Kiran’s

SSC ALGEBRA TRIGONOMETRY GEOMETRY MENSURATION Useful for ➼ ➼ ➼ ➼

SSC CGL Tier-I Exam SSC CGL Tier-II Exam SSC CAPFs Exam SSC 10+2 DEO/LDC/PA/SA Exam and other SSC and Competitive Exams FOR FRANCHISEE CONTACT :

RU-67, PITAMPURA, DELHI-110034, Ph : 27345258, 27342249, Fax : 27345258

CONTENTS ■ ALGEBRA ................................................ ATGE–3 ■ SEQUENCE & SERIES............................ ATGE–35 ❖ Solved Objective Questions on ALGEBRA & SEQUENCE & SERIES Asked in Various Exams (e.g. FCI Assistant, SSC CPOs SI & SSC Tier-I Exam) Conducted by SSC in 2011, 2012, 2013, 2014 & 2015 on New Syllabus ............................................................................................... ATGE–51 ■ TRIGONOMETRY

........................... ATGE–73–96

❖ Solved Objective Questions on TRIGONOMETRY Asked in Various Exams (e.g. FCI Assistant, SSC CPOs SI & SSC Tier-I Exam) Conducted by SSC in 2011, 2012, 2013, 2014 & 2015 on New Syllabus ....ATGE–97 ■ GEOMETRY ........................................ ATGE–121 ❖ Solved Objective Questions on GEOMETRY Asked in Various Exams (e.g. FCI Assistant, SSC CPOs SI & SSC Tier-I Exam) Conducted by SSC in 2011, 2012, 2013, 2014 & 2015 on New Syllabus .............ATGE–157 ■ Solved Objective Questions on MENSURATION Asked in Various Exams (e.g. FCI Assistant, SSC CPOs SI & SSC Tier-I Exam) Conducted by SSC in 2011, 2012, 2013, 2014 & 2015 on New Syllabus ............ ATGE–185 © KIRAN INSTITUTE OF CAREER EXCELLENCE PVT. LTD. (KICX) NEW EDITION The copyright of this book is entirely with the Kiran Institute of Career Excellence Pvt. Ltd. The reproduction of this book or a part of this will be punishable under the Copyright Act. All disputes subject to Delhi jurisdiction. Every possible effort has been made to ensure that the information contained in this book is accurate at the time of going to press, and the publishers and authors cannot accept responsibility for any errors or omissions, however caused. No responsibility for loss or damage occasioned to any person acting, or refraining from action, as a result of the material in this publication can be accepted by the editor, the publisher or any of the authors. Compiled by : Think Tank of PRATIYOGITA KIRAN, KIRAN PRAKASHAN & KICX Assistance : ● Govind Pd. Singh ● Rakesh Kumar ● Sanket Sah Design & Layout by : KICX COMPUTER SECTION, New Delhi.

ALGEBRA

ALGEBRA

ALGEBRA Examples : (i) 3x + 5 is a polynomial in x of degree 1.

IMPORTANT POINTS Constant : A symbol having a fixed numerical value is called a constant. Examples : 8, – 7,

5 , 9

etc. are all constants.

Variables : A symbol which may be assigned different numerical vlaues is known as a variable. Example : We know that area of circle is given by the formula A = r2 where r is the radius of the circle. Here, is constant while A and r are variables. Algebraic Expressions : A combination of constants and variables, connected by some or all of the operations +, –, × and ÷, is known as an algebraic expression. Terms Of An Algebraic Expression The several parts of an algebraic expression separated by + or – operations are called the terms of the expression.

3 xy 7 (ii) x3 + 3x2y + 3xy2 + y3 + 7 Polynomials : An algebraic expression in which the variables involved have only non-negative integral powers is called a polynomial. General Form : p(x) = a0 + a1x + a2x2 + ... + anxn is a polynomial in variable x, where a0, a1, a2, a3 ... an are real numbers and n is non-negative integer. Examples : (i) 6x3 – 4x2 + 7x – 3 is a polynomial in one variable x. Examples : (i) 5 + 9x – 7x2y +

(ii) 9y5 + 6y4 + 7y3 + 10y2 – 8y +

2 is a polynomial 7

in one variable y. (iii) 3 + 2x2 – 6x2y + 5xy2 is a polynomial in two variables x and y. (v) 5 + 8x5/2 + 7x3 is an expression but not a polynomial since it contains a term containing x5/2 where

5 is not a non-negative integer.. 2 Coefficients : In the polynomial 6x3 – 5x2 + 5x – 7 we say that coefficients of x3, x2 and x are 6, –5 and 5 respectively and we also say that –7 is the constant term in it. Degree Of A Polynomial In One Variable : In case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial.

(ii) 4y2 –

7 y 2

5 is a polynomial in y of degree 2.

Degree Of A Polynomial In Two Or More Variables : In case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of polynomial. Examples : (i) 7x2 – 5x2y2 + 3xy + 7y + 9 is a polynomial in x and y of degree 4. (ii) 4x3y3 – 5xy2 + 2x4 – 7 is a polynomial in x and y of degree 6. Polynomials Of Various Degrees (1) Linear Polynomial : A polynomial of degree 1 is called a linear polynomial. Examples : (i) 2x + 7 is a linear polynomial in x. (ii) 2x + y + 7 is a linear polynomial in x and y. (2) Quadratic Polynomial : A polynomial of degree 2 is called a quadratic polynomial. Examples : (i) x2 + 2x + 3 is a quadratic polynomial in x. (ii) xx + yz + zx is a quadratic polynomial in x, y and z. (3) Cubic Polynomial : A polynomial of degree 3 is called a cubic polynomial. Examples : (i) 5x3 – 3x2 + 7x + 7 is a cubic polynomial in x and y. (ii) 4x2y + 7xy2 + 7 is a cubic polynomial in x and y. (4) Biquadratic Polynomial : A polynomial of degree 4 is called a biquadratic polynomial. Examples : (i) x4 – 7x3 + 15x2 + 7x – 9 is a biquadratic polynomial in x. (ii) x2y2 + xy3 + y4 – 8xy + 2y2 + 9 is a biquadratic polynomial in x and y. Number of Terms In A Polynomial (i) Monomial : A polynomial containing one nonzero term is called a monomial. Example : 5, 3x,

7 xy are all monomials. 4

(ii) Binomial : A polynomial containing two nonzero terms is called a binomial.

ATGE-3

ALGEBRA

ALGEBRA

Examples : (7 + 5x), (x – 7y), (5x2y + 3yz) are all binomials. (iii) Trinomial : A polynomial containing three nonzero terms is called a trinomial. Examples : (8 + 5x + x2), 3x – 5xy + 7y2 are all trinomials. Constant Polynomial A polynomial containing one term only, consisting of a constant is called a constant polynomial. Examples : 7, – 5,

7 etc. are all constant polynomials. 9

Clearly, the degree of a non-zero constant polynomial is zero. Zero Polynomial A polynomial consisting of one term, namely zero only, is called a zero polynomial. The degree of a zero polynomial is not defined. Zeros Of A Polynomial Let p(x) be a polynomial. If p(a) = 0, then we say that a is a zero of the polynomial p(x). Finding the zeros of a polynomial p(x) means solving the equation p(x) = 0. Example : If f(t) = 3t2 – 10t + 6, find f(0). Solution : f(t) = 3t2 – 10t + 6 f(0) = 3 × 02 – 10 × 0 + 6 = 6 Example : If p(x) = 2x2 – 5x + 4, find p(2) Solution : p(x) = 2x2 – 5x + 4 p(2) = (2 × 22 – 5 × 2 + 4) = 8 – 10 + 4 = 2 Example : Find a zero of polynomial p(x) = x – 7 Solution : p(x) = x – 7 Now, p(x) = 0 x–7=0 x=7 7 is a zero of polynomial p(x) Addition And Difference Of Two Polynomials Addition of two polynomials is determined by arranging terms of same degrees with signs and adding the co-efficients. The operation of subtraction is similar to the operation of addition. Only difference is that the signs of the polynomial to be subtracted are changed and then operation of addition is performed. Example : If p (x) = x4 – 5x3 + 3x + 9 and q (x) = 2x4 – 3x3 + 5x – 4 Then, p (x) + q (x) = (x4 – 5x3 + 3x + 9) + (2x4 – 3x3 + 5x – 4) = (x4 + 2x4) + (–5x3 – 3x3) + (3x + 5x) + (9 – 4) = 3x4 – 8x3 + 8x + 5

= (x4 – 5x3 + 3x + 9) – (2x4 – 3x3 + 5x – 4) = (x4 – 5x3 + 3x + 9) + (– 2x4 + 3x3 – 5x + 4) = (x4 – 2x4) + (–5x3 + 3x3) + (3x – 5x) + (9 + 4) = – x4 – 2x3 – 2x + 13 For the sake of convenience, p (x) = x4 – 5x3 + 3x + 9 q (x) = 2x4 – 3x3 + 5x – 4 – – + – + p (x) – q(x) = – x4 – 2x3 – 2x + 13 Multiplication Of Two Polynomials To determine the product of two polynomials, the distributive law of multiplication is used first and then grouping is made of terms of same degrees for addition and subtraction. x3 – 6x2 + x + 1 x2 – 3x + 2 x5 – 6x4 + x3 + x2 – 3x4 + 18x3 – 3x2 – 3x + 2x3 – 12x2 + 2x + 2 x5 – 9x4 + 21x3 – 14x2 – x + 2 Remember : (–x) × (–x) = + x2 (+x) × (–x) = – x2 (x) × (x) = x2 etc. Or,

(–) × (–) = + (–) × (+) = – (+) × (–) = – (+) × (+) = + Division Of Polynomial By Another Polynomial Let p(x) and q(x) be two polynomials and q(x) 0. If we find two polynomials g(x) and r(x) such that p (x) = g (x) q (x) + r (x) i.e. Dividend = Divisor × Quotient + Remainder Where degree of r(x) < degree of g(x), then we say that on dividing p(x) by q(x), the quotient is g(x) and remainder is r(x). If remainder r(x) = zero, we say that q(x) is a factor of p(x). Let’s take a few examples to illustrate the method of division of a polynomial by a polynomial of lesser degree. Example % Divide p (x) = x3 + 3x2 – 12x + 4 by g (x) = x – 2. Solution : x – 2

For the sake of convenience, the above operation can be written in the following form : p (x) = x4 – 5x3 + 3x + 9 q (x) = 2x4 – 3x3 + 5x – 4 p (x) + q (x) = 3x4 – 8x3 + 8x + 5 and, p (x) – q (x)

ATGE-4

x 2 + 5x – 2 x 3 + 3x 2 – 12x + 4 x 3 – 2x 2 – + 5x 2 – 12x + 4 5x 2– 10x – + – 2x + 4 – 2x + 4 + –

ALGEBRA

ALGEBRA

Note : It is to be noted that the degree of q(x) is less than that of p(x) and polynomial of higher degree is always divided by a polynomial of lower degree. The operation of division ends when the remainder is either zero or the degree of remainder is less than that of divisor. In the above example, the quotient is x2 + 5x – 2 and remainder is zero. As the remainder is zero, (x – 2) is a factor of x3 + 3x2 – 12x + 4. Example % Divide p (x) = x3 – 14x2 + 37x – 60 by g(x) = x – 2.

(iii) If a polynomial p(x) is divided by (ax + b), then remainder is the value of p(x) at x =

b i.e. p a

[  ax + b = 0

x x x

x x x

x

x x

x x

remainder is the value of p(x) at x =

Remainder Theorem Let f(x) be a polynomial of degree n > 1, and let a be any real number. When f (x) is divided by (x – a), then the remainder is f (a). Proof : Suppose that when f(x) is divided by (x – a), the quotient is g(x) and the remainder is r (x). Then,degree r(x) < degree (x – a) degree r(x) < 1 degree r(x) = 0 [  degree of (x – a) = 1] r(x) is constant, equal to r (say). Thus, when f(x) is divided by (x – a), then the quotient is g(x) and the remainder is r. f(x) = (x – a) . g(x) + r ... (i) Putting x = a in (i), we get r = f(a). Thus, when f(x) is divided by (x – a), then the remainder is f (a). Remarks (i) If a polynomial p(x) is divided by (x + a), the remainder is the value of p(x) at x = –a i.e. p (–a) [ x + a = 0 x = –a] (ii) If a polynomial p(x) is divided by (ax – b), the

[  ax – b = 0

x=

b ] a

x 3 + x 2 – 2x 0 x 3 – 3x 2 + 2x + 5 Solution : x – 1 4 x – x3 – + x 3 – 3x 2 + 2x + 5 x 3 – x2 – + – 2x 2 + 2 x + 5 – 2x 2 + 2 x + – 5

0, then (x – 2) is not a factor of

b i.e. p a

FG IJ H K

b b i.e. p a a

x4 +

Here, quotient = x2 – 12x + 13 and remainder = – 34

remainder is the value of p(x) at x =

b ] a

Example % Let p (x) = x4 – 3x2 + 2x + 5. Find remainder when p (x) is divided by (x – 1).

x x

Since remainder x3 – 14x2 + 37x – 60.

x=–

(iv) If a polynomial p(x) is divided by b – ax, the

[  b – ax = 0 Solution : x

FG b IJ H aK

FG b IJ . HaK x=

b ] a

Here, remainder = 5 Find the value of p(1) from the above example. p(1) = 1 – 3 × 1 + 2 × 1 + 5 = 5 Thus, remainder obtained on dividing p(x) by (x – 1) is same as p (1). Factor Theorem Let p(x) be a polynomial of degree greater than or equal to 1 and a be a real number such that p(a) = 0, then (x – a) is a factor of p(x). Conversely, if (x – a) is a factor of p(x), then p(a) =0 Proof : First, let p(x) be a polynomial of degree greater than or equal to one and a be a real number such that p(a) = 0. then we have to show that (x – a) is a factor of p(x). Let q(x) be the quotient, when p(x) is divided by (x – a). By remainder theorem, p(x) when divided by (x – a) gives remainder equal to p(a). p(x) = (x – a) q(x) + p(a) p(x) = (x – a) q(x) [  p(a) = 0] (x – a) is a factor of p(x) Conversely, Let (x – a) is a factor of p(x). Then, we have to prove that p(a) = 0. Now (x – a) is a factor of p (x) p(x), when divided by (x – a) gives remainder zero. But by Remainder theorem,

ATGE-5

ALGEBRA

ALGEBRA

p(x) when divided by (x – a) gives the remainder equal to p(a). p(a) = 0 Remarks (i) (x + a) is a factor of a polynomial iff (if and only if) p (–a) = 0 (ii) (ax – b) is a factor of a polynomial iff p

FG b IJ HaK

=0

(iii) (ax + b) is a factor of a polynomial p(x) iff

p

FG b IJ H aK

=0

(iv) (x – a) (x – b) are factors of a polynomial p(x) iff p(a) = 0 and p(b) = 0

SOLVED EXAMPLES When the polynomial f (x) = x4 + 2x3 – 3x2 + x – 1 is divided by (x – 2) what will be the remainder? (1) 21 (2) 22 (3) 23 (4) 29 Sol. (1) Here, x – 2 = 0 x=2 By Remainder Theorem, when polynomial f(x) is divided by (x – 2), the remainder is f(2). f (2) = 24 + 2 × 23 – 3 × 22 + 2 – 1 ¹Remember : x has been replaced by 2º = 16 + 16 – 12 + 2 – 1 = 21 Remainder = 21 2. When f(x) = x3 – 3x2 + 4x + 50 is divided by (x + 3), the remainder is (1) 16 (2) – 16 (3) 15 (4) – 15 1.

Sol. (2) Divisor = x + 3 x+3=0 x=–3 By Remainder Theorem, Remainder = f (–3) = [(–3)3 – 3(–3)2 + 4(–3) + 50] = (– 27 – 27 – 12 + 50) = – 16 3. When polynomial f(x) = 4x3 – 12x2 + 14x – 3 is divided by 2x – 1, the remainder is : 2 1 (2) (1) 3 3 3 2 (4) (3) 2 7 Sol. (2) Here, 2x – 1 = 0

x=

1 2

4.

Sol. (1) f (x) = 2x3 + ax2 + 3x – 5 g(x) = x3 + x2 – 2x + a By Remainder Theorem, f (2) = (2 × 23 + a × 22 + 3 × 2 – 5) = 17 + 4a Again, g(2) = (23 + 22 – 2 × 2 + a) = 8 + a 17 + 4a = 8 + a 3a = – 9 a=–3 5. If the polynomial f(x) = x4 – 2x3 + 3x2 – ax + b is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. The values of a and b are : (1) a = 8, b = 7 (2) a = 5, b = 8 (3) a = 8, b = 5 (4) a = 6, b = 8 Sol. (2) f (x) = x4 – 2x3 + 3x2 – ax + b f (1) = 1 – 2 + 3 – a + b = 2 – a + b [x – 1 = 0 x = 1] f(–1) = 1 + 2 + 3 + a + b =6+a+b [x + 1 = 0 x = – 1] 2–a+b=5 b–a=3 ... (i) and, 6 + a + b = 19 a + b = 13 ... (ii) By adding equations (i) and (ii), 2b = 16 b=8 From equation (ii), a + b = 13 a = 13 – 8 = 5 6. The factor of polynomial f(x) = x4 + x2 – 17x + 15 is (1) x – 3 (2) x – 5 (3) x + 3 (4) x + 1 Sol. (1) By Factor Theorem, If f (a) = 0, then (x – a) is a factor of f(x). f (3) = 33 + 32 – 17 × 3 + 15 = 27 + 9 – 51 + 15 = 0 (x – 3), is a factor of f(x). Remember : f(5) 0; f(–3) 0; f(–1) 0 7. For what value of a, (x – a) is a factor of f(x) = x5 – a2x3 + 2x + a – 3 ? (1) – 1 (2) 1 (3) – 2 (4) 2 Sol. (2) (x – a), is a factor of polynomial x5 – a2x3 + 2x + a – 3

By Remainder Theorem,

f (a) = 0 a5 – a5 + 2a + a – 3 = 0

Remainder

LM F 1 I 3 MN GH 2 JK I 3 7 3J K 2

1 =f( )= 4 2

=

FG 1 H2

3

12

FG 1 IJ 2 H 2K

If polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 2x + a are divided by (x – 2), the same remainders are obtained. Find the value of a. (1) –3 (2) 3 (3) –4 (4) –9

14

FG 1 IJ H 2K

3

OP PQ

8.

ATGE-6

3a = 3 a=1 For what value of a, (x + a) is a factor of polynomial f(x)= x3 + ax2 – 2x + a + 6 ? (1) 2 (2) 3 (3) – 2 (4) – 3

ALGEBRA

ALGEBRA

Sol. (3) Here, x + a = 0 f (–a) = 0

x=–a

(–a)3 + a (–a)2 – 2 (–a) + a + 6 = 0 3a = –6 a = –2 9. For what value of k, (2x4 + 3x3 + 2kx2 + 3x + 6) is exactly divisible by (x + 2) ? (1) 1 (2) 2 (3) – 2 (4) – 1 Sol. (4) Here, x + 2 = 0 x=–2 By Factor Theorem, f (–2) = 0 2(–2)4 + 3 (–2)3 + 2k(–2)2 + 3 (–2) + 6 = 0 32 – 24 + 8k – 6 + 6 = 8k + 8 = 0 8k = – 8 k=–1 10. For what values of a and b, (x3 – 10x2 + ax + b) is exactly divisible by (x – 1) and (x – 2) ? (1) a = 23, b = –14 (2) a = – 23, b = 14 (3) a = 21, b = –14 (4) a = – 21, b = 15 Sol. (1) f (x) = x3 – 10x2 + ax + b By Factor Theorem, f (1) = 1 – 10 + a + b = a + b – 9 [ x – 1 = 0 x = 1] f(1) = 0 a+b=9 ...(i) f(2) = 8 – 40 + 2a + b = 2a + b – 32 f(2) = 0 2a + b = 32 ...(ii) From equation (ii) – equation (i), a = 23 From equation (i), b = 9 – 23 = – 14

FACTORISATION To express a sgiven polynomial as the product of polynomials, each of degree less than that of the given polynomial such that no such a factor has a factor of lower degree, is called factorisation. Examples : (i) x2 – 16 = (x – 4) (x + 4) (ii) x2 – 3x + 2 = (x – 2) (x – 1) Formulae for Factorisation (i) (x + y)2 = x2 + y2 + 2xy (ii) (x – y)2 = x2 + y2 – 2xy (iii) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (iv) (x + y)3 = x3 + y3 + 3xy (x – y) (v) (x – y)3 = x3 – y3 – 3xy (x – y) (vi) x4 + x2y2 + y4 = (x2 + xy + y2) (x2 – xy + y2) Thus, for factorisation, we have (i) x2 – y2 = (x – y) (x + y) (ii) x3 + y3 = (x + y) (x2 – xy + y2) (iii) x3 – y3 = (x – y) (x2 + xy + y2) METHODS OF FACTORISATION Method 1 : When each term of an expression has a common factor, we divide each term by this factor and take it out as a multiple as shown below :

Example : 36a3b – 60a2bc = 12a2b (3a – 5c) Example : x (x – y)3 + 3x2y (x – y) = x (x – y) [(x – y)2 + 3xy] = x (x – y) [x2 + y2 – 2xy + 3xy] = x (x – y) (x2 + y2 + xy) Method 2 : Sometimes in a given expression it is not possible to take out a common factor directly. However the terms of the expression are grouped in such a manner that we may have a common factor. This can now be factorised as discussed above. Example : Factorise : 6ab – b2 + 12ac – 2bc Sol. : 6ab – b2 + 12ac – 2bc = (6ab + 12ac) – (b2 + 2bc) = 6a (b + 2c) – b (b + 2c) = (b + 2c) (6a – b) Example : Factorise : x2 + 18x + 81 Sol. x2 + 18x + 81 = x2 + 18x + 92 = x2 + 2 × 9 × x + 9 2 = (x + 9)2 Example : Factorise : 64x2 – 16x + 1 Sol. : 64x2 – 16x + 1 = (8x)2 – 16x + 12 = (8x)2 – 2. (8x).1 + 12 = (8x – 1)2 Example : Factorise : 81 – 64x2 Sol. 81 – 64x2 = 92 – (8x)2 = (9 + 8x) (9 – 8x) Mothod 3 : Factorisation of Quadratic Trinomials Case I : Polynomial of the form x2 + bx + c. We find integers p and q such that p + q = b and pq = c. Then, x2 + bx + c = x2 + (p + q)x + pq = x2 + px + qx + pq = x (x + p) + q (x + p) = (x + q) (x + p) Case II : Polynomial of the form ax2 + bx + c. In this case, we find integers p and q such that p + q = b and pq = ac.

ATGE-7

Then, ax2 + bx + c = ax2 + (p + q)x +

pq a

= a2x2 + apx + aqx + pq = ax (ax + p) + q (ax + p) = (ax + p) (ax + q) Example : Factorise : x2 + 9x + 18 Sol. : We try to split 9 into two parts whose sum is 9 and product is 18. Clearly, 6 + 3 = 9 and 6 × 3 = 18 x2 + 9x + 18 = x2 + 6x + 3x + 18 = x (x + 6) + 3 (x + 6) = (x + 3) (x + 6) Example : Factorise : x2 + 5x – 24 Sol. : We try to split 5 into two parts whose sum is 5 and product is –24. Clearly, 8 + (–3) = 5 and 8 × (–3) = –24 x2 + 5x – 24 = x2 – 3x + 8x – 24 2 (x – 3x) + (8x – 24)

ALGEBRA

ALGEBRA

= x (x – 3) + 8 (x – 3) = (x – 3) (x + 8) Example : Factorise : x2 – 4x – 21

Method 4 : Factorisation of forms x3 – y3 and x3 + y3 : Remember these formulae :

Sol. : We try to split –4 into two parts whose sum is –4 and product is –21. Clearly, (–7) + 3 = –4 and (–7) × 3 = –21 x2 – 4x – 21 = x2 – 7x + 3x – 21 = x (x – 7) + 3 (x – 7) = (x – 7) (x + 3) Example : Factorise : 6x2 + 7x – 3 Sol. : Here, 6 × –3 = – 18 So, we try to split 7 into two parts whose sum is 7 and product is –18. Clearly, 9 + (–2) = 7 and 9 × (–2) = –18 6x2 + 7x – 3 = 6x2 + 9x – 2x – 3 = 3x (2x + 3) – (2x + 3) = (3x – 1) + (2x + 3) Example : Factorise : 2x2 – 7x – 39.

x3 – y3 = (x – y) (x2 + xy + y2) x3 + y3 = (x + y) (x2 – xy + y2) Example : Factorise : x3 – 27y3 Sol. : x3 – 27y3 = (x)3 – (3y)3 = (x – 3y) {(x)2 + x × 3y + (3y)2} = (x – 3y) (x2 + 3xy + 9y2) Example : 8x3 + 27 Sol. : 8x3 + 27 = (2x)3 + (3)3 = (2x + 3) {(2x)2 – 2x × 3 + (3)2} = (2x + 3) (4x2 – 6x + 9) Method 5 : Factorisation of x3 + y3 + z3 – 3xyz Theorem : prove that

Sol. : Here, 2 × (–39) = –78 So, we try to split –7 into two parts whose sum is –7 and product is –78. Clearly, (–13) + 6 = –7 and (–13) × 6 = –78 2x2 – 7x – 39 = 2x2 – 13x + 6x – 39 = x (2x – 13) + 3 (2x – 13) = (x + 3) + (2x – 13) Example : Factorise : 9x2 – 22x + 8. Sol. : Here, 9 × 8 = 72 So, we try to split –22 into two parts whose sum is –22 and product is –72. Clearly, (–18) + (–4) = –22 and (–18) × (–4) = –72 9x2 – 22x + 8 = 9x2 – 18x – 4x + 8 = 9x (x – 2) – 4 (x – 2) = (x – 2) (9x – 4) Example : Factorise :

3 6 3 = 18 So, we try to split 11 into two parts whose sum is 11 and product is 18. Clearly, 9 + 2 = 11 and 9 × 2 = 18

=

3 x (x

= (x

6 3 =

3 3)

2(x

3 3 )( 3 x

2)

3x 2

9x

2x

=

1 (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) 2

=

1 ( x – y )2 2

(y

z )2

(z

x )2

Theorem : If x + y + z = 0 then prove that x3 + y3 + z3 = 3xyz Sol : First Method :

3 x 2 + 11x + 6 3

Sol. : Here,

3 x 2 11x

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) Proof : x3 + y3 + z3 – 3xyz = (x3 + y3) + z3–3xyz = [(x + y)3 – 3xy (x + y)] + z3 – 3xyz = u3 – 3xyu + z3 – 3xyz] tgka (x + y) = u = (u3 + z3) – 3xy (u + z) = (u + z) (u2 – uz + z2) – 3xy (u + z) = (u + z) (u2 + z2 – uz – 3xy) = (x + y + z) [(x + y)2 + z2 – (x +y) z – 3xy] = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) Note : x2 + y2 + z2 – xy – yz – zx

6 3

3 3)

Example : Factorise : 4x2 – 11x + 7 Sol. : Here, a = 4, b = – 11, c = 7 ac = 4 × 7 = 28 and – 11 = – 7 – 4 4x2 – 11x + 7 = 4x2 – 4x – 7x + 7 = 4x (x – 1) – 7 (x – 1) = (x – 1) (4x – 7)

ATGE-8

x+y+z=0 x+y=–z Cubing both sides (x + y)3 = (–z)3 x3 + y3 + 3xy (x + y) = – z3 x3 + y3 + 3xy (–z) = –z3 [Putting the value of (x + y)] x3 + y3 – 3xyz = –z3 x3 + y3 + z3 = 3xyz Second Method : We know that x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) — Formula = 0 × (x2 + y2 + z2 – xy – yz – zx) (x + y + z = 0) =0 x3 + y3 + z3 – 3xyz = 0 x3 + y3 + z3 = 3xyz

ALGEBRA

ALGEBRA

Example : Factorise : a3 – 8b3 + 64c3 + 24abc Sol. : a3 – 8b3 + 64c3 + 24abc = a3 + (–2b)3 + (4c)3 – 3a × (–2b) (4c) = x3 + y3 + z3 – 3xyz Where, a = x, –2b = y and 4c = z = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = [a + (–2b) + 4c] [a2 + (–2b)2 + (4c)2 – a (–2b) – (–2b) (4c) – a(4c)] = (a – 2b + 4c) (a2 + 4b2 + 16c2 + 2ab + 8bc – 4ac) Example : Factorise : a3 – b3 + 1 + 3ab Sol. a3 – b3 + 1 + 3ab = a3 + (–b)3 + (1)3 – 3a (–b) × 1 = (a – b + 1) (a2 + b2 + 1 + ab + b – a) Example : Factorise : (p – q)3 + (q – r)3 + (r – p)3 Sol. : If (p – q) = x, (q – r) = y and r – p = z, then, x + y + z = p – q + q – r + r – p = 0 x3 + y3 + z3 = 3xyz = 3 (p – q) (q – r) (r – p) Example : If p = 2 – a, find the value of, a3 + 6ap + p3 – 8 Sol. p = 2 – a a + p + (–2) = 0 x+y+z=0 Where, a = x, p = y and –2 = z x3 + y3 + z3 = 3xyz a3 + p3 + (–2)3 = 3 × a × p × (–2) a3 + 6ap + p3 – 8 = 0

OTHER SOLVED EXAMPLES

FG a H

1. Factorise : x 2 Sol. x 2

FG a H

2 = x

IJ K

1 x a

ax

1 = x (x 1 a

FG H

x a

1 a

1 (x a

a)

a)

IJ K

2. Factorise : x 2

Sol. x 2

1

1

x a

FG x H

= (x – a)

IJ K

1 x a

IJ K

x

F a 2 1I GH a JK

1

1

= x2

x a

FG H

1 a

IJ K

2 1 = x

xa

= x (x

a)

1 (x a

a ) = (x

a) x

FG H

x.

1 a 1 a

1

IJ K

3. Factorise : x6 – 1 Sol. x6 – 1 = (x3)2 – (1)2 = (x3 – 1) (x3 + 1) = (x – 1) (x2 + x + 1) (x + 1) (x2 – x + 1)

4. Factorise : x5 – x Sol. x5 – x = x(x4 – 1) = x[(x2)2 – 1] = x(x2 + 1) (x2 – 1) = x(x2 + 1) (x + 1) (x – 1) 5. Factorise : x2 – y2 – 2y – 1 Sol. x2 – y2 – 2y – 1 = x2 – (y2 + 2y + 1) = x2 – (y + 1)2 = (x + y + 1) (x – y – 1) 2x (a 2 a

6. Factorise : 4 x 2 Sol. 4 x 2

2 = 4x

2x

F a 2 1I GH a JK

1

2x a

1

2xa

= 2x (2x – a) +

FG H

= (2x – a) 2x

1)

1

1 (2x – a) a 1 a

IJ K

7. Factorise : x2 + ab – (a2 – 3ab + 4b2) Sol. x2 + ab – a2 + 3ab – 4b2 = x2 – a2 + 4ab – 4b2 = x2 – (a2 – 4ab + 4b2) = x2 – (a – 2b)2 = (x + a – 2b) (x – a + 2b) 8. Factorise : 4x2 + 12xy + 5y2 Sol. 4x2 + 12xy + 9y2 – 4y2 = (2x)2 + 2 × 2x × 3y + (3y)2 – 4y2 = (2x + 3y)2 – (2y)2 = (2x + 3y + 2y) (2x + 3y – 2y) = (2x + 5y) (2x + y) 9. Factorise : 4x4 + 7x2y2 + 16y4 Sol. 4x4 + 7x2y2 + 16y4 = 4x4 + 16x2y2 + 16y4 – 9x2y2 = (2x2 + 4y2)2 – (3xy)2 = (2x2 + 4y2 + 3xy) (2x2 + 4y2 – 3xy) 10. Factorise : x4 + 4y4 Sol. x4 + 4x2y2 + 4y4 – 4x2y2 = (x2 + 2y2)2 – (2xy)2 = (x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy) 11. Factorise : x2 + 2x – 899 Sol. x2 + 2x – 899 = x2 + 31x – 29x – 899 = x (x + 31) – 29 (x + 31) = (x + 31) (x – 29) 12. Factorise : 6x3 – 31x2 + 35x Sol. 6x3 – 31x2 + 35x = 6x3 – 21x2 – 10x2 + 35x = 3x2 (2x – 7) – 5x (2x – 7) = (2x – 7) (3x2 – 5x) = (2x – 7) x (3x – 5) = x(2x – 7) (3x – 5)

ATGE-9

ALGEBRA

ALGEBRA

13. Factorise : 3 – 7x + 4x2 Sol. 3 – 7x + 4x2 = 3 – 3x – 4x + 4x2 = 3 (1 – x) – 4x (1 – x) = (1 – x) (3 – 4x) 14. Factorise : 9x4 + 4y4 – 13x2y2 Sol. 9x4 + 4y4 – 13x2y2 = 9x4 – 13x2y2 + 4y4 = 9x4 – 9x2y2 – 4x2y2 + 4y4 = 9x2 (x2 – y2) – 4y2 (x2 – y2) = (x2 – y2) (9x2 – 4y2) = (x – y) (x + y) (3x – 2y) (3x + 2y) 15. Factorise : (2a + 3b)2 – 14 (2a + 3b) (3a – b) – 32 (3a – b)2 Sol. Let (2a + 3b) = x and (3a – b) = y Expression = x2 – 14xy – 32y2 = x2 – 16xy + 2xy – 32y2 = x (x –16y) + 2y (x – 16y) = (x – 16y) (x + 2y) = [(2a + 3b) – 16 (3a – b)] [2a + 3b) + 2 (3a – b)] = (19b – 46a) (8a + b) 16. Factorise : (2x – 3y)2 – 10x + 15y – 6 Sol. (2x – 3y)2 – 5 (2x – 3y) – 6 Let (2x – 3y) = a = a2 – 5a – 6 = a2 – 6a + a – 6 = a (a – 6) + 1 (a – 6) = (a – 6) (a + 1) = (2x – 3y – 6) (2x – 3y + 1) (Putting the value of a) 17. Factorise : 7 (a + b)2 + 48 (a + b) ab – 7a2b2 Sol. 7 (a + b)2 + 48 (a + b) ab – 7a2b2 Let (a + b) = x Expression = 7x2 + 48xab – 7a2b2 = 7x2 + 49xab – xab – 7a2b2 = 7x (x + 7 ab) – ab(x + 7ab) = (x + 7ab) (7x – ab) Putting the value of x, = (a + b + 7ab) [7 (a + b) – ab] = (a + b + 7ab) (7a + 7b – ab)

Fa 18. Factorise : GH b

Fa Sol. GH b =

FG a Hb

b a

b a

IJ 4 2FG a 2 K H b2 b a

IJ K

4

FG a Hb

2

IJ 4 2FG a 2 K H b2

I2 F a J GH b a2K b2

b a

IJ FG a K Hb

a b a x and b a b = x4 – 2x2y2 + y4 – c2

Let

I2 F a J GH b a2 K b2

2

b a

b a

b a y

IJ 4 K

– c2

IJ K

– c2

– c2

IJ FG a K Hb 2

IJ 4 K

b a

b a

4

= (x2 – y2)2 – c2

LF a = MGH b MN

b a

IJ 2 FG a K Hb

b a

IJ 2 OP K PQ

2

– c2

(Putting the value of x, y)

La 2 = M 2 MN b

b2 a2

2

a b

b a

a2

b2

b2

a2

2

a b

b a

OP2 PQ – c

2

= (4)2 – c2 = (4 + c) (4 – c) 19. Factorise : 27x3 – 125y3 Sol. 27x3 – 125y3 = (3x)3 – (5y)3 = (3x – 5y) [(3x)2 + 3x × 5y + (5y)2] = (3x – 5y) (9x2 + 15xy + 25y2) 20. Factorise : 64 – x6 Sol. 64 – x6 = (8)2 – (x3)2 = (8 + x3) (8 – x3) = (23 + x3) (23 – x3) = (2 + x) [(2)2 – 2 × x + x2] (2 – x) (22 + 2 × x + x2) = (2 + x) (4 – 2x + x2) (2 – x) (4 + 2x + x2) 21. Factorise : (x + y)3 + (x – y)3 Sol. Let (x + y) = a, (x – y) = b Expression = a3 + b3 = (a + b) (a2 – ab + b2) = (x + y + x – y) [(x + y)2 – (x + y) (x – y) + (x – y)2] (Putting the value of a, b) = (2x) [x2 + y2 + 2xy – (x2 – y2) + x2 + y2 – 2xy] 2x (x2 + y2 + 2xy – x2 + y2 + x2 + y2 – 2xy) = 2x (x2 + 3y2) 22. Factorise : 512 xy10 + xy10 Sol. 512 xy10 + xy10 = xy [512y9 + x9] = xy [83y9 + x9] = xy [(8y3)3 + (x3)3] = xy (8y3 + x3) (64y6 – 8y3x3 + x6) = xy [(2y)3 + x3] (64y6 – 8y3x3 + x6) = xy (2y + x) (4y2 – 2xy + x2) (64y6 – 8x3y3 + x6) 23. Factorise : x12 – y12 Sol. x12 – y12 = (x6)2 – (y6)2 = (x6 + y6) (x6 – y6) = (x6 + y6) (x3 + y3) (x3 – y3) = [(x2)3 + (y2)3] (x3 + y3) (x3 – y3) = (x2 + y2) (x4 – x2y2 + y4) (x + y) (x2 – xy + y2) × (x – y) (x2 + xy + y2) 24. Factorise : 8x3 + 27y3 – 6x – 9y Sol. 8x3 + 27y3 – 6x – 9y = (2x)3 + (3y)3 – 3 (2x + 3y) = (2x + 3y) [(2x)2 – 2x × 3y + (3y)2] – 3 (2x + 3y) = (2x + 3y) (4x2 – 6xy + 9y2) – 3 (2x + 3y) = (2x + 3y) (4x2 – 6xy + 9y2 – 3)

ATGE-10

ALGEBRA

ALGEBRA

25. x3 – 8y3 – 2x2y + 4y2x Sol. x3 – 8y3 – 2x2y + 4y2x = x3 – (2y)3 – 2xy (x – 2y) = (x – 2y) [x2 + 2xy + (2y)2] – 2xy (x – 2y) = (x – 2y) (x2 + 2xy + 4y2) – 2xy (x – 2y) = (x – 2y) (x2 + 2xy + 4y2 – 2xy) = (x – 2y) (x2 + 4y2)

Sol. p (x)

q (x)

G.C.D & L.C.M OF POLYNOMIALS Greatest Common Divisor/Highest Common Factor (GCD/HCF) : The GCD of two polynomials p(x) and q(x) is that common divisor which has the highet degree among all common divisors and the coefficient of the highest degree term be positive. Example : What is the HCF of (x + 4)2 (x – 3)3 and (x – 1) (x + 4) (x – 3)2 ? Sol. p(x) = (x + 4)2 (x – 3)2 q(x) = (x – 1) (x + 4) (x– 3)2 We see that (x + 4) (x – 3)2 is such a polynomial that is a common divisor and whose degree is highest among all common divisors. HCF = (x + 4) (x – 3)2 Lowest Common Multiple (LCM) : The LCM of two polynomials p(x) and q(x) is a polynomial of lowest degree of which p(x) and q(x) both are multiples. Example : Find the LCM of (x – 1) (x + 2)2 and (x – 1)3 (x + 2). Sol. p(x) = (x – 1) (x + 2)2 q(x) = (x – 1)3 (x + 2) We make a polynomial by taking each factor of p(x) and q(x). If a factor is common in both, then we take that factor which has highest degree in p(x) and q(x). LCM = (x – 1)3 (x + 2)2 Relation Between Two Polynomials And Their HCF And LCM p(x) = 8 (x3 – 3x + 2) and q(x) = 14 (x2 + x – 2) Now, p(x) = 8 (x2 – 3x + 2) = 2 × 2 × 2 × (x – 2) (x – 1) q(x) = 14 (x2 + x – 2) = 2 × 7 × (x + 2) (x – 1) HCF = 2 (x – 1) LCM = 2 × 2 × 2 × 7 × (x – 1) (x – 2) (x + 2) = 56 (x – 1) (x – 2) (x + 2) HCF × LCM = 112 (x – 1)2 (x – 2) (x + 2) Also, p(x) × q(x) = 112 (x – 1)2 (x – 2) (x + 2) HCF × LCM = p(x) × q(x) HCF of polynomials × LCM of same polynomials = Product of same polynomials. Example : Find the HCF of following pair of polynomials : p (x) = (x2 – 9) (x – 3) q (x) = x2 + 6x + 9

= = = = = = =

(x2 – 9) (x – 3) (x + 3) (x – 3) (x – 3) (x + 3) (x – 3)2 x2 + 6x + 9 x2 + 3x + 3x + 9 x (x + 3) + 3 (x + 3) = (x + 3) (x + 3) (x + 3)2

HCF of p(x) and q(x) = x + 3 Example : Find the LCM of p (x) = (x + 3) (x – 2)2 and q (x) = (x – 2) (x – 6). Sol. p (x) = (x + 3) (x – 2)2 q (x) = (x – 2) (x – 6) HCF of p (x) and q (x) = (x – 2) LCM of p (x) and q (x)

b g q bx g

p x

HCF

bx 3g bx – 2g2 bx – 2g bx – 6g b x – 2g = (x + 3) (x – 2)2 (x – 6) Example : Find the LCM of 7x3 + 2x2 – 16x – 32 and 3x2 – 2x – 8. Sol. p (x) = 7x3 + 2x2 – 16x – 32 p (2) = 7 (2)3 + 2 (2)2 – 16 (2) – 32 = 56 + 8 – 32 – 32 = 0 (x – 2), is a factor of q (x). x – 2)7x 3 + 2x2 – 16x – 32 (7x 2+ 16x + 16 7x 3 – 14x 2 – + 16x 2 – 16x 16x 2 – 32x – + 16x – 32 16x – 32 – + ×

p (x) = (x – 2) (7x2 + 16x + 16) q (x) = 3x2 – 2x – 8 = 3x2 + 4x – 6x – 8 = x (3x + 4) – 2 (3x + 4) = (x – 2) (3x + 4) LCM of p (x) and q(x) = (x – 2) (3x + 4) (7x2 + 16x + 16) Example : The HCF of two polynomials is (x + 3) and their LCM is x3 – 7x + 6. If one of the polynomial is x2 + 2x – 3, then find the second polynomial. Sol. HCF = (x + 3) LCM = x3 –7x + 6 p (x) = x2 + 2x – 3 q (x) = ?

ATGE-11

We know that p (x) × q (x) = HCF × LCM q (x) =

HCF LCM p x

bg

bx 3g ex 3 – 7x 6j x2

2x – 3

ALGEBRA

ALGEBRA

b x 3g e x 3 – 7 x 6j bx – 1g bx 3g

x

3

– 7x 6 x –1

Now, x – 1)x3 – 7x + 6 (x2 + x – 6 x 3 –x 2 – + x2 – 7x + 6 x2 – x – + – 6x + 6 – 6x + 6 + – ×

Second polynomial q (x) = x2 + x – 6

SOLVED OBJECTIVE TYPE QUESTIONS The LCM of x3 – 1, x4 + x2 + 1 and x4 – 5x2 + 4 is : (1) (x – 1) (x + 1) (x – 2) (2) (x – 1) (x + 1) (x + 2) (3) (x2 – 1) (x2 – 4) (4) (x2 – 1) (x2 – 4) (x2 + x + 1) (x2 + 1 – x) Sol. (4) (i) x3 – 1 = (x – 1) (x2 + x + 1) (ii) x4 + x2 + 1 = x2 + 2x2 + 1 – x2 = (x2 + 1)2 – x2 = (x2 + 1 – x) (x2 + 1 + x) (iii)x4 – 5x2 + 4 = x4 – 4x2 – x2 + 4 = x2 (x2 – 4) – 1 (x2 – 4) = (x2 – 4) (x2 – 1) = (x – 2) (x + 2) (x – 1) (x + 1) LCM 1.

= (x – 1) (x + 1) (x – 2) (x + 2) (x2 + x + 1) (x2 + 1 – x) 2. What is the LCM of x2 – 1, x2 + 4x + 3 and x2 – 3x + 2 ? (1) (x + 3) (x + 1) (2) (x2 – 1) (x – 2) (x + 3) (3) (x – 1) (x – 2) (x – 3) (4) (x2 – 1) (x + 2) (x + 3) Sol. (2) (i) x2 – 1 = x2 – (1)2 = (x – 1) (x + 1) (ii) x2 + 4x + 3 = x2 + 3x + x + 3 = x (x + 3) + 1 (x + 3) = (x + 3) (x + 1) (iii)x2 – 3x + 2 = x2 – 2x – x + 2 = x (x – 2) – 1 (x – 2) = (x – 2) (x – 1) LCM = (x – 1) (x + 1) (x – 2) (x + 3) 3. What is the HCF of x2 – x – 12, x2 – 7x + 12 and 2x2 – 11x + 15 ? (1) 1 (2) (x – 3) (3) 2x – 5 (4) x – 4

(ii) x2 – 7x + 12 = x2 – 3x – 4x + 12 = x (x – 3) – 4 (x – 3) = (x – 3) (x – 4) (iii)2x2 – 11x + 15 = 2x2 – 6x – 5x + 15 = 2x (x – 3) – 5 (x + 3) = (x – 3) (2x – 5) HCF = 1 as no term is common. 4. If the LCM and HCF of two quadratic polynomials are x3 – 7x + 6 and (x – 1) respectively, find the polynomials. (1) (x2 – 3x + 2), (x2 + 2x + 3) (2) (x2 + 3x – 2), (x2 – 2x + 3) (3) (x2 – 3x + 2), (x2 + 2x – 3) (4) (x2 + 3x + 2), (x2 + 2x + 3) Sol. (3) Putting x – 1 = 0 i.e. x = 1 in (x – 1) respectively. Remainder = (+1)3 – 7 (1) + 6 = 1 – 7 + 6 = 0 (x – 1) is a factor of expression x3 – 7x + 6 Now, x3 – 7x + 6 = x2 (x – 1) + x (x – 1) – 6 (x – 1) = (x – 1) (x2 + x – 6) = (x – 1) [x2 + 3x – 2x – 6] = (x – 1) [x (x + 3) – 2 (x + 3)] = (x – 1) (x – 2) (x + 3) LCM = x3 – 7x + 6 = (x – 1) (x – 2) (x + 3) and their HCF = (x – 1)  (x – 1) is common in both. First expression = (x – 1) (x – 2) = x2 – 3x + 2 and second expression = (x – 1) (x + 3) = x2 + 2x – 3 5. The HCF of two expressions is x – 7 and their LCM is x3 – 10x2 + 11x + 70. If one of them is x2 – 5x – 14, find the other. (1) x2 – 12x + 35 (2) x2 + 12x – 35 2 (3) x – 14x + 35 (4) x2 + 14x – 35 Sol. (1) p(x) × q(x) = HCF × LCM (x2 – 5x – 14) × q(x) = (x – 7) (x3 – 10x2 + 11x + 70)

Sol. (2) (i) x2 – x – 12 = x2 – 4x + 3x – 12 = x (x – 4) + 3 (x – 4) = (x – 4) (x + 3)

q(x) =

(x

(x

2

10x 2 5x

11x

70)

14)

Putting x = 5 in x3 – 10x2 + 11x + 70 Remainder = (5)3 – 10 (5)2 + 11 × 5 + 70 = 125 – 250 + 55 + 70 = 0 (x – 5) is a factor. Now, x3 – 10x2 + 11x + 70 = x2 (x – 5) – 5x (x – 5) – 14 (x – 5) = (x – 5) (x2 – 5x – 14) Second expression = = (x – 7) [x – 5) = x2 – 12x + 35

ATGE-12

7)( x 3

(x

5)(x 2

7)( x (x

2

5x

5x 14)

14)

ALGEBRA

ALGEBRA

6.

The HCF and LCM of two expressions are respectively (x + 3) and (x3 – 7x + 6). If one of the polynomials is x2 + 2x – 3, the other polynomial is (1) x2 – x + 6 (2) x2 – 2x + 6 2 (3) x – 2x + 8 (4) x2 + x – 6 Sol. (4) p(x) × q(x) = HCF × LCM (x2 + 2x – 3) × q(x) = (x + 3) × (x3 – 7x + 6) (x

Second expression =

(x =

3)(x 1)(x ( x 1)(x

2)( x 3)

3)( x 3 x2

2x

7x

6)

3

3)

= (x + 3) (x – 2) = x2 + x – 6 7. If HCF and LCM of two terms x and y are a and b respectively and x + y = a + b, then x2 + y2 = ? (1) a2 – b2 (2) 2a2 + b2 2 2 (3) a + b (4) a2 + 2b2 Sol. (3)  p(x) × q(x) = LCM × HCF x×y=b×a xy = ab and, x + y = a + b (Given) Squaring both sides, (x + y)2 = (a + b)2 or, x2 + y2 + 2xy = a2 + b2 + 2ab or, x2 + y2 + 2ab = a2 + b2 + 2ab (  xy = ab) x2 + y2 = a2 + b2 8. If common factor of x2 + bx + c and x2 + mx + n is (x + a), then the value of a is :

ALGEBRAIC IDENTITIES An algebraic identity is an algebraic equation which is true for all values of the variable (s).

Important Formulae 1.

(a + b)2 = a2 + 2ab + b2

2.

(a – b)2 = a2 – 2ab + b2

3.

(a + b)2 = (a – b)2 + 4ab

4.

(a – b)2 = (a + b)2 – 4ab

5.

(a + b)3 = a3 + b3 + 3ab (a + b)

6.

(a – b)3 = a3 – b3 – 3ab (a – b)

7.

(a3 + b3) = (a + b) (a2 – ab + b2)

8.

(a3 – b3) = (a – b) (a2 + ab + b2)

9.

(a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

10.

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) If a + b + c = 0, then a3 + b3 + c3 = 3abc

OBJECTIVE QUESTIONS BASED ON THESE FORMULAE 1-

If

1

x

5 , find the value of x 2

x

(1) 527

(2) 227

(3) 427

(4) 327 1

x

Sol. (1)

x

5

(1)

c n b m

(2)

c n b m

Squaring both sides,

(3)

c 1 b m

(4)

c n m b

F GH

Sol. (1) Since (x + a) is common factor of x2 + bx + c and x2 + mx + n therefore, remainder = 0 when both polynomials are divided by (x + a)  x+a=0 or, x = – a For first polynomial, remainder = (–a)2 + b (–a) + c 0 = a2 – ab + c or, a2 = ab – c ....(i) For second polynomial, remainder = (–a)2 + m (–a) + n 0 = a2 – ma + n or, a2 = ma – n ....(ii) Comparing values of a2 from equations (i) and (ii), ab – c = ma – n ab – ma = c – n a (b – m) = c – n a=

c n b m

I 2 b5g2 J xK

1

x

x

1 x

2 x

x

1 x

2

1 x x2 x2 x2

ATGE-13

1

25

x

25

x

1 x

Squaring both sides again,

FG x H

1 .

x2

IJ 2 b23g2 K 1 x2 1 x2 1 x2

2

x

529

527

1 x 2

529

25

2

23

ALGEBRA 2-

ALGEBRA 1

If x 2 1

x3

7, x

x2 3

x (1) 19 (3) 38

FG x H

1 x

IJ 2 K

1

2

x

x2

2

x

IJ K

9

1

x3

x

1

x3

FG x 2 H

1

3-

3

9

1 x 1 x

2

1 x

IJ K

2

2

IJ K

51 x4

1

x4

If x

x4

FG x H

(3)2 1 x

1 x

(1) 16.625 (3) 1.6256

2

x2

2

49

49

4.25

1

x2

18.0625

x2

= 18.0625 – 2

x4 1

= 16.0625

x4

7 then find the value of x 2

1 x2

.

(2) 52 (4) 25 1 x

7

Squaring both sides,

x

9

2

1 x

2

1 x

2

1 x2

IJ 2 b7g2 K 1

x2

9 2

x

7

x2

1

2

1

Sol. (1)  x

IJ 2 b7g2 x2K

x4

2

(1) 51 (3) 15

.

1

x4

1 x4

1 x

If x

FG x 2 H

1

6.25

IJ 2 b4.25g2 x2K

x2

x4

1 x

1

Squaring both sides again,

x4

x

6.25

x2

x4

3

x

2

1

x4

3 , find the value of x 4

FG x H

FG x H

4-

27

(2) 43 (4) 47

Sol. (4) 

x

IJ K

18

(1) 42 (3) 45

2

1 x

27

27

x3 1 x

If x

FG H

1 x x

x

3

x3

x3

1

x4

3

3

(2.5)2

Squaring both sides again,

3

IJ 3 b3g3 K

1 x

IJ 2 K x2

x2

Cubing both sides

FG x H

1 x

x2

1 x

=7+2=9 1 x

2.5

Squaring both sides, (2) 28 (4) 18

FG x H

1 x

Sol. (2)  x

.

FG x H

Sol. (4) 

0, then find the value of

49

6-

2

1 x2 1

x2

x

2

49

1 x

49

= 49 + 2 = 51

x2 1

2 If x

2

= 38 then find the value of x

(1) + 6 (3) ± 6 47

2.5 , then find the value of x 4

Sol. (3)  1 x4

.

(2) 16.0625 (4) 160.625

FG x H

1 x

IJ 2 K

x2

1 x

2

= 38 – 2 = 36 

ATGE-14

(2) – 6 (4) ± 9

FG x H

1 x

IJ K

36 = ± 6

2

1 x

x

1 . x

ALGEBRA 7-

ALGEBRA 1 4 = 5, then find the value of x x

If x

(1) 727 (3) 722 1 x

IJ K

1

x3

.

(2) 772 (4) 277

FG x H

Sol. (1) 

1 x4

x

5

1 x

IJ 2 = (5) K 1

x2

x

2

2

1

x2

x

1 x

10- If

25

Squaring both sides again

FG x 2 H

IJ 2 x2K 1

x4 1

x4

1 x4 1

x4

x4

x2

FG x H

FG x H

= 729

x2

9-

2 x

IJ FG x K H 2

1 x

2 x

IJ K

2

=2

4

x

2 (Formula) 3

3

1 x3 1 x3

3

x

3 2

FG H

1 x x

1 x

1 =3 x

1 x

IJ K

2

= (3)2 1 x

2

1

x2

1 =4 x

x2 1

x2

x2

2

x

FG x 2 H ATGE-15

IJ 2 x2K 1

1 x

+2=9 =9–2=7

Squaring again,

= (4)3

=8

=8–6=2

(2) 46 (4) 49

x2

3

IJ K

1 4 = 3, then find the value of x x

Sol. (3)  x

FG x H

1 x

8

(1) 45 (3) 47

(2) 76 (4) 57

IJ 3 K

x

11- If x

Cubing both sides,

FG x H

IJ K

Squaring both sides,

(1) 75 (3) 67

.

= (2)3 1

x3

2 2 if x = 3. x x (2) – 1 (4) 2

2 =± 1 =±1 x

Sol. (2)  x

1 x

IJ 3 K

x3

1 = 4 find the value of x 3 x

If x

1 x x3

= (3)2 – 8 =9–8 =1 x–

FG x H

= 727

(1) 1 (3) ± 1 Sol. (3) 

1

= 729 – 2

Find the value of x

1 x3

Cubing both sides,

+ 2 = 729

x4

x4

8-

2

= 2, then find the value of x 3 (2) 4 (4) 16

Sol. (1) 

= (27)2

1

x4

IJ K

= 64

= 64 + 12 = 76

x3 1 x

IJ K

= 64

(1) 2 (3) 8

= 25 + 2 = 27

x2

1

FG x H

IJ K

1 x

– 3 × 4 = 64

x3

x3

2

1 x

3 x

1

x3

FG H

1 x x

x

FG H

1 x

x3

Squaring both sides,

FG x H

3

3

= (7)2

9

1 x4

.