Polynomials: Types, Equations, Degrees & Theorems

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Polynomials are expressions that contain constants, variables, exponents, and terms. Polynomials are important mathematical expressions in algebra. They have several applications in mathematics as well as physics. Several operations such as addition, subtraction, and multiplication can be performed on polynomials. Based on that, the division algorithm, factor theorem, and remainder theorem are proved.


Polynomials

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Polynomials are defined as expressions that contain constants, variables, exponents, and terms. The term polynomials is derived from the words 'poly' meaning 'many' and 'nomial' meaning terms. Thus, a polynomial is something that has many terms.

The general expression of polynomials is

a0xn + a1xn-1 + a2xn-2 + … an

Where,

n is a non-negative integer

x is a variable

a0, a1, a2, …, an are real numbers

Some examples of Polynomials are:

x2 + 6x + 9;

2x + 5;

x2 + 4y + 2;

x3 + y3 + 3xy(x + y);

w + x + y + z;

The video below explains this:

Polynomials Detailed Video Explanation:

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Types Of Polynomials

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Every expression in a polynomial is known as a term of that polynomial. For example, in the polynomial x2 + 4y + 2, the expressions x2, 4y, and 2 are each term of that polynomial.

A polynomial should have at least one term. No polynomials can have zero terms. The terms of a polynomial can be constants, variables, or a mixture of constants and variables.

  • Monomials: Polynomials containing a single term are called monomials. Example: 2x
  • Binomials: Polynomials containing two terms are called binomials. Example: 2x + y
  • Trinomials: Polynomials containing three terms are called trinomials. Example: 2x2 + 4y + 2

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Important Terms Related to Polynomial

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Some of the important terms related to polynomials are as follows:

  • Coefficients Of Polynomials: Every term in polynomials has a coefficient. In the polynomial x2 + 4y + 2, the terms are x2, 4y, and 2. The coefficient of x2 is 1 and of y is 4. Thus, for a polynomial ax2 + bx + c, a and b are coefficients where at least one of the coefficients is non-zero. 
  • Constants Of Polynomials: In polynomials, the real numbers without any variables are called constants. For example, in the polynomial x2 + 4y + 2, 2 is the constant. A polynomial having no variables is called a constant polynomial. 32 is a constant polynomial.
  • Exponents Of Polynomials: Exponents of polynomials are defined as the power to which the variables are raised. Example: In polynomial x2, 2 is the exponent.

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Degree Of A Polynomial

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The degree of a polynomial is defined as the highest power of the polynomial. For example, in the polynomial x2 + 4y + 2, 2 is the degree of that polynomial. And in the polynomial x3 + y3 + 3xy(x + y), 3 is the degree of the polynomial. For a constant polynomial, the degree of the polynomial is zero.

Also Read: Permutations and Combinations 

Based on the degree of polynomials, the polynomials can be classified as linear, quadratic, cubic, etc.

  • Linear Polynomials: The polynomials having the degree of polynomial one are called linear polynomials. Example: 2x + 5
  • Quadratic Polynomials: The polynomials having the degree of polynomial two are called quadratic polynomials. Example: x2 + 6x + 9
  • Cubic Polynomials: The polynomials having the degree of polynomial three are called cubic polynomials. Example: x3 + y3 + 3xy(x + y)
  • Polynomials In One Variable: Polynomials in one variable are defined as the polynomials that have only one type of variable in all of its terms. Example: 3x + 2, 5x2 + 3x + 7, etc.

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Zeros Of Polynomials

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In a polynomial p(x), a real number 'a' is called zero of that polynomial if p(a) = 0.

Zeros of polynomials follow the rules of algebraic identities.

Sum Of Zeros Of Polynomials

For a quadratic polynomial ax2 + bx + c = 0, if α and β are the two zeros of that polynomial then the sum of the zeros of the polynomial is given as,

α + β = -ba = −coefficient of x . coefficient of x2

For a cubic polynomial ax3 + bx2 + cx + d = 0, if α, β, and γ are the three zeros of that polynomial then the sum of the zeros of the polynomial is given as,

α + β + γ = -ba = −coefficient of x2 . coefficient of x3

Also,

αβ + βγ + γα = ca = coefficient of x . coefficient of x3

Also Read: Types of Probability

Product Of Zeros Of Polynomials

For a quadratic polynomial ax2 + bx + c = 0, if α and β are the two zeros of that polynomial then the product of the zeros of the polynomial is given as,

αβ = ca = constant term . coefficient of x2

For a cubic polynomial ax3 + bx2 + cx + d = 0, if α, β, and γ are the three zeros of that polynomial then the product of the zeros of the polynomial is given as,

αβγ = -da = -constant term coefficient of x3


Division Algorithm

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It is known that,

Dividend = Divisor x Quotient + Remainder

Therefore, for two polynomials p(x) and g(x) where g(x) ≠ 0, the relationship between these two polynomials and their remainder r(x) and quotient q(x) is given as,

p(x) = g(x) × q(x) + r(x)

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Remainder Theorem

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Remainder Theorem states that if any polynomial p(x) has the degree of polynomial greater than or equal to one and if this polynomial is divided by a linear polynomial (x - a) then the remainder of this division is p(a).

Proof: Let p(x) be any polynomial with the degree of polynomial greater than or equal to 1.

Suppose, p(x) is divided by (x – a), and the quotient is q(x) and the remainder is r(x).

Therefore, by Division Algorithm,

p(x) = (x – a) x q(x) + r(x) …(1)

Since the degree of (x – a) is 1 and the degree of r(x) is less than the degree of (x – a), the degree of r(x) = 0.

This means that r(x) is a constant. Let us assume this constant to be r.

Thus, for every value of x, r(x) = r.

Therefore, we can write equation (1) as,

p(x) = (x – a) x q(x) + r

If one assumes that x = a, then the above equation can be written as

p(a) = (a – a) x q(a) + r = r

This proves the theorem.

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Factorisation Of Polynomials

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From the Division Algorithm, it is known that p(x) = g(x) × q(x) + r(x)

But if the remainder is zero i.e. r(x) = 0 then

p(x) = g(x) × q(x)

This proves that a single polynomial can be expressed as a product of two or more polynomials.

The process of obtaining two or more polynomials from a single polynomial such that their products give the original polynomial is known factorisation of polynomials.

Example: 6p2 + 3p = 3p x (2p + 1)

Here 3p and (2p + 1) are factors of 6p2 + 3p.

Also Read: Complex numbers and Quadratic equations


Factor Theorem

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Factor Theorem states that “for a polynomial p(x), (x - c) is its factor if and only if p(c) = 0”.

Also, if (x - c) is a factor of a polynomial p(x) then p(c) = 0.

Proof: It is known that Remainder Theorem gives us the equation

p(x) = (x – a) x q(x) + p(a)

If one assumes that p(a) = 0, then the above equation can be written as,

p(x) = (x – a) x q(x)

This proves that (x - a) is a factor of p(x).

Also,

If (x - a) is a factor of p(x), then

p(x) = (x – a) x g(x)

Where,

 g(x) is a polynomial which is a factor of p(x).

If one assumes that x = a then the above equation can be written as,

p(x) = (a – a) x g(x)

p(x) = 0

This proves the Factor Theorem.

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Methods Of Factorisation Of Polynomials

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Factorisation of polynomials can be done with a number of methods. Some of them are:

  1. Factorisation by dividing the polynomial by its Highest Common Factor (HCF).
  2. Factorisation by grouping the terms in the polynomial.
  3. Factorisation by use of identities.

Important Identities:

  • (x + y)2 = x2 + 2xy + y2
  • (x – y)2 = x2 – 2xy + y2
  • x2 – y2 = (x + y) (x – y)
  • (x + a) (x + b) = x2 + (a + b)x + ab
  • (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
  • (x + y)3 = x3 + y3 + 3xy(x + y)
  • (x – y)3 = x3 – y3 – 3xy(x – y)
  • x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

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Things to Remember

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  • Polynomials are defined as expressions that contain constants, variables, exponents, and terms.

The general expression of polynomials is

a0xn + a1xn-1 + a2xn-2 + … an

  • Every polynomial has at least one constant, coefficient, and exponent.
  • The degree of a polynomial is the highest power to which the variable term is raised in the polynomial.
  • Dividend = Divisor x Quotient + Remainder
  • The Remainder Theorem states that if any polynomial p(x) has the degree of polynomial greater than or equal to one and if this polynomial is divided by a linear polynomial (x - a) then the remainder of this division is p(a).
  • Factor Theorem states that for a polynomial p(x), (x - c) is its factor if and only if p(c) = 0. Also, if (x - c) is a factor of a polynomial p(x) then p(c) = 0.

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Sample Questions

Ques. Define a polynomial with an example. What is a linear polynomial? [3 Marks]

Ans. A polynomial is defined as an expression that contains constants, variables, exponents, and terms. An example of a polynomial is 3x2 + 4y.

A polynomial having a degree of polynomial one is called a linear polynomial.

Ques. Factorize 6y2 + 17y + 5 by dividing the middle term. [3 Marks]

Ans. The given polynomial is

6y2 + 17y + 5

The middle term can be divided as 17y = (15 + 2)y

= 6y2 + (2 + 15)y + 5

= 6y2 + 2y + 15y + 5

= 2y(3y + 1) + 5(3y + 1)

= (3y + 1)(2y + 5)

Ques. Find the value of 'p' from the polynomial x2 + 3x + p, if one of the zeroes of the polynomial is 2. [3 Marks]

Ans. The given polynomial is

x2 + 3x + p

2 is one of the zeros of the given polynomial.

It is known that

If p(a) is a zero of the polynomial p(x) then p(a) = 0

So let’s assume x2 + 3x + p = 0

If 2 is the zero of the above polynomial then it should satisfy the polynomial.

So substitute x = 2 in x2 + 3x + p = 0

22 + 3(2) + p = 0

4 + 6 + p = 0

10 + p = 0

p = -10

∴ The value of 'p' from the polynomial x2 + 3x + p is -10.

Ques. Find out the zeros of the polynomial (x - 3)2 - 4. [3 Marks]

Ans. The given polynomial is (x - 3)2 - 4.

Expanding the bracket in the given polynomial we get the equation

x2 + 9 - 6x - 4

x2 - 6x + 5

The degree of the polynomial is two so it is clearly a quadratic polynomial.

It is known that quadratic polynomials have two zeros of polynomials.

Let's assume x2 - 6x + 5 = 0 to solve and get the zeros of the polynomial.

x2 - 6x + 5 = 0

x2 - x - 5x + 5 = 0

x(x - 1) - 5(x - 1) = 0

(x - 1)(x - 5) = 0

So either (x - 1) = 0 or (x - 5) = 0

Therefore, x = 1 or x = 5

So, the two zeros of the polynomial (x - 3)2 - 4 are 1 and 5.

Ques. Check whether x = -16 is zero of the polynomial p(a) = 6a + 1. [3 Marks]

Ans. It is known that

If p(x) is a zero of the polynomial p(a) then p(x) = 0

Substituting x = -16 in polynomial 6a + 1

6-16+ 1 = -1 + 1 = 0

Since, p(a) = 0, x = -16 is a zero of the polynomial p(a) = 6a + 1

Ques. Find the value of polynomial 2x2 + 5x + 1 at x = 3. [3 Marks]

Ans. The given polynomial is

2x2 + 5x + 1

To find its value at x = 3 one must substitute x = 3 in the above polynomial.

Substituting x = 3 one gets

2(3)2 + 5(3) + 1

= 2(9) + 15 + 1

= 18 + 16

= 34

∴ The value of polynomial 2x2 + 5x + 1 at x = 3 is 34.

Ques. Factorise the polynomial 8x3 + 27y3 + 36x2y + 54xy2. [2 Marks]

Ans. The given polynomial is

8x3 + 27y3 + 36x2y + 54xy2

It can be written as

(2x)3 + (3y)3 + 3(4x2)(3y) + 3(2x)(9y2)

= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2

= (2x + 3y)3

This is done by using the identity (x + y)3 = x3 + y3 + 3xy(x + y).

Ques. : Evaluate 105 × 106 without direct multiplication. [3 Marks]

Ans. 105 × 106 = (100 + 5) × (100 + 6)

= (100)2 + (5 + 6)(100) + (5 × 6)

(This step is done by using the identity (x + a) (x + b) = x2 + (a + b)x + ab)

= 10000 + 1100 + 30

= 11130

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CBSE X Related Questions

  • 1.
    The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

      • 0
      • 1
      • 3
      • 2

    • 2.
      A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.


        • 3.
          A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

            • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
            • \(314\) \(\text{cm}^{2}\)
            • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
            • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

          • 4.
            Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
            Reason (R) : Sum of the any two irrational numbers is always irrational.

              • 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.

            • 5.
              PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

                • \(a^2 + (a + 2)^2 = (2b)^2\)
                • \(b^2 = a + 4\)
                • \(2a^2 + 1 = b^2\)
                • \(b^2 = a + 1\)

              • 6.
                Prove that \(2 + 3\sqrt{5}\) is an irrational number given that \(\sqrt{5}\) is an irrational number.

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