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(1) Algebraic Expressions :Â Any expression containing constants, variables, and the operations like addition, subtraction, etc. is called as an algebraic expression.
For example:Â 5x, 2x â€“ 3, x^{2} + 1, etc. are some algebraic expressions.
(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^{2} + 2a + 5, etc. are polynomials in one variable.
(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.
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(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.
For example:Â p(x) = 7x^{2} + 7x + 7, t(r) = r^{3} + 2r + 1, etc.
(9) Monomials : The expressions which have only one term are called as monomials.
For Example:Â p(x) = 3x, q(a) = 2a^{2}, etc. are some monomials.
(10) Binomials : The expressions which have two terms are called as binomials.
For example:Â r(x) = x + 10, c(z) = 7z^{2 }+ z etc. are some binomials.
(11) Trinomials :Â The expressions which have three terms are called as trinomials.
For example:Â p(x) = 7x^{2} + x + 7, d(t) = t^{3} â€“ 3t + 4, etc. are some 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^{5} â€“ x^{3} + 7 is 5.
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^{2}â€“ 9, a^{2} + 7, etc. are some quadratic polynomials.
(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 Example:Â x^{3}â€“ 9x +2, a^{3} + a^{2} + a + 7, etc. are some cubic polynomials.
(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.
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(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^{2} + 5a + 1 at a = 3.
(i) Here, p(a) = 3a^{2} + 5a + 1.
(ii) Now, substituting a = 3, we get,
(iii) p(3) = 3 x (3)^{2} + 5 x 3 + 1 = 27 + 15 + 1 = 43
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.
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(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^{2} + x â€“ 1 by x + 1.
(i) Let, p(x) = 3x^{2} + x â€“ 1 and g(x) = x + 1.
(ii) Performing divisions on these polynomials, we get,(iii) Now, we can re-write p(x) as 3x^{2} + x â€“ 1 = (x + 1) (3x -2) +1.
For Example:Â Find remainder on dividing x^{3} + 3x^{2} + 3x + 1 by 2x + 5.Thus, remainder obtained on dividing x^{3} + 3x^{2} + 3x + 1 by 2x + 5 is -27/8.
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(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^{3} + x^{2} + x + 1.
(i) As per Factor Theorem, (x + 1) is factor of p(x) = x^{3} + x^{2} + x + 1, if p(-1) = 0.
(ii) Therefore, p(-1) =(-1)^{3} + (-1)^{2} + (-1) + 1 = -1 + 1 -1 + 1 = 0.
(iii) Thus, (x + 1) is factor of p(x) = x^{3} + x^{2} + x + 1.
For Example:Â Find value of k, if (x â€“ 1) is factor of p(x) = kx^{2} â€“ 3x + k.
(i) As per Factor theorem, here, p(1) = 0.
(ii) So, k(1)^{2} â€“ 3(1) + k = 0.
(iii) k â€“ 3 + k = 0
(iv) 2k â€“ 3 = 0
(v) k = 3/2.
For Example:Â Factorise 2y^{3} + y^{2} â€“ 2y â€“ 1.
(i) On using trial and error method, we get,
(ii) p(1) = 2(1)^{3} + (1)^{2} â€“ 2(1) â€“ 1 = 2 + 1 â€“ 2 -1 = 0.
(iii) Thus, (y â€“ 1) is factor of 2y^{3} + y^{2} â€“ 2y â€“ 1.
(iv) Now, using division method, we get,(v) Thus, p(y) = 2y^{3} + y^{2} â€“ 2y â€“ 1
Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (2y^{2} + 3y + 1)
Â Â Â Â Â Â Â Â Â Â Â Â = (y â€“ 1) (2y^{2} + 2y + y + 1)
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)^{ 2} = (a^{2} + 2ab + b^{2})
(ii) (a â€“ b)^{ 2} = (a^{2} - 2ab + b^{2})
(iii) a^{2} â€“ b^{2} = (a + b) (a â€“ b)
(iv) (x + a) (x + b) = x^{2} + (a + b)x + ab
(v) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
(vi) (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)
(vii) (a â€“ b)^{ 3} = a^{3} - b^{3} - 3ab (a - b) = a^{3} â€“ 3a^{2}b + 3ab^{2} - b^{3
}(viii) a^{3} + b^{3 }+ c^{3} - 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)
For Example:Â Use suitable identity to find (x + 2) (x â€“ 3).
(i) We know the identity, (x + a) (x + b) = x^{2 }+ (a + b)x + ab
(ii) Using the identity, (x + 2) (x â€“ 3) = x^{2} + (2 â€“ 3)x + (2)(-3)Â = x^{2} â€“ x â€“ 6.
For Example:Â Evaluate (102 x 107) without multiplying directly.
We know the identity, (x + a) (x + b) = x^{2 }+ (a + b)x + ab
(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^{2} + (2 + 7)100 + (2)(7)Â = 10000 + 900 + 14 = 10914
For Example:Â Factorise (a + b + c)^{2} = 4a^{2} + 16b^{2} + 64c^{2} + 16ab + 64bc + 32ca.
(i) We know the identity, (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
(ii) Now, 4a^{2} + 16b^{2} + 64c^{2} + 16ab + 64bc + 32ca
Â Â Â Â Â Â Â = (2a)^{2} + (4b)^{2} + (8c)^{2} + 2(2a)(4b) + 2(4b)(8c) + 2(8c)(2a).
Â Â Â Â Â Â Â = (2a + 4b + 8c)^{2
}Â Â Â Â Â Â Â = (2a + 4b + 8c) (2a + 4b + 8c)
For Example:Â Write (x â€“ 2/3y)^{3 }in expanded form.
(i) We know the identity, (a â€“ b)^{ 3} = a^{3} - b^{3} - 3ab (a - b)
(ii) (x â€“ 2/3y)^{3 }= x^{3} â€“ (2/3y)^{3} â€“ 3(x)(2/3y) (x â€“ 2/3y)
Â Â Â Â Â Â Â Â Â = x^{3} â€“ 8/27y^{3} â€“ 2xy (x â€“ 2/3y)
Â Â Â Â Â Â Â Â Â = x^{3} â€“ 8/27y^{3} â€“ 2x^{2}y + 4/3 xy^{2}
For example:Â Factorise 8a^{3} + 27b^{3 }+ 64c^{3} - 72abc.
(i) We know the identity, a^{3} + b^{3 }+ c^{3} - 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)
(ii) So, 8a^{3} + 27b^{3 }+ 64c^{3} - 72ab
Â Â Â Â Â Â = (2a)^{3} + (3b)^{3 }+ (4c)^{3} â€“ 3(2a)(3b)(4c)
Â Â Â Â Â Â = (2a +3b + 4c) ((2a)^{2} + (3b)^{2} + (4c)^{2} â€“ (2a)(3b) â€“(3b)(4c) -(4c)(2a))
Â Â Â Â Â Â = (2a + 3b + 4c) (4a^{2} + 9b^{2} + 16c^{2} - 6ab - 12bc - 8ca)
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how x+y is a algrebaic expression ?
only there is a variable but there is no constant
then how it will be algrebaic expressions
hey!
x + y is an Algebraic expression because an algebraic expression by definition means that it can have a constant, a variable and symbols like subtraction, multiplication , etc. but that dosent mean that it should be called an algebraic expression only when it has a constant that is visible . like if we add 0 or subtract 0 from the expression x+y the answer remains unchanged. just like '7' is also considered an algebraic expression because there is a hidden constant which is x to the power 0. which makes the answer unchanged.
Very helpful for students.
Best definitions for easy learning of students
very helpful
very helpful for students
It was really helpful.
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