Polynomials: Definition, Types and Examples

Polynomial is an expression which comprises of variables and coefficient involving the basic arithmetic operations of addition, subtraction, and Multiplication and the exponential negative exponential of variables. 

The origin of the word polynomial is from the Greek language. ’Poly – many’ and ‘nomial’ – terms. So as a whole the word polynomial refers to many terms/multiple terms.

Also Read:- Class 10 Degree of polynomial

There are several features of a polynomial: -

  • Notation
  • Degree
  • Terms
  • Types

Notation: - Polynomials are denoted by p(x)where x is the variable and p is the denotation.

Example: - 5x2+3x-2


Degree of an Polynomial

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Degree of a polynomial is defined as the highest power of a monomial present in a polynomial expression. So, a polynomial equation with consists of the highest exponential power is known as the degree of a polynomial.

Polynomials are categorized into 5 categories based on their degree: -

Polynomial

Degree

Example

Constant

0

3

Linear Polynomial 

1

4x+1

Quadratic Polynomial 

2x2+6x+2

Cubic Polynomial 

3

5x3+ 7x2 +2x+3

Quartic polynomial 

4

8x4+2x3+x2+2x+1

The video below explains this:

Polynomials Detailed Video Explanation:


Terms of a Polynomial 

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A polynomial is made up of 2 terms. Mainly the constant and the variable.

Constant: - The term whose value is pre-defined and does not change in any case is known as a constant. 

Example: - 2,6,8

Variable: - The term whose value is not predefined and is bound to change with the requirements of the problem.

Example: - x,y

Also Read:- Complex Numbers and Quadratic Equations


Types of Polynomials

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There are Constituently 3-types of polynomials: -

Monomial

A polynomial that comprises a single term/mono term is known as a monomial expression. The single term in this monomial expression should be non zero.

Example: - 7x

Binomial 

A polynomial expression that consists of precisely two terms is known as Binomial expression. It can also be understood as an arithmetic operation between two monomial expressions.

Example: - 3x+8, 5x-9

Trinomial 

A polynomial that consists of exactly 3 terms is known as a trinomial expression. It can also be interpreted as an arithmetic operation between one Binomial and Monomial term or an operation between 3 monomial terms.

Example: - 4x2+3x+1 

  • Zeroes of Polynomial

Zeros of a polynomial are known as the points where the polynomial becomes zero as a whole. A polynomial having a value zero is called a zero polynomial. Highest power associated with variable x is the degree of a polynomial.

Also Read: 


Solving Polynomial Equations

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Polynomial equations require basic knowledge of Arithmetic and Algebra operations and some basic properties that are specific to polynomial equations.

Algebraic Properties of Polynomials: -

Division Algorithm-If a polynomial P(x) is divided by a polynomial f(x) results in quotient Q(x) with the remainder an (x), then,

P(x) = f(x) • Q(x) + a(x)

Bezout’s Theorem-

Polynomial P(x) is divisible by binomial (x – y) if and only if P(y) = 0.

Remainder Theorem-

If P(x) is divided by (x – y) with remainder r, then P(y) = r.

Factor Theorem-

A polynomial P(x) divided by f(x) results in q(x) with zero remainders if and only if f(x) is a factor of P(x)

Intermediate Value Theorem-

If P(x) is a polynomial, and P(x) is not equal to P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].

Arithmetic property-

The addition, subtraction, and multiplication of polynomials P and Q give an output in the form of a polynomial where,

Degree(P ± Q) ≤ Degree(P or Q)

Degree(P × Q) = Degree(P) + Degree(Q)

Fundamental Theorem of Algebra-

Every non-constant single-variable polynomial which has complex coefficients has at least one complex root

Also Read:- Roots of Polynomials


Solving Quadratic polynomial equations

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Quadratic polynomial equations are solved using polynomial factorization.

The equation has to be rearranged according to the degree so it becomes easy to factorize it further.

Equating the equation to zero is a necessary step as well.

Example:   Arrange the polynomial in the decreasing order of degree and then make them equal to zero

⇒ x3 + 3x2 -6x – 18 = 0

Now, take the common terms.

x2(x+3) – 6(x+3) =0

⇒ (x2-6)(x+3)=0

So, the solutions will be x =-3 and

x2 = 6

Or, x = √6

Polynomial Operations: -

There are mainly 4 polynomial operations: -

  • Addition of Polynomials – Polynomials of the same degree and variable are added. The result is also a polynomial
  • Subtraction of Polynomials – Polynomials of the same degree and variable are subtracted. The result is also a polynomial
  • Multiplication of Polynomials- Two or more polynomial when multiplied always result in the form of a polynomial of higher degree

Example – 

6x ×(2x+5y)–3y × (2x+5y) ——— Using distributive law of multiplication

⇒ (12x2+30xy) – (6yx+15y2) ——— Using distributive law of multiplication

⇒12x2+30xy–6xy–15y2 ————— as XY = yx

Thus, (6x−3y)×(2x+5y)=12x2+24xy−15y2

Division of Polynomials – 

  1. Write the polynomial expression in descending order.
  2. Check the highest power and divide the terms by it.
  3. Use result from step 2 as the division symbol.
  4. Now subtract it and bring it down the next term.
  5. Repeat steps 2 to 4 until you don’t have any more terms to carry down.
  6. You get the final answer, including the remainder, which will be in the fraction form (last subtract term).

Sample Linear Polynomial Problem with a solution

Question: (2x-6) Solve this polynomial for the value of x (2 marks)

Answer: -  Let us assume the equation to be zero.

    2x-6=0

Eventually, if the RHS is zero then the LHS value will shift to RHS and will change the sign

    2x=6

    x=6/2

     x=3

Hence the equation is solved using basic arithmetic operations


Previous Year Question Polynomial

Question: If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. (CBSE 2012)

Solution: Here a = 3, b = -k, c = 6
Sum of the zeroes, (α + β) = −ba = 3 …. (given)
⇒ −(−k)3 = 3
⇒ k = 9

Question: If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. (CBSE 2017 D)

Solution: p(x) = (k2 – 14) x2 – 2x – 12
Here a = k2 – 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 …[Given]
⇒ −ba = 1
⇒ −(−2)k2−14 = 1
⇒ k2 – 14 = 2
⇒ k2 = 16
⇒ k = ±4

Question: If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial. (CBSE2016 D)

Solution: Quadratic polynomial is x2 – Sx + P = 0
⇒ x2 – (-6)x + 5 = 0
⇒ x2 + 6x + 5 = 0

Question: Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other. (CBSE 2017 OD)

Solution: Let α and 1α be the zeroes of P(x).
P(a) = ax2 + bx + c …(given)
Product of zeroes = ca
⇒ α × 1α = ca
⇒ 1 = ca
⇒ a = c (Required condition)
Coefficient of x2 = Constant term

Question: If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, find the value of p and q (CBSE 2012)

Solution: We have, 2x2 – 5x – 3 = 0
= 2x2 – 6x + x – 3
= 2x(x – 3) + 1(x – 3)
= (x – 3) (2x + 1)
Zeroes are:
x – 3 = 0 or 2x + 1 = 0
⇒ x = 3 or x = −12
Since the zeroes of required polynomial is double of given polynomial.
Zeroes of the required polynomial are:
3 × 2, (−12 × 2), i.e., 6, -1
Sum of zeroes, S = 6 + (-1) = 5
Product of zeroes, P = 6 × (-1) = -6
Quadratic polynomial is x2 – Sx + P
⇒ x2 – 5x – 6 …(i)
Comparing (i) with x2 + px + q
p = -5, q = -6

Also Read: 

CBSE X Related Questions

  • 1.
    Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)


      • 2.
        Solve the linear equations \(3x + y = 14\) and \(y = 2\) graphically.


          • 3.
            Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

              • \(2\pi r^3\)
              • \(3\pi r^3\)
              • \(5\pi r^3\)
              • \(4\pi r^3\)

            • 4.
              Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

                • \(10^{\circ}\)
                • \(60^{\circ}\)
                • \(45^{\circ}\)
                • \(100^{\circ}\)

              • 5.
                If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

                  • \(4 \text{ cm}\)
                  • \(4\sqrt{2} \text{ cm}\)
                  • \(8 \text{ cm}\)
                  • \(2\sqrt{2} \text{ cm}\)

                • 6.
                  Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
                  Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

                    • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
                    • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
                    • Assertion (A) is true, but Reason (R) is false.
                    • Assertion (A) is false, but Reason (R) is true.

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