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Class 10th Chapters
1. Real Numbers 2. Polynomials 3. Pair of Linear Equations in Two Variables
4. Quadratic Equations 5. Arithmetic Progressions 6. Triangles
7. Coordinate Geometry 8. Introduction to Trigonometry 9. Some Applications of Trigonometry
10. Circles 11. Constructions 12. Areas Related to Circles
13. Surface Areas And Volumes 14. Statistics 15. Probability

Content On This Page
Terms Related to Polynomials Value and Zero of a Polynomial Geometrical Representation of a Polynomial
Relationship Between Zeroes and Coefficient of a Polynomial Division Algorithm for Polynomials


Chapter 2 Polynomials (Concepts)

This chapter marks a significant deepening of our study of Polynomials, extending the foundational concepts established in Class 9. We transition to a more analytical exploration, focusing primarily on the properties and behavior of quadratic (degree 2) and, to some extent, cubic (degree 3) polynomials. We will begin by reinforcing our understanding of core definitions – what constitutes a polynomial (an algebraic expression with variables having only non-negative integer exponents), identifying its terms and their coefficients, and determining the degree of a polynomial (the highest power of the variable). Recall also the pivotal concept of the zeroes of a polynomial $P(x)$: these are the specific values of $x$ for which the polynomial evaluates to zero, i.e., $P(x)=0$.

A crucial advancement in this chapter is grasping the geometrical interpretation of the zeroes. The zeroes of a polynomial $P(x)$ have a direct visual correspondence on its graph, $y = P(x)$. Specifically, the real zeroes of $P(x)$ are precisely the x-coordinates of the points where the graph of the polynomial intersects or touches the x-axis. For a simple linear polynomial $P(x) = ax+b$ ($a \neq 0$), its graph is a straight line that intersects the x-axis at exactly one point, corresponding to its single zero $x = -\frac{b}{a}$. The graphical representation of a quadratic polynomial, $P(x) = ax^2 + bx + c$ ($a \neq 0$), is a distinctive curve called a parabola. The parabola opens upwards if the leading coefficient $a > 0$ and downwards if $a < 0$. The relationship between the parabola and the x-axis visually reveals the nature of the quadratic's zeroes:

Understanding this graphical connection provides valuable intuition about polynomial solutions.

The analytical core of the chapter lies in uncovering and utilizing the profound relationship between the zeroes and the coefficients of a polynomial. For a quadratic polynomial $P(x) = ax^2 + bx + c$ ($a \neq 0$), let its zeroes be denoted by the Greek letters $\alpha$ (alpha) and $\beta$ (beta). There exist two fundamental relationships:

These relationships are incredibly useful. They allow us to verify potential zeroes and, importantly, to construct a quadratic polynomial if we know its zeroes or simply their sum and product.

These relationships extend to higher-degree polynomials as well. For a cubic polynomial $P(x) = ax^3 + bx^2 + cx + d$ ($a \neq 0$), let its zeroes be $\alpha$, $\beta$, and $\gamma$ (gamma). The relationships are:

While used less frequently for direct construction, these confirm the deep structural connection between a polynomial's roots and its coefficients.

Finally, another vital tool explored is the Division Algorithm for Polynomials. Analogous to Euclid's division lemma for integers, this algorithm states that for any two polynomials $p(x)$ (dividend) and $g(x)$ (divisor), where $g(x) \neq 0$, we can always find unique polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that: $$ p(x) = g(x)q(x) + r(x) $$ where either the remainder $r(x)$ is zero, or the degree of the remainder $r(x)$ is strictly less than the degree of the divisor $g(x)$. This algorithm is used for polynomial long division. A key application arises when we know some zeroes of a polynomial. If $k$ is a zero, then $(x-k)$ is a factor. If we know one or more zeroes, we can divide the original polynomial $p(x)$ by the product of the factors corresponding to the known zeroes. The quotient $q(x)$ obtained will be a polynomial of lower degree, whose zeroes are the remaining zeroes of $p(x)$. This technique is particularly crucial for finding all zeroes of cubic or higher-degree polynomials when some are provided or easily found.



Basic Terms Related to Polynomials

In algebra, we work with expressions formed by combining variables and constants using operations like addition, subtraction, multiplication, and division. Examples include $x+2$, $3y^2-5$, $\frac{z}{4}+1$, etc. A polynomial is a specific and very important type of algebraic expression that adheres to certain rules regarding the powers of the variables involved.


What is a Polynomial?

The word "Polynomial" is derived from two words: "Poly" meaning many, and "nomial" meaning term. So, a polynomial is an algebraic expression that consists of many terms. However, there are specific rules for an expression to be a polynomial.

A polynomial in one variable (let's say $x$) is an algebraic expression where the terms are formed by a constant (called a coefficient) multiplied by the variable raised to a non-negative integer power. Non-negative integers are the set of whole numbers: $\{0, 1, 2, 3, ...\}$.


General Form of a Polynomial

The standard way to write a polynomial in the variable $x$ is:

$$p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_2 x^2 + a_1 x + a_0$$

Let's break down each part of this form:


The Constant Term

The constant term in a polynomial is the term that does not contain a variable. It's a standalone number.

In the general form $p(x) = a_n x^n + ... + a_1 x + a_0$, the constant term is $a_0$.

This is because it can be thought of as $a_0 x^0$. Since any non-zero number raised to the power of 0 is 1 ($x^0 = 1$), the term is simply $a_0 \times 1 = a_0$.

Examples:


Identifying Polynomials: Examples

The most important rule to remember is that the exponents of the variable must be whole numbers (0, 1, 2, ...). An expression is NOT a polynomial if:

The following table clarifies this with examples:

Expression Is it a Polynomial? Reason
$3x^2 + 5x - 7$ Yes All exponents of $x$ (2, 1, and 0) are non-negative integers.
$y^5 - \frac{1}{3}y^2 + \sqrt{5}$ Yes All exponents of $y$ (5, 2, and 0) are non-negative integers. Coefficients can be fractions or roots.
$10$ Yes This is a constant polynomial, which can be written as $10x^0$. The exponent (0) is a non-negative integer.
$\sqrt{x} + 4$ No The term $\sqrt{x}$ is $x^{1/2}$. The exponent $1/2$ is a fraction, not an integer.
$\frac{1}{y} - 2$ No The term $\frac{1}{y}$ is $y^{-1}$. The exponent $-1$ is negative.
$z^{3/2} + 5z$ No The term $z^{3/2}$ has an exponent of $3/2$, which is not an integer.
$\frac{x^2+2x+1}{x-1}$ No This is a rational expression because the variable $x$ appears in the denominator.

Degree of a Polynomial

The degree of a polynomial in one variable is the highest power of the variable in the polynomial, provided that the coefficient of the term with the highest power is not zero.


Classification of Polynomials

Polynomials can be classified into different types based on three main criteria: the number of variables they contain, their degree, and the number of terms they have.


1. Classification Based on Number of Variables

This is the most fundamental classification, distinguishing polynomials by how many different variables they involve.

Polynomial in One Variable

A polynomial that contains only one type of variable (like $x$ or $y$ or $z$, but not a mix). Most of the polynomials studied in early algebra are of this type.

Examples:

Polynomial in Two or More Variables

A polynomial that contains two or more different variables.

Examples:

Note on Degree: For polynomials with more than one variable, the degree of a term is the sum of the exponents of the variables in that term. The degree of the entire polynomial is the highest degree of any of its terms. For example, in $7x^3y^2 + 2xy - 5$, the degree of the first term is $3+2=5$, so the degree of the polynomial is 5.


2. Classification Based on Degree

The degree of a polynomial in one variable is the highest exponent of the variable in any of its terms. This is one of the most common ways to classify polynomials.

Type of Polynomial Degree General Form Example
Constant Polynomial 0 $p(x) = c$ $p(x) = 7$ (can be written as $7x^0$)
Linear Polynomial 1 $ax + b$ (where $a \neq 0$) $5x - 2$
Quadratic Polynomial 2 $ax^2 + bx + c$ (where $a \neq 0$) $2x^2 - 3x + 1$
Cubic Polynomial 3 $ax^3 + bx^2 + cx + d$ (where $a \neq 0$) $x^3 + 2x^2 - 5x + 4$
Biquadratic Polynomial (or Quartic) 4 $ax^4 + bx^3 + cx^2 + dx + e$ (where $a \neq 0$) $3y^4 - 7y^3 + y - 9$

Polynomials of degree 5 are called quintic, degree 6 are sextic, and so on, but these names are less commonly used.


3. Classification Based on Number of Terms

Polynomials are also named based on how many non-zero terms they contain. This classification can apply to polynomials of any degree or number of variables.

Monomial

A polynomial with exactly one term. (Mono- means one).

Binomial

A polynomial with exactly two terms. (Bi- means two).

Trinomial

A polynomial with exactly three terms. (Tri- means three).

An expression with more than three terms is usually just called a polynomial rather than having a specific name.


Putting It All Together: Combined Classification

The three classification systems (by variables, degree, and terms) are independent of each other. This means any single polynomial fits into one category from each system. To fully describe a polynomial, you can combine these classifications.

Here are some examples showing how one polynomial can have multiple classifications:

Polynomial Classification Full Description
$5x^2 + 3x - 1$ By Variables: One Variable ($x$) It is a quadratic trinomial in one variable.
By Degree: Degree 2 (Quadratic)
By Terms: 3 Terms (Trinomial)
$y^3 - 8$ By Variables: One Variable ($y$) It is a cubic binomial in one variable.
By Degree: Degree 3 (Cubic)
By Terms: 2 Terms (Binomial)
$7m^4$ By Variables: One Variable ($m$) It is a biquadratic monomial in one variable.
By Degree: Degree 4 (Biquadratic)
By Terms: 1 Term (Monomial)
$x+y$ By Variables: Two Variables ($x, y$) It is a linear binomial in two variables.
By Degree: Degree 1 (Linear)
By Terms: 2 Terms (Binomial)

Equal Polynomials

Two polynomials are said to be equal if they are identical in their structure. This means they must satisfy two conditions:

Sometimes, a polynomial may need to be simplified or expanded to its standard form before it can be compared with another.

Condition for Equality

Let's consider two polynomials, $p(x)$ and $q(x)$:

$p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$

$q(x) = b_n x^n + b_{n-1} x^{n-1} + ... + b_1 x + b_0$

Then, $p(x) = q(x)$ if and only if $a_i = b_i$ for all $i$ from 0 to $n$. In simpler terms, $a_n = b_n$, $a_{n-1} = b_{n-1}$, and so on, down to $a_0 = b_0$.

Example 1. Find the values of $A$, $B$, and $C$ if the following polynomials are equal:

$p(x) = (A+2)x^2 - 7x + 4$

$q(x) = 5x^2 + Bx + C$

Answer:

Solution

For two polynomials to be equal, we must equate the coefficients of the corresponding powers of $x$.

1. Equating coefficients of $x^2$:

$A + 2 = 5$

$A = 5 - 2$

$A = 3$

2. Equating coefficients of $x$:

$-7 = B$

$B = -7$

3. Equating the constant terms:

$4 = C$

$C = 4$

Therefore, for $p(x)$ to be equal to $q(x)$, the values must be A = 3, B = -7, and C = 4.

Example 2. Are the polynomials $p(x) = (x-2)^2 + 5$ and $q(x) = x^2 - 4x + 9$ equal?

Answer:

Solution

To compare the polynomials, we must first express $p(x)$ in its standard (expanded) form.

Given $p(x) = (x-2)^2 + 5$.

Using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$p(x) = (x^2 - 2(x)(2) + 2^2) + 5$

$p(x) = (x^2 - 4x + 4) + 5$

$p(x) = x^2 - 4x + 9$

Now, we compare the simplified form of $p(x)$ with $q(x) = x^2 - 4x + 9$.

Since the simplified form of $p(x)$ is identical to $q(x)$, the two polynomials are equal.



Value and Zero of a Polynomial

A polynomial is a function of its variable(s). Just like other functions, we can evaluate a polynomial for specific numerical values of the variable(s). When the value of the polynomial becomes zero for a particular input value of the variable, that input value holds special significance and is called a 'zero' of the polynomial.


Value of a Polynomial

The value of a polynomial $p(x)$ at a specific real number, say $x = a$, is the result obtained when we substitute the value $a$ for the variable $x$ throughout the polynomial expression. This value is denoted by $p(a)$.

To find the value of $p(x)$ at $x=a$, simply replace every instance of $x$ in the polynomial expression with the number $a$ and then simplify the resulting numerical expression.


Example 1. Find the value of the polynomial $p(x) = 2x^2 - 3x + 4$ at $x = -1$.

Answer:

Given:

The polynomial $p(x) = 2x^2 - 3x + 4$.

The value of the variable at which to evaluate is $x = -1$.

To Find:

The value of the polynomial $p(x)$ at $x = -1$, i.e., $p(-1)$.

Solution:

To find $p(-1)$, we substitute $x = -1$ into the polynomial expression $p(x) = 2x^2 - 3x + 4$:

$p(-1) = 2(-1)^2 - 3(-1) + 4$

Now, perform the operations following the order of operations (PEMDAS/BODMAS): exponents first, then multiplication, then addition and subtraction.

$p(-1) = 2(1) - (-3) + 4$

($(-1)^2 = 1$, $-3 \times -1 = +3$)

$p(-1) = 2 + 3 + 4$

$p(-1) = 9$

... (1)

The value of the polynomial $p(x) = 2x^2 - 3x + 4$ at $x = -1$ is 9.

Answer: $p(-1) = 9$.


Zero of a Polynomial

A zero of a polynomial $p(x)$ is a specific real number $a$ such that when substituted for $x$, the value of the polynomial becomes 0. In other words, $a$ is a zero of $p(x)$ if $p(a) = 0$.

The zeros of a polynomial are also referred to as the roots of the polynomial equation $p(x) = 0$. Finding the zeros of a polynomial is a fundamental problem in algebra.

The number of zeros a polynomial can have is related to its degree:


Example 2. Check if $x=2$ is a zero of the polynomial $p(x) = x^2 - 4x + 4$.

Answer:

Given:

The polynomial $p(x) = x^2 - 4x + 4$.

The value to check is $x = 2$.

To Check:

If $x=2$ is a zero of the polynomial $p(x)$.

Solution:

To check if $x=2$ is a zero of $p(x)$, we need to evaluate the polynomial at $x=2$. If the value of the polynomial at $x=2$, i.e., $p(2)$, is 0, then $x=2$ is a zero.

Substitute $x = 2$ into the polynomial expression $p(x) = x^2 - 4x + 4$:

$p(2) = (2)^2 - 4(2) + 4$

Perform the operations:

$p(2) = 4 - 8 + 4$

($(2)^2=4$, $-4 \times 2 = -8$)

$p(2) = (4 + 4) - 8$

$p(2) = 8 - 8$

$p(2) = 0$

... (1)

Since $p(2) = 0$, the value of the polynomial at $x=2$ is 0.

By the definition of a zero of a polynomial, if $p(a)=0$, then $a$ is a zero of $p(x)$. In this case, $p(2)=0$, so 2 is a zero of the polynomial $p(x) = x^2 - 4x + 4$.

Answer: Yes, $x=2$ is a zero of the polynomial $p(x) = x^2 - 4x + 4$.

Note that $x^2 - 4x + 4$ is a perfect square trinomial, $(x-2)^2$. Setting $(x-2)^2 = 0$ gives $x-2 = 0$, so $x=2$ is the only zero (a repeated zero) of this polynomial.



Geometrical Meaning of the Zeros of a Polynomial

The relationship between algebra and geometry is profound. Any polynomial $p(x)$ can be represented geometrically by its graph in a coordinate plane. The graph of the equation $y = p(x)$ provides a visual representation of the polynomial's behavior. Most importantly, the real zeros of the polynomial are directly visible on its graph.


Real Zeros and the X-Axis

By definition, a real number 'a' is a zero of a polynomial $p(x)$ if $p(a) = 0$. When we plot the graph of $y = p(x)$, the points on the graph have coordinates $(x, y)$ where the y-value is given by the polynomial, i.e., $y = p(x)$.

If 'a' is a zero, then $p(a) = 0$, which means the point $(a, 0)$ lies on the graph. Any point with a y-coordinate of 0 lies on the x-axis.

Therefore, the conclusion is fundamental:

The real zeros of a polynomial are the x-coordinates of the points where its graph intersects or touches the x-axis.


Graph of a Linear Polynomial

A linear polynomial is of the form $p(x) = ax + b$, where $a \neq 0$. The graph of $y = ax + b$ is always a straight line.

A non-horizontal straight line will always cross the x-axis at exactly one point. The x-coordinate of this intersection is the single zero of the linear polynomial, which can be found algebraically as $x = -\frac{b}{a}$.

Graph of a linear polynomial (straight line) intersecting the x-axis at one point

The graph shows a straight line crossing the x-axis at one point, indicating it has exactly one zero.


Graph of a Quadratic Polynomial

A quadratic polynomial is of the form $p(x) = ax^2 + bx + c$, where $a \neq 0$. The graph of $y = ax^2 + bx + c$ is a U-shaped curve called a parabola.

Depending on its position, a parabola can interact with the x-axis in three possible ways, corresponding to the number of real zeros:

Case 1: Two distinct real zeros.

The graph intersects the x-axis at two different points.

Graph of a quadratic polynomial (parabola) intersecting the x-axis at two points

Case 2: One real zero (or two equal zeros).

The graph touches the x-axis at exactly one point (the vertex of the parabola is on the axis).

Graph of a quadratic polynomial (parabola) touching the x-axis at one point

Case 3: No real zeros.

The graph is entirely above or entirely below the x-axis and never intersects it. The zeros in this case are complex numbers.

Graph of a quadratic polynomial (parabola) not intersecting the x-axis

Graph of a Cubic Polynomial and Higher Degrees

A cubic polynomial ($p(x) = ax^3 + ...$) has a more complex, S-shaped curve. It can intersect the x-axis at one, two, or three points. Therefore, a cubic polynomial can have 1, 2, or 3 real zeros.

Example graphs of cubic polynomials intersecting the x-axis at 1, 2, or 3 points

In general, for a polynomial $p(x)$ of degree $n$, its graph can intersect the x-axis at at most $n$ points. This means a polynomial of degree $n$ can have at most $n$ real zeros.


Finding Information from a Graph

By observing the graph of a polynomial, we can deduce key information about it, primarily the number of real zeros and an estimate of its minimum degree.


Method to Find the Number of Zeros

This is a direct observation. The number of real zeros of the polynomial is equal to the number of times its graph intersects or touches the x-axis. Simply count these points.

Example A:

A parabola intersecting the x-axis at two distinct points.

The graph intersects the x-axis at 2 distinct points. Therefore, the polynomial has 2 real zeros.

Example B:

An S-shaped curve crossing the x-axis at three distinct points.

The graph intersects the x-axis at 3 distinct points. Therefore, the polynomial has 3 real zeros.

Example C:

A curve touching the x-axis at one point.

The graph touches the x-axis at exactly 1 point. Therefore, the polynomial has 1 real zero (which is a repeated root).

Example D:

A curve that does not intersect the x-axis at all.

The graph never intersects or touches the x-axis. Therefore, the polynomial has 0 real zeros.


Method to Estimate the Minimum Degree

The exact degree of a polynomial cannot always be determined from its graph, but we can find the minimum possible degree using two rules. You must check both and take the higher result.

Rule 1 (Based on Zeros): If the graph has $m$ real zeros, the degree of the polynomial, $n$, must be at least $m$.

Degree $(n) \geq$ Number of Real Zeros $(m)$

Rule 2 (Based on Turns): A "turn" on a graph is a point where it changes direction from increasing to decreasing (a peak) or vice versa (a valley). If the graph has $k$ turns, the degree, $n$, must be at least $k+1$.

Degree $(n) \geq$ Number of Turns $(k) + 1$


Applying the Degree Estimation Method to Examples

Let's apply these two rules to the previous examples to see how they work together.

Example A Analysis:

A parabola intersecting the x-axis at two distinct points.

Applying the Rules:

Both rules suggest the degree is at least 2. Therefore, the minimum possible degree is 2.

Example B Analysis:

An S-shaped curve crossing the x-axis at three distinct points.

Applying the Rules:

Both rules suggest the degree is at least 3. Therefore, the minimum possible degree is 3.

Example C Analysis:

A curve touching the x-axis at one point.

Applying the Rules:

Here, the rules give different minimums (1 and 2). We must choose the higher value to satisfy both conditions. Therefore, the minimum possible degree is 2. This shows why the 'turns' rule is crucial; relying only on zeros would give an incorrect estimate.

Example D Analysis:

A curve that does not intersect the x-axis at all.

Applying the Rules:

Again, we must choose the higher value. The minimum degree must be at least 2. Therefore, the minimum possible degree is 2. This is another clear example of why the 'turns' rule is essential, especially when a graph has few or no real zeros.


Worked Examples

Example 1. For the polynomial represented by the graph below, find the number of zeros and the minimum possible degree.

A parabola opening upwards, intersecting the x-axis twice.

Answer:

Solution

1. Number of Zeros:

By inspection, the graph intersects the x-axis at 2 distinct points.

So, the number of zeros is 2.

2. Minimum Degree:

Rule 1 (Zeros): Since there are 2 zeros, the degree $n \geq 2$.

Rule 2 (Turns): The graph has one "valley", so it has 1 turn. The degree $n \geq 1 + 1 = 2$.

Both rules suggest the minimum degree is 2. The polynomial is likely a quadratic polynomial.

Example 2. For the polynomial represented by the graph below, find the number of zeros and the minimum possible degree.

An S-shaped curve intersecting the x-axis three times.

Answer:

Solution

1. Number of Zeros:

The graph intersects the x-axis at 3 distinct points.

So, the number of zeros is 3.

2. Minimum Degree:

Rule 1 (Zeros): Since there are 3 zeros, the degree $n \geq 3$.

Rule 2 (Turns): The graph has one peak and one valley, for a total of 2 turns. The degree $n \geq 2 + 1 = 3$.

Both rules suggest the minimum degree is 3. The polynomial is likely a cubic polynomial.

Example 3. For the polynomial represented by the graph below, find the number of zeros and the minimum possible degree.

A W-shaped curve intersecting the x-axis twice.

Answer:

Solution

1. Number of Zeros:

The graph intersects the x-axis at 2 distinct points.

So, the number of zeros is 2.

2. Minimum Degree:

Rule 1 (Zeros): Since there are 2 zeros, the degree $n \geq 2$.

Rule 2 (Turns): The graph has two valleys and one peak, for a total of 3 turns. The degree $n \geq 3 + 1 = 4$.

We must take the higher of the two results (2 and 4). Therefore, the minimum possible degree is 4. The polynomial is likely a biquadratic (quartic) polynomial.

Example 4. For the polynomial represented by the graph below, find the number of zeros and the minimum possible degree.

A complex curve crossing the x-axis four times.

Answer:

Solution

1. Number of Zeros:

The graph crosses the x-axis at 4 distinct points.

So, the number of zeros is 4.

2. Minimum Degree:

Rule 1 (Zeros): Since there are 4 zeros, the degree $n \geq 4$.

Rule 2 (Turns): The graph has two peaks and one valley, for a total of 3 turns. The degree $n \geq 3 + 1 = 4$.

Both rules suggest the minimum degree is 4. The polynomial is likely a biquadratic (quartic) polynomial.



Relationship Between Zeros and Coefficients of a Polynomial

For polynomials of degree 1, 2, or 3, there are specific relationships that connect the values of their zeros to the values of their coefficients. These relationships are derived directly from the structure of the polynomial and are very useful in solving problems related to finding sums/products of zeros or constructing polynomials.


For a Quadratic Polynomial

Let the quadratic polynomial be $p(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are coefficients with $a \neq 0$. A quadratic polynomial has at most two real zeros. Let's denote these zeros as $\alpha$ (alpha) and $\beta$ (beta).

If $\alpha$ and $\beta$ are the zeros of $p(x)$, this means that when $x = \alpha$ or $x = \beta$, the value of the polynomial is zero ($p(\alpha) = 0$ and $p(\beta) = 0$). From the Factor Theorem, if $x=a$ is a zero of $p(x)$, then $(x-a)$ is a factor of $p(x)$. Therefore, if $\alpha$ and $\beta$ are the zeros, then $(x - \alpha)$ and $(x - \beta)$ are factors of $p(x)$.

So, $p(x)$ can be written as a product of these factors and a constant $k$:

$p(x) = k(x - \alpha)(x - \beta)$

Expand the right side of the equation:

$p(x) = k(x^2 - \beta x - \alpha x + \alpha\beta)$

$p(x) = k(x^2 - (\alpha + \beta)x + \alpha\beta)$

So, we have $ax^2 + bx + c = k(x^2 - (\alpha + \beta)x + \alpha\beta)$. Since this equality holds for all values of $x$, the coefficients of corresponding powers of $x$ on both sides must be proportional. By comparing the leading coefficients, we can see that $k$ must be equal to $a$.

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

Distribute $a$ on the right side:

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

Now, compare the coefficients of the terms with the same power of $x$ on both sides of the equation:

From the coefficient of $x$, we can find the sum of the zeros:

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

... (Sum of Zeros)

From the constant term, we can find the product of the zeros:

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

... (Product of Zeros)

These two formulas establish the relationship between the zeros and coefficients of a quadratic polynomial.

Converse: If the sum of the zeros ($\alpha + \beta = S$) and the product of the zeros ($\alpha\beta = P$) of a quadratic polynomial are known, then the polynomial can be written in the form $k(x^2 - Sx + P)$ for any non-zero constant $k$. If the leading coefficient is specified as $a$, the polynomial is $a(x^2 - Sx + P)$.


Example 1. Find the sum and product of the zeros of the quadratic polynomial $6x^2 - 7x - 3$.

Answer:

Given:

The quadratic polynomial $p(x) = 6x^2 - 7x - 3$.

To Find:

The sum and product of its zeros.

Solution:

The given polynomial is in the standard form $ax^2 + bx + c$. By comparing the coefficients:

$a = 6$

$b = -7$

$c = -3$

Let the zeros of the polynomial be $\alpha$ and $\beta$.

Using the relationship for the sum of zeros:

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

Substitute the values of $b$ and $a$:

$\alpha + \beta = -\frac{(-7)}{6}$

$\alpha + \beta = \frac{7}{6}$

... (1)

Using the relationship for the product of zeros:

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

Substitute the values of $c$ and $a$:

$\alpha\beta = \frac{-3}{6}$

$\alpha\beta = -\frac{1}{2}$

... (2)

Answer: The sum of the zeros is $\frac{7}{6}$ and the product of the zeros is $-\frac{1}{2}$.


Example 2. Find a quadratic polynomial whose zeros are 2 and -6.

Answer:

Given:

The zeros of a quadratic polynomial are 2 and -6.

To Find:

A quadratic polynomial with these zeros.

Solution:

Let the zeros of the quadratic polynomial be $\alpha = 2$ and $\beta = -6$.

First, calculate the sum of the zeros:

$S = \alpha + \beta = 2 + (-6) = 2 - 6 = -4$

... (1)

Next, calculate the product of the zeros:

$P = \alpha\beta = (2)(-6) = -12$

... (2)

A quadratic polynomial with sum of zeros $S$ and product of zeros $P$ can be written in the form $k(x^2 - Sx + P)$, where $k$ is any non-zero constant.

Using the calculated sum ($S = -4$) and product ($P = -12$) from equations (1) and (2), and choosing the simplest case where $k=1$, the polynomial is:

$p(x) = 1(x^2 - (-4)x + (-12))$

$p(x) = x^2 + 4x - 12$

... (3)

Other quadratic polynomials with these zeros can be obtained by choosing different values for $k$ (e.g., if $k=2$, $p(x) = 2(x^2 + 4x - 12) = 2x^2 + 8x - 24$). The simplest polynomial with leading coefficient 1 is $x^2 + 4x - 12$.

Answer: A quadratic polynomial whose zeros are 2 and -6 is $x^2 + 4x - 12$.


For a Cubic Polynomial

Let the cubic polynomial be $p(x) = ax^3 + bx^2 + cx + d$, where $a, b, c$, and $d$ are coefficients with $a \neq 0$. A cubic polynomial has at most three real zeros. Let's denote these zeros as $\alpha$ (alpha), $\beta$ (beta), and $\gamma$ (gamma).

If $\alpha, \beta$, and $\gamma$ are the zeros of $p(x)$, then $(x - \alpha)$, $(x - \beta)$, and $(x - \gamma)$ are factors of $p(x)$.

So, $p(x)$ can be written in the form:

$p(x) = k(x - \alpha)(x - \beta)(x - \gamma)$

where $k$ is a constant (which must be equal to $a$). Expand the right side:

$p(x) = a(x - \alpha)(x^2 - (\beta + \gamma)x + \beta\gamma)$

Now multiply $(x - \alpha)$ by the quadratic factor:

$p(x) = a[x(x^2 - (\beta + \gamma)x + \beta\gamma) - \alpha(x^2 - (\beta + \gamma)x + \beta\gamma)]$

$p(x) = a[x^3 - (\beta + \gamma)x^2 + \beta\gamma x - \alpha x^2 + \alpha(\beta + \gamma)x - \alpha\beta\gamma]$

Group terms by powers of $x$:

$p(x) = a[x^3 - (\beta + \gamma + \alpha)x^2 + (\beta\gamma + \alpha\beta + \alpha\gamma)x - \alpha\beta\gamma]$

$p(x) = a[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma]$

So, we have

$$ax^3 + bx^2 + cx + d = a x^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \beta\gamma + \gamma\alpha)x - a(\alpha\beta\gamma)$$

Comparing the coefficients of corresponding powers of $x$ on both sides:

From these comparisons, we get the relationships between the zeros and coefficients of a cubic polynomial:

Converse: If the sum of zeros ($S_1 = \alpha + \beta + \gamma$), the sum of products of zeros taken two at a time ($S_2 = \alpha\beta + \beta\gamma + \gamma\alpha$), and the product of zeros ($S_3 = \alpha\beta\gamma$) are known, then a cubic polynomial can be written in the form $k(x^3 - S_1 x^2 + S_2 x - S_3)$ for any non-zero constant $k$. If the leading coefficient is specified as $a$, the polynomial is $a(x^3 - S_1 x^2 + S_2 x - S_3)$.


Example 1. Verify the relationship between the zeros and the coefficients for the cubic polynomial $p(x) = 2x^3 - 5x^2 - 14x + 8$, given that its zeros are $4, -2$, and $\frac{1}{2}$.

Answer:

Solution

Given Polynomial: $p(x) = 2x^3 - 5x^2 - 14x + 8$

By comparing this with the general form $ax^3 + bx^2 + cx + d$, we get:

$a = 2, b = -5, c = -14, d = 8$

Given Zeros: $\alpha = 4, \beta = -2, \gamma = \frac{1}{2}$

Now, we verify each relationship.

(i) Sum of the zeros:

$\alpha + \beta + \gamma = 4 + (-2) + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2}$

From the formula, the sum should be:

$-\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$

Since $\frac{5}{2} = \frac{5}{2}$, the relationship is verified.

(ii) Sum of the products of zeros taken two at a time:

$\alpha\beta + \beta\gamma + \gamma\alpha = (4)(-2) + (-2)(\frac{1}{2}) + (\frac{1}{2})(4)$

$= -8 - 1 + 2 = -7$

From the formula, this sum should be:

$\frac{c}{a} = \frac{-14}{2} = -7$

Since $-7 = -7$, the relationship is verified.

(iii) Product of the zeros:

$\alpha\beta\gamma = (4)(-2)(\frac{1}{2}) = -4$

From the formula, the product should be:

$-\frac{d}{a} = -\frac{8}{2} = -4$

Since $-4 = -4$, the relationship is verified. All three relationships hold true.

Example 2. Find a cubic polynomial whose zeros are $3, -1$, and $-2$.

Answer:

Solution

Let the zeros be $\alpha = 3, \beta = -1, \gamma = -2$.

We first calculate the sum of zeros ($S_1$), the sum of products of zeros taken two at a time ($S_2$), and the product of zeros ($S_3$).

Sum of zeros ($S_1$):

$\alpha + \beta + \gamma = 3 + (-1) + (-2) = 3 - 3 = 0$

Sum of products of zeros taken two at a time ($S_2$):

$\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-2) + (-2)(3)$

$= -3 + 2 - 6 = -7$

Product of zeros ($S_3$):

$\alpha\beta\gamma = (3)(-1)(-2) = 6$

The cubic polynomial is given by the form $p(x) = k(x^3 - S_1 x^2 + S_2 x - S_3)$.

Substituting the values of $S_1, S_2$, and $S_3$:

$p(x) = k(x^3 - (0)x^2 + (-7)x - (6))$

$p(x) = k(x^3 - 7x - 6)$

For the simplest polynomial, we can take $k=1$.

Thus, the required cubic polynomial is $x^3 - 7x - 6$.

Example 3. Two zeros of the cubic polynomial $p(x) = x^3 - 2x^2 - 5x + 6$ are $3$ and $-2$. Find the third zero.

Answer:

Solution

Given Polynomial: $p(x) = x^3 - 2x^2 - 5x + 6$

The coefficients are $a = 1, b = -2, c = -5, d = 6$.

Let the three zeros be $\alpha, \beta$, and $\gamma$.

We are given two zeros: $\alpha = 3$ and $\beta = -2$. We need to find the third zero, $\gamma$.

We can use any of the three relationships. The simplest one to use here is the sum of the zeros.

Relationship for the sum of zeros:

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

Substitute the known values:

$3 + (-2) + \gamma = -\frac{-2}{1}$

Simplify the equation:

$1 + \gamma = 2$

Solve for $\gamma$:

$\gamma = 2 - 1 = 1$

Therefore, the third zero of the polynomial is 1.

Alternate Solution (Using Product of Zeros)

We can also use the relationship for the product of zeros:

$\alpha\beta\gamma = -\frac{d}{a}$

Substitute the known values:

$(3)(-2)\gamma = -\frac{6}{1}$

$-6\gamma = -6$

$\gamma = \frac{-6}{-6} = 1$

This confirms that the third zero is 1.



Division Algorithm for Polynomials

Just as you can divide one number by another to get a quotient and a remainder (for instance, $25 \div 4$ gives a quotient of 6 and a remainder of 1, so $25 = 4 \times 6 + 1$), a similar process exists for polynomials. The Division Algorithm for Polynomials is a fundamental theorem that formalizes this procedure.


Statement of the Division Algorithm

The Division Algorithm for Polynomials states that for any two polynomials, a dividend $p(x)$ and a non-zero divisor $g(x)$, there exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that:

$p(x) = g(x) \cdot q(x) + r(x)$

The crucial condition is that the remainder, $r(x)$, must either be the zero polynomial ($r(x) = 0$) or its degree must be strictly less than the degree of the divisor, $g(x)$.

Condition: $r(x) = 0$ or degree($r(x)$) < degree($g(x)$)

This condition is what tells us when to stop the long division process.


The Method of Polynomial Long Division

To find the quotient $q(x)$ and remainder $r(x)$, we use a method similar to long division for numbers. Here are the steps:

  1. Arrange: Write both the dividend and the divisor in standard form (descending powers of the variable). If any term is missing, insert it with a coefficient of 0. For example, write $x^3 - 2$ as $x^3 + 0x^2 + 0x - 2$.
  2. Divide: Divide the first term of the dividend by the first term of the divisor. This gives the first term of the quotient.
  3. Multiply: Multiply the entire divisor by this first term of the quotient.
  4. Subtract: Write this product below the dividend and subtract it to get a new polynomial. (Be very careful with signs!).
  5. Repeat: Treat the new polynomial as the new dividend and repeat steps 2-4. Continue this process until the degree of the remainder is less than the degree of the divisor.

Worked Examples

Example 1. Divide the polynomial $p(x) = x^3 - 3x^2 + 5x - 3$ by the polynomial $g(x) = x^2 - 2$. Find the quotient and remainder, and verify the result.

Answer:

Solution

Step 1: Arrange. Both polynomials are already in standard form.

$p(x) = x^3 - 3x^2 + 5x - 3$

$g(x) = x^2 + 0x - 2$

Steps 2-5: Perform Long Division.

$\begin{array}{r} x-3\phantom{} \\ x^2-2{\overline{\smash{\big)}\,x^3-3x^2+5x-3\phantom{}}} \\ \underline{-~\phantom{(}(x^3\phantom{-3x^2}-2x)}\phantom{-3)} \\ -3x^2+7x-3\phantom{)} \\ \underline{-~\phantom{()}(-3x^2\phantom{+7x}+6)} \\ 7x-9\phantom{)} \end{array}$

The division stops because the degree of the remainder ($7x-9$, degree 1) is less than the degree of the divisor ($x^2-2$, degree 2).

From the division, we find:

Quotient, $q(x) = x - 3$

Remainder, $r(x) = 7x - 9$

Verification

According to the division algorithm, $p(x) = g(x) \cdot q(x) + r(x)$.

RHS = $(x^2 - 2)(x - 3) + (7x - 9)$

= $(x^3 - 3x^2 - 2x + 6) + (7x - 9)$

= $x^3 - 3x^2 - 2x + 7x + 6 - 9$

= $x^3 - 3x^2 + 5x - 3$

= $p(x)$ = LHS

Thus, the result is verified.

Example 2. Divide $p(x) = x^4 - 3x^2 + 4x + 5$ by $g(x) = x^2 - x + 1$.

Answer:

Solution

Step 1: Arrange. The dividend $p(x)$ is missing the $x^3$ term, so we add $0x^3$.

$p(x) = x^4 + 0x^3 - 3x^2 + 4x + 5$

$g(x) = x^2 - x + 1$ (already in standard form)

Steps 2-5: Perform Long Division.

$\begin{array}{r} x^2+x-3\phantom{x^2+x} \\ x^2-x+1{\overline{\smash{\big)}\,x^4+0x^3-3x^2+4x+5\phantom{}}} \\ ~\phantom{(...........}\underline{-(x^4-x^3+x^2)}\phantom{.......+5)} \\ x^3-4x^2+4x\phantom{+5} \\ \underline{-~\phantom{(}(x^3-x^2+x)}\phantom{+5)} \\ -3x^2+3x+5\phantom{)} \\ \underline{-~\phantom{()}(-3x^2+3x-3)} \\ 8\phantom{)} \end{array}$

The division stops because the degree of the remainder (8, degree 0) is less than the degree of the divisor ($x^2-x+1$, degree 2).

Answer: The quotient is $q(x) = x^2 + x - 3$ and the remainder is $r(x) = 8$.

Example 3. Check whether the polynomial $g(t) = t^2 - 3$ is a factor of the polynomial $p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$.

Answer:

Solution

$g(t)$ is a factor of $p(t)$ if and only if the remainder is 0 when we divide $p(t)$ by $g(t)$.

Step 1: Arrange.

$p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$

$g(t) = t^2 + 0t - 3$

Steps 2-5: Perform Long Division.

$\begin{array}{r} 2t^2+3t+4\phantom{2t^2+} \\ t^2-3{\overline{\smash{\big)}\,2t^4+3t^3-2t^2-9t-12\phantom{}}} \\ \underline{-~\phantom{(}(2t^4\phantom{+3t^3}-6t^2)}\phantom{-9t-12)} \\ 3t^3+4t^2-9t\phantom{-12} \\ \underline{-~\phantom{(}(3t^3\phantom{+4t^2}-9t)}\phantom{-12)} \\ 4t^2+0t-12\phantom{)} \\ \underline{-~\phantom{()}(4t^2\phantom{+0t}-12)} \\ 0\phantom{)} \end{array}$

Since the remainder $r(t) = 0$, the divisor $g(t)$ divides the dividend $p(t)$ exactly.

Answer: Yes, $g(t) = t^2 - 3$ is a factor of $p(t) = 2t^4 + 3t^3 - 2t^2 - 9t - 12$.


Application: Finding Zeros of a Polynomial

The division algorithm is a powerful tool for finding the zeros of a polynomial when one or more zeros are already known.

Example 4. Find all the zeros of the polynomial $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$, if you know that two of its zeros are $\sqrt{2}$ and $-\sqrt{2}$.

Answer:

Solution

Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros, the corresponding factors are $(x - \sqrt{2})$ and $(x + \sqrt{2})$.

Therefore, their product must also be a factor of $p(x)$.

$g(x) = (x - \sqrt{2})(x + \sqrt{2}) = x^2 - (\sqrt{2})^2 = x^2 - 2$.

Now, we divide $p(x)$ by $g(x) = x^2 - 2$ to find the other factors.

$\begin{array}{r} 2x^2-3x+1\phantom{2x^2} \\ x^2-2{\overline{\smash{\big)}\,2x^4-3x^3-3x^2+6x-2\phantom{}}} \\ \underline{-~\phantom{(}(2x^4\phantom{-3x^3}-4x^2)}\phantom{+6x-2)} \\ -3x^3+x^2+6x\phantom{-2} \\ \underline{-~\phantom{()}(-3x^3\phantom{+x^2}+6x)}\phantom{-2)} \\ x^2+0x-2\phantom{)} \\ \underline{-~\phantom{()}(x^2\phantom{+0t}-2)} \\ 0\phantom{)} \end{array}$

The quotient is $q(x) = 2x^2 - 3x + 1$. The remaining zeros of $p(x)$ are the zeros of $q(x)$.

To find the zeros of $q(x)$, we set it to zero and solve:

$2x^2 - 3x + 1 = 0$

We can factor this by splitting the middle term:

$2x^2 - 2x - x + 1 = 0$

$2x(x - 1) - 1(x - 1) = 0$

$(2x - 1)(x - 1) = 0$

This gives two solutions: $2x-1=0 \implies x=\frac{1}{2}$ and $x-1=0 \implies x=1$.

Answer: The zeros of the polynomial are the two given zeros and the two we just found. All the zeros are $\sqrt{2}, -\sqrt{2}, 1$, and $\frac{1}{2}$.