\frac{5}{a}$. Double-check your negative signs carefully.
• Forgetting to Divide by a: Students often memorize the rules simply as $-b$ and $c$. Remember that this shortcut only works if the leading coefficient $a$ equals 1. If $a \neq 1$, you must divide every term by $a$.
• Misapplying the Formula to Missing Terms: If a polynomial reads $P(x) = 2x^2 - 9$, the linear $x$ term is missing. This means the coefficient $b = 0$. Do not accidentally mistake the constant term -9 for b.