**(1) Algebraic Expressions :Â Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
**

(2) Polynomials :Â **The expression which contains one or more terms with non-zero coefficient is called a polynomial. A polynomial can have any number of terms.**

** For example:**Â 10, a + b, 7x + y + 5, w + x + y + z, etc. are some polynomials.

(3) Polynomials in One Variable :Â The expression which contains only one type of variable in entire expression is called a polynomial in one variable.

** For example:Â **2x, a

(4) Term : A term is either a single number or variable and it can be combination of numbers and variable. They are usually separated by different operators like +, -, etc.

** For example:**Â Consider an expression 6x - 7. Then, the terms in this expression are 6x and -7.

(5) Coefficient : The number multiplied to variable is called as coefficient.

** For example:Â **The coefficient of the term 2x will be 2.

(6) Constant Polynomials :Â An expression consisting of only constants is called as constant polynomial.

** For Example:Â **7, -27, 3, etc. are some constant polynomials.

(7) Zero Polynomial : The constant polynomial 0 is called as zero polynomial.

**
(8) Denoting Polynomials in One Variable:
**Let us take an example to understand it:

If the variable in a polynomial is x, then we can denote the polynomial by p(x) or q(x) etc.

(9) Monomials : The expressions which have only one term are called as monomials.

** For Example:Â **p(x) = 3x, q(a) = 2a

(10) Binomials : The expressions which have two terms are called as binomials.

** For example:Â **r(x) = x + 10, c(z) = 7z

**
(11) Trinomials :Â **The expressions which have three terms are called as trinomials.

(12) Degree of polynomial : The highest power of the variable in a polynomial is called as the degree of the polynomial.

** For Example:Â **The degree of p(x) = x

Note: The degree of a non-zero constant polynomial is zero.

(13) Linear polynomial : A polynomial of degree one is called a linear polynomial.

** For Example:Â **2x â€“ 7, s + 5, etc. are some linear polynomials.

(14) Quadratic polynomial :Â A polynomial having highest degree of two is called a quadratic polynomial. In general, a quadratic polynomial can be expressed in the form ax^{2} + bx + c, where aâ‰ 0 and a, b, c are constants.

** For Example:Â **x

(15) Cubic polynomial : A polynomial having highest degree of three is called a cubic polynomial. In general, a quadratic polynomial can be expressed in the form ax^{3} + bx^{2} + cx + d, where aâ‰ 0 and a, b, c, d are constants.

**For E xample:Â **x

(16) General expression of polynomial : A polynomial in one variable x of degree n can be expressed as a_{n} x^{n }+ a_{n-1 }x^{n-1 }+ ..... + a_{1} x + a_{0}, where a_{n} â‰ 0 and a_{0}, a_{1}, .... a_{n }are constants.

(17) Zeroes of a Polynomial : The value of variable for which the polynomial becomes zero is called as the zeroes of the polynomial.

** For Example:Â **Consider p(x) = x + 2. Find zeroes of this polynomial.

(i) If we put x = -2 in p(x), we get,

(ii) p(-2) = -2 + 2 = 0.

(iii) Thus, -2 is a zero of the polynomial p(x).

**
(18) Some Note-worthy Points:
**(i) A non-zero constant polynomial has no zero.

(ii) A linear polynomial has one and only one zero.

(iii) A zero of a polynomial might not be 0 or 0 might be a zero of a polynomial.

(iv) A polynomial can have more than one zero.

(19) Some Examples:

** For Example:Â **Find value of polynomial 3a

(i) Here, p(a) = 3a

(ii) Now, substituting a = 3, we get,

(iii) p(3) = 3 x (3)

** For Example:Â **Check whether at x = -1/7 is zero of the polynomial p(x) = 7x + 1.

(i) Given, p(x) = 7x + 1.

(ii) Now, substituting x = -1/7, we get,

(iii) p(-1/7) = 7(-1/7) + 1 = -1 + 1 = 0.

(iv) Here, p(-1/7) is zero. Thus, -1/7 is zero of the given polynomial.

** For Example:Â **Find zero of the polynomial p(x) = 2x+ 2.

(i) Equating p(x) to zero, we get,

(ii) p(x) = 0

(iii) 2x + 2 = 0

(iv) 2x = -2 i.e. x = -1.

(v) Thus, x = -1 is a zero of the given polynomial.

(20) Remainder Theorem:

*Statement*:Â Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x â€“ a, then the remainder is p(a).

** Proof :
**(i) Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x â€“ a, the quotient is q(x) and the remainder is r(x), i.e., p(x) = (x â€“ a) q(x) + r(x)

(ii) Since the degree of (x â€“ a) is 1 and the degree of r(x) is less than the degree of (x â€“ a), the degree of r(x) = 0. This means that r(x) is a constant, say r.

(iii) So, for every value of x, r(x) = r.

(iii) Therefore, p(x) = (x â€“ a) q(x) + r

(iv) In particular, if x = a, this equation gives us

(v) p(a) = (a â€“ a) q(a) + r = r, which proves the theorem.

In other words, If p(x) and g(x) are two polynomials such that degree of p(x) â‰¥ degree of g(x) and g(x)â‰ 0, then there exists two polynomials q(x) and r(x) such thatÂ p(x) = g(x)q(x) + r(x), where, q(x) represents the quotient and r(x) represents remainder when p(x) is divided by g(x).

** For Example:Â **Divide 3x

(i) Let, p(x) = 3x

(ii) Performing divisions on these polynomials, we get,(iii) Now, we can re-write p(x) as 3x

** For Example:Â **Find remainder on dividing x

(21) Factorisation of Polynomials:

**(i) Factor Theorem:**Â **If p(x) is a polynomial of degree n â‰¥ 1 and a is any real number, then
**(a) x â€“ a is a factor of p(x), if p(a) = 0

(b) p(a) = 0, if x â€“ a is a factor of p(x)

** For Example:Â **Check whether (x + 1) is factor of p(x) = x

(i) As per Factor Theorem, (x + 1) is factor of p(x) = x

(ii) Therefore, p(-1) =(-1)

(iii) Thus, (x + 1) is factor of p(x) = x

** For Example:Â **Find value of k, if (x â€“ 1) is factor of p(x) = kx

(i) As per Factor theorem, here, p(1) = 0.

(ii) So, k(1)

(iii) k â€“ 3 + k = 0

(iv) 2k â€“ 3 = 0

(v) k = 3/2.

** For Example:Â **Factorise 2y

(i) On using trial and error method, we get,

(ii) p(1) = 2(1)

(iii) Thus, (y â€“ 1) is factor of 2y

(iv) Now, using division method, we get,(v) Thus, p(y) = 2y

Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (2y

Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (2y

Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (2y (y + 1) + 1 (y + 1))

Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (y + 1) (2y + 1)

**
(22) Algebraic Identities:
**(i) (a +b)

(ii) (a â€“ b)

(iii) a

(iv) (x + a) (x + b) = x

(v) (a + b + c)

(vi) (a + b)

(vii) (a â€“ b)

** For Example:Â **Use suitable identity to find (x + 2) (x â€“ 3).

(i) We know the identity, (x + a) (x + b) = x

(ii) Using the identity, (x + 2) (x â€“ 3) = x

** For Example:Â **Evaluate (102 x 107) without multiplying directly.

We know the identity, (x + a) (x + b) = x

(i) Here, we can write, 102 as (100 + 2) and 107 as (100 + 7). So, x = 100, a = 2 and b = 7.

(ii) Using the identity, (102 x 107) = 100

** For Example:Â **Factorise (a + b + c)

(i) We know the identity, (a + b + c)

(ii) Now, 4a

Â Â Â Â Â Â Â = (2a)

Â Â Â Â Â Â Â = (2a + 4b + 8c)

** For Example:Â **Write (x â€“ 2/3y)

(i) We know the identity, (a â€“ b)

(ii) (x â€“ 2/3y)

Â Â Â Â Â Â Â Â Â = x

Â Â Â Â Â Â Â Â Â = x

** For example:Â **Factorise 8a

(i) We know the identity, a

(ii) So, 8a

Â Â Â Â Â Â = (2a)

Â Â Â Â Â Â = (2a +3b + 4c) ((2a)

Â Â Â Â Â Â = (2a + 3b + 4c) (4a

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Very helpful notes for everyone thanks from jaat

(x+1/x)=8 then find the value of (x-1/x)=?

(x+1/x)=8 then , the value of (x-1/x)= -6

Is this a correct answer ?

Good Question talha

The answer is 2 * (15)^(1/2), i.e, 2 * root 15

The answer is verified

Nice..........

very popola

how to take print of notes

Very very helpful guys #D_C