Polynomials Notes

POLYNOMIALS For example: 5, –9, CONTENTS  Introduction  Constants & Variables  Algebraic Expression  Factors & Coefficients  Degree of a Polynomial  Types of Polynomial & Polynomial in one variable  Remainder Theorem  Values & Zeroes of a Polynomial  Geometric Meaning of the Zeroes of a Polynomial   Relation between Zeroes & Coefficients Formation of Quadratic Polynomial 3 7 , , , etc. 8 15  Variable : It is a symbol whose value changes according to the situation. For example : x, y, z, ax, a + x, 5y, – 7x, etc. ALGEBRAIC EXPRESSION (a) An algebraic expression is a collection of terms separated by plus (+) or minus (–) sign. For example : 3x + 5y, 7y – 2x, 2x – ay + az, etc. (b) The various parts of an algebraic expression that are separated by ‘+’ or ‘–’ sign are called terms. For example : Algebraic No. of expression terms Terms (i) –32x 1 –32x (ii) 2x + 3y 2 2x and 3y (iii) ax – 5y + cz 3 ax, –5y and cz (iv) 3 y xy + – +9 7 8 x 4 3 y xy , ,– 8 x 7 and 9 & so on. INTRODUCTION Algebra is that branch of mathematics which treats the relation of numbers. CONSTANTS AND VARIABLES In algebra, two types of symbols are used: constants and variable (literals).  Constant : It is a symbol whose value always remains the same, whatever the situation be. Types of Algebraic Expressions : (i) Monomial : An algebraic expression having only one term is called a monomial. For ex. 8y, –7xy, 4x2, abx, etc. ‘mono’ means ‘one’. (ii) Binomial : An algebraic expression having two terms is called a binomial. For ex. 8x + 3y, 8x + 3, 8 + 3y, a + bz, 9 – 4y, 2x2 – 4z, 6y2 – 5y, etc. ‘bi’ means ‘two’. (iii) Trinomial : An algebraic expression having three terms is called a trinomial. For ex. ax – 5y + 8z, 3x2 + 4x + 7, 9y2 – 3y + 2x, etc. ‘tri means ‘three’. Edubull (iv) Multinomial : An algebraic expression having two or more terms is called a multinomial. (ii) The power of term –8x7 = 7 (iii) The power of 3x = 1 Since, the greatest power is 7, therefore degree of the polynomial 5x2 – 8x7 + 3x is 7 FACTORS AND COEFFICIENTS (b) The degree of polynomial :  Factor : Each combination of the constants and variables, which form a term, is called a factor. (i) 4y3 – 3y + 8 is 3 (ii) 7p + 2 is 1(p = p1) For examples : (i) 7, x and 7x are factors of 7x, in which 7 is constant (numerical) factor and x is variable (literal) factor. (iii) 2m – 7m8 + m13 is 13 and so on. EXAMPLES  Ex.2 (ii) In –5x2y, the numerical factor is –5 and literal factors are : x, y, xy, x2 and x2y.  Coefficient : Any factor of a term is called the coefficient of the remaining term. For example : Sol. Find which of the following algebraic expression is a polynomial. (i) 3x2 – 5x (ii) x + (iii) (iv) z5 – z +8 It is a polynomial. (ii) In –5x2y; 5 is coefficient of –x2y; –5 is coefficient of x2y. (ii) x + Write the coefficient of : 1 = x1 + x–1 x It is not a polynomial. (i) x2 in 3x3 – 5x2 + 7 (iii) (ii) xy in 8xyz Sol. 3 (i) 3x2 – 5x = 3x2 – 5x1 (i) In 7x ; 7 is coefficient of x Ex. 1 y–8 1 x y – 8 = y1/2 – 8 (iii) –y in 2y2 – 6y + 2 Since, the power of the first term ( y ) is (iv) x0 in 3x + 7 1 , which is not a whole number. 2 (i) –5 (ii) 8z (iv) z5 – (iii) 6 (iv) Since x0 3x + 7 = 3x + 7x0 Ex.3 (ii) 2y12 + 3y10 – y15 + y + 3 The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial. (iii) x (iv) 8 For example : (i) The power of term 5x2 = 2 Find the degree of the polynomial : (i) 5x – 6x3 + 8x7 + 6x2  DEGREE OF A POLYNOMIAL (a) In polynomial 5x2 – 8x7 + 3x : z + 8 = z5 – z1/3 + 8 Since, the exponent of the second term is 1/3, which in not a whole number. Therefore, the given expression is not a polynomial. = 1, Therefore coefficient of x0 is 7. 3 Sol. (i) Since the term with highest exponent (power) is 8x7 and its power is 7.    The degree of given polynomial is 7. Note : General form of a polynomial in one variable x of degree 'n' is anxn + an–1xn–1 + an–2xn–2 + ….+ a2x2 + a1x + a0, an  0, where an, an–1,… a2, a1, a0 all are constants. (ii) The highest power of the variable is 15    degree = 15. (iii) x = x1  degree is 1. (iv) 8 = 8x0  degree = 0  for linear ax + b, for quadratic ax2 + bx + c, for cubic  TYPES OF POLYNOMIALS 3 a0 a0 2 ax + bx + cx + d, a  0 REMAINDER THEOREM  (A) Based on degree : (i) Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a) . If degree of polynomial is Examples 1. One Linear x + 3, y – x + 2, 2. Two Quadratic 1 2x –7, x2+y2 –2xy, x2 +1+ 3y 3 3 x –3 2 (ii) If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a. (iii) (x – a) is a factor of polynomial p(x) if p(a) = 0 (iv) (x + a) is a factor of polynomial p(x) if p(–a) = 0 (v) (x – a) (x – b) is a factor of polynomial p(x), 3. Three Cubic 4. Four 3 2 2 3 x + 3x –7x+8, 2x +5x +7, if p(a) = 0 and p(b) = 0. EXAMPLES  bi-quadratic x4 + y4 + 2x2y2, x4 + 3,… Ex.4 (B) Based on Terms : If number of terms in polynomial is (a) x – 1 Examples 1. One Monomial 7x, 5x9, 7 16 x , xy, …… 3 2. Two Binomial 2 + 7y6, y3 + x14, 7 + 5x9,… 3. Three Trinomial x3 –2x + y, x31+y32+ z33,….. Note : (1) Degree of constant polynomials (Ex.5, 7, –3, 8/5, …) is zero. (2) Degree of zero polynomial (zero = 0 = zero polynomial) is not defined.  POLYNOMIAL IN ONE VARIABLE If a polynomial has only one variable then it is called polynomial in one variable. 3 Ex. P(x) = 2x + 5x – 3 7 5 Cubic trinomial 3 Q(x) = 7x – 5x – 3x + x + 3 polynomial of degree 7 R(y) = y 2 S(t) = t + 3 Linear, monomial Quadratic Binomial Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by Sol. (b) x + 2 (c) x + 1 2 Let p(x) = 4x3 – 3x2 + 2x – 4 (a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1) p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4 =4×1–3×1+2×1–4 =4–3+2–4=–1 (b) When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2). p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4 = 4 × (–8) – 3 × 4 – 4 – 4 = – 32 – 12 – 8 = – 52  1 2 (c) When p(x) is divided by,  x   then by  remainder theorem, remainder will be the required  1  2  1  2 3  1  2 2  1  2 p    = 4   – 3   + 2   – 4  1  8 = 4 ×   – 3 × 1 1 –2× –4 4 2 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x). For example : (i) For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0 1 3 1 3 – – 1– 4 = – – 5 2 4 2 4 =– x = 2 or simply 2 is a zero of the polynomial p(x) = x – 2. 2  3  20 25 = = 4 4 (ii) For the polynomial g(u) = u2 – 5u + 6; g(3) = (3)2 – 5 × 3 + 6 = 9 – 15 + 6 = 0  VALUES OF A POLYNOMIAL For a polynomial f(x) =    If for x = a, the value of the polynomial p(x) is 0 3x2   – 4x + 2. = u2 – 5u + 6. To find its value at x = 3; Also, g(2) = (2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 0 replace x by 3 everywhere.  2 is also a zero of the polynomial So, the value of f(x) = 3x2 – 4x + 2 at x = 3 is g(u) = u2 – 5u + 6 f(3) = 3 × 32 – 4 × 3 + 2 (a) Every linear polynomial has one and only one zero. = 27 – 12 + 2 = 17. (b) A given polynomial may have more than one zeroes. Similarly, the value of polynomial f(x) = 3x2 – 4x + 2, (c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n. (i) at x = –2 is f(–2) = 3(–2)2 –4(–2) + 2 = 12 + 8 + 2 = 22 For example : (ii) at x = 0 is f(0) = 3(0)2 – 4(0) + 2 If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a polynomial is 8; largest number of zeroes it can have is 8. =0–0+2=2 2 (iii) at x = = Ex.5 1 1 1 1 is f   = 3   – 4   + 2 2 2 2     2 (d) A zero of a polynomial need not be 0. 3 3 –2+2= 4 4 For example : If f(x) = x2 – 4, then f(2) = (2)2 – 4 = 4 – 4 = 0 Find the value of the polynomial 5x – 4x2 + 3 at: (i) x = 0 Sol. 3 is a zero of the polynomial g(u) Here, zero of the polynomial f(x) = x2 – 4 is 2 which itself is not 0. (ii) x = –1 Let p(x) = 5x – 4x2 (e) 0 may be a zero of a polynomial. + 3. For example : If f(x) = x2 – x, (i) At x = 0, p(0) = 5 × 0 – 4 × (0)2 + 3 then f(0) = 02 – 0 = 0 =0–0+3=3 Here 0 is the zero of polynomial (ii) At x = –1, p(–1) = 5(–1) – 4(–1)2 + 3 f(x) = x2 – x. = –5 – 4 + 3 = – 6  ZEROES OF A POLYNOMIAL EXAMPLES  Ex.6 Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases : Edubull (i) p(x) = 3x + 1, x = – (i) For the zero of polynomial p(x) = x + 5 1 3 (ii) p(x) = (x + 1) (x – 2), x = – 1, 2 m  (ii) p(x) = 0   1 (v) p(x) = 2x + 1, x = 2 Sol.  (i) p(x) = 3x + 1  1  3   p    = 3 × –    x = –   1 + 1 = –1 + 1 = 0 3 1 is a zero of p(x) = 3x + 1. 3  (iii) p(x) = x2   p(0) = 02 =0  x = 0 is a zero of the given polynomial    polynomial. 1 2 2 . 3 2 is zero of p(x) = 3x – 2 3 For example : The graph of y = 3x + 4 is a straight line passing through (0, 4) and (2, 10). x 0 2 y  3x  4 4 10 Po int s A B y m is a zero of the given  (v) p(x) = 2x + 1 p   = 2 ×  Let us consider linear polynomial ax + b. The graph of y = ax + b is a straight line. =–m+m=0  x = –   GEOMETRIC MEANING OF THE ZEROES OF  A POLYNOMIAL  m  m  =   + m      5 is a zero of p(x) = 2x + 5. 2  3x = 2 and x = (iv) p(x) = x + m p    5 2 (iii) p(x) = 0  3x – 2 = 0  p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0  x = –1 and x = 2 are zeroes of the given polynomial.  2x + 5 = 0   2x = –5 and x = x = and, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0    x= (ii) p(x) = (x + 1) (x – 2)       x = –5 x = –5 is a zero of the polynomial p(x) = x + 5. (iii) p(x) = x2, x = 0 (iv) p(x) = x + m, x = –  x + 5 = 0 p(x) = 0 1 +1 2 B(2, 10) A(0, 4) = 1 + 1 = 2 0    x = 1 is not a zero of the given 2 polynomial. Ex.7 Find the zero of the polynomial in each of the following cases : (i) p(x) = x + 5 (ii) p(x) = 2x + 5 (iii) p(x) = 3x – 2 Sol. To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0. x' O x (i) Let us consider the graph of y = 2x – 4 intersects the x-axis at x = 2. The zero 2x – 4 is 2. Thus, the zero of the polynomial 2x – 4 is the x-coordinate of the point where the graph y = 2x – 4 intersects the x-axis. Edubull x 2 (iv) Now let us consider one more polynomial –x2 + 2x + 8. Graph of this polynomial intersects the x-axis at the points (4, 0), (–2, 0). Zeroes of the polynomial – x2 + 2x + 8 are the x-coordinates of the points at which the graph intersects the x-axis. The shape of the graph of the given quadratic polynomial is  and the curve is known as parabola. 0 y  2x  4 0  4 Po int s A B y A O x'  2 1 0 x x –1 –2 –3 –4 B y Po int s 0 A 5 B 1 2 3 4 8 9 8 7 0 C D E F G y y' (ii) A general equation of a linear polynomial is ax + b. The graph of y = ax + b is a straight b  , 0 .  a  line which intersects the x-axis at  Zero of the polynomial ax + b is the xcoordinate of the point of intersection of the graph with x-axis. (iii) Let us consider the quadratic polynomial x2 – 4x + 3. The graph of x2 – 4x + 3 intersects the x-axis at the point (1, 0) and (3, 0). Zeroes of the polynomial x2 – 4x + 3 are the x-coordinates of the points of intersection of the graph with x-axis. x' D 9 E 8 C F 7 6 B 5 4 3 2 A –2 1 G –4 –3 –1O 1 2 3 4 x y' 5 The zeroes of a quadratic polynomial ax2 + bx + c he x-coordinates of the points where the graph of y = ax2 + bx + c intersects the x-axis. y  x 2  4x  3 0  1 0 3 8 Po int s A B C D E Cubic polynomial : Let us find out geometrically how many zeroes a cubic has. x 1 2 3 4 The shape of the graph of the quadratic polynomials is  and the curve is known as parabola. y Let consider cubic polynomial x3 – 6x2 + 11x – 6. x 0 0.5 1 1.5 2 y  x 3  6x 2  11x  6  6  1.875 0 0.375 0 Po int s D C A x' O –1 –2 y' 1 2 B 3 4 x A B C D E 2.5  0.375 F 3 3.5 4 0 1.875 6 G H I Case 1 : The graph of the cubic equation intersects the x-axis at three points (1, 0), (2, 0) and (3, 0). Zeroes of the given polynomial are the x-coordinates of the points of intersection with the x-axis. at n points. Therefore, a polynomial p(x) of degree n has at most n zeroes. y I 6 5 4 3 2 1 x' EXAMPLES  Ex.8 D E C 1 –2 –1 0 –1 B –2 –3 –4 –5 –6 A H G 3 2 4 5 6 Which of the following correspond to the graph to a linear or a quadratic polynomial and find the number of zeroes of polynomial. x F y y (i) (ii) x' x' x O y' x O y' y y y' (iii) Case 2 : The cubic equation x3 – x2 intersects the xaxis at the point (0, 0) and (1, 0). Zero of a polynomial x3 – x2 are the x-coordinates of the point where the graph cuts the x-axis. y (iv) x' O y' y (v) O x (vi) x O y' y' y y x' (vii) y' x O y' x' x x' x' y x' O x Zeroes of the cubic polynomial are 0 and 1. x O (viii) O y' Case 3 : y = x3 x x' y' y y Cubic polynomial has only one zero. (ix) y x' O x y' In brief : A cubic equation can have 1 or 2 or 3 zeroes or any polynomial of degree three can have at most three zeroes. Remarks : In general, polynomial of degree n, the graph of y = p(x) passes x-axis at most x' O y' x (x) x' O x y' Sol. (i) The graph is a straight line so the graph is of a linear polynomial. The number of zeroes is one as the graph intersects the x-axis at one point only. (ii) The graph is a parabola. So, this is the graph of quadratic polynomial. The number of zeroes is zero as the graph does not intersect the x-axis. (iii) Here the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as Edubull So if ax2 + bx + c, a  0 is a quadratic polynomial and ,  are two zeroes of polynomial b c then     – ,   a a the graph intersects the x-axis at one point only (two coincident points). (iv) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is two as the graph intersects the x-axis at two points. (v) The polynomial is linear as the graph is straight line. The number of zeroes is zero as the graph does not intersect the x-axis. (vi) The polynomial is quadratic as the graph is a parabola. The number of zeroes is 1 as the graph intersects the x-axis at one point (two coincident points) only.  EXAMPLES  Ex.9 Find the zeroes of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeroes and its coefficients. Sol. We have, 6x2 – 13x + 6 = 6x2 – 4x – 9x + 6 = 2x (3x – 2) –3 (3x – 2) = (3x – 2) (2x – 3) (vii)The polynomial is quadratic as the graph is a parabola. The number of zeroes is zero, as the graph does not intersect the x-axis. So, the value of 6x2 – 13x + 6 is 0, when (3x – 2) = 0 or (2x – 3) = 0 i.e., (viii) Polynomial is neither linear nor quadratic as the graph is neither a straight line nor a parabola is one as the graph intersects the xaxis at one point only. When x= 2 3 or 3 2 Therefore, the zeroes of 6x2 – 13x + 6 are 3 2 and . 3 2 (ix) Here, the polynomial is quadratic as the graph is a parabola. The number of zeroes is one as the graph intersects the x-axis at one point only (two coincident points). Sum of the zeroes (x) The polynomial is linear as the graph is a straight line. The number of zeroes is one as the graph intersects the x-axis at only one point. = RELATIONSHIP BETWEEN THE ZEROES AND THE COEFFICIENTS OF A POLYNOMIAL. = 3 2 13 (13) coefficient of x + = = = 2 3 6 6 coefficient of x 2 Product of the zeroes     3 2 6 cons tan t term × = = 2 3 6 coefficient of x 2 P(x) = 2x2 – 16x + 30. Ex.10 Find the zeroes of the quadratic polynomial 4x2 – 9 and verify the relation between the zeroes and its coefficients. Now, 2x2 – 16x + 30 = (2x – 6) (x – 3) Sol. Consider quadratic polynomial We have, 4x2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3) = 2 (x – 3) (x – 5) The zeroes of P(x) are 3 and 5. So, the value of 4x2 – 9 is 0, when 2x – 3 = 0 or 2x + 3 = 0 i.e., when x = 3 3 or x = – . 2 2 Sum of the zeroes =3+5=8=  coefficient of x  (16) =–  2 2  coefficient of x  Product of the zeroes = 3 × 5 = 15 = 30 constan t term = 2 coefficient of x 2 Therefore, the zeroes of 4x2 – 9 are 3 3 &– . 2 2 Sum of the zeroes = 3 3 (0) coefficient of x – =0= = 2 2 4 coefficient of x 2 Edubull Product of the zeroes 3  3 2  2 =     = 9 cons tan t term = 4 coefficient of x 2 Ex.11 Find the zeroes of the quadratic polynomial 9x2 – 5 and verify the relation between the zeroes and its coefficients. Sol. Ex. 13 Prove relation between the zeroes and the coefficient of the quadratic polynomial ax2 + bx + c. Let  and  be the zeroes of the polynomial ax2 + bx + c Sol. We have, 9x2 – 5 = (3x)2 – ( 5 )2 = (3x – 5 ) (3x + 5 )   =  b  b 2  4ac 2a ....(1)   b  b 2  4ac 2a ....(2) So, the value of 9x2 – 5 is 0, when 3x – 5 = 0 or 3x + 5 = 0 i.e., when x = 5 3 or x =  5 . 3 = By adding (1) and (2), we get += Sum of the zeroes = 5 5 (0) coefficient of x – =0= = 3 3 9 coefficient of x 2 Product of the zeroes = Ex.12 If  and  are the zeroes of ax2 + bx + c, a  0 then verify the relation between the zeroes and its coefficients. 2b b coefficient of x =– = 2a a coefficient of x 2 ax2 + bx + c is – ax2 + bx + c. Therefore, (x – ), (x – ) are the factors of the polynomial ax2 + bx + c.  + bx + c = k (x – ) (x – ) ax2 + bx + c = k {x2 – ( + ) x + } ax2 + bx + c = kx2 – k ( + ) x + k...(1) Comparing the coefficients of x2, x and constant terms of (1) on both sides, we get a = k, b = – k ( + ) and c = k  +=– += b a b c and  = k k and  = Sum of the zeroes = c a [ k = a] b coefficient of x = a coefficient of x 2 b a By multiplying (1) and (2), we get   b  b 2  4ac    b  b 2  4ac     =    Since  and  are the zeroes of polynomial ax2  b  b 2  4ac  b  b 2  4ac + 2a 2a Hence, sum of the zeroes of the polynomial  5  5    = 5 = cons tan t term =  3   3  9 coefficient of x 2     Sol. c cons tan t term = a coefficient of x 2 Product of the zeroes = = = =   2a ( b) 2  (b 2  4ac) 2 4a 4ac 4a 2 = 2   2a = b 2  b 2  4ac 4a 2 c a cons tan t term coefficient of x 2 Hence, product of zeroes = c a Edubull In general, it can be proved that if , ,  are the zeroes of a cubic polynomial ax3 + bx2 + cx + d, then ++= b a Note, On putting x = – 2 in the cubic polynomial = 2(–2)3 + (–2)2 – 5 (–2) + 2 = – 16 + 4 + 10 + 2 = 0 d a Hence, b c d , and are meaningful because a  0. a a a Ex.14 find the zeroes of the quadratic polynomial x2 – 2x – 8 and verify a relationship between zeroes and its coefficients. Sol. = 2(1)3 + (1)2 – 5(1) + 2 = 2 + 1 – 5 + 2 = 0 2x3 + x2 – 5x + 2 c  +  +  = a  = 2x3 + x2 – 5x + 2 x2 – 2x – 8 = x2 – 4x + 2x – 8 polynomial. Sum of the zeroes of p(x) = 1 1  coefficient of x 2 +1–2=– = 2 2 coefficient of x 3 Sum of the products of two zeroes taken at a time = x (x – 4) + 2 (x – 4) = (x – 4) (x + 2) = 1 1 ×1+ × (–2) + 1 × (–2) 2 2 So, the value of x2 – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2. = 1 5 coefficient of x –1–2=– = 2 2 coefficient of x 3 So, the zeroes of x2 – 2x – 8 are 4, – 2. Product of all the three zeroes Sum of the zeroes =4–2=2= (2) coefficient of x = 1 coefficient of x 2 Product of the zeroes = 4 (–2) = –8 = 8 cons tan t term = 1 coefficient of x 2 the coefficients. 2x3 + x2 – 5x + 2 ; 1 , 1, – 2 2 Here, the polynomial p(x) is 2x3 + x2 – 5x + 2 Value of the polynomial 2x3 + x2 – 5x + 2 when x = 1/2 1 2 3 1 2 2 1 2 = 2   +   – 5   +2 = 1 2 =   × (1) × (–2) = –1 = Ex.15 Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and Sol. 1 , 1, – 2 are the zeroes of the given 2 1 1 5 + – +2= 0 4 4 2 So, 1/2 is a zero of p(x). On putting x = 1 in the cubic polynomial (2) constant term = 2 coefficient of x 3  SYMMETRIC FUNCTIONS OF ZEROS OF A  QUADRATIC POLYNOMIAL.   Symmetric function : An algebraic expression in  and , which remains unchanged, when  and  are interchanged is known as symmetric function in  and . For example, 2 +  2 and 3 +  3 etc. are symmetric functions. Symmetric function is to be expressed in terms of ( + ) and . So, this can be evaluated for a given quadratic equation.  Some useful relations involving  and  : 1. 2 + 2 = ( + )2 – 2 2. ( – )2 = ( + )2 – 4 3. 2 – 2 = (+ ) ( – ) = ( + ) Edubull = ( + ) [(2 + 2 + 2) – 3] (  ) 2  4 = ( + ) [( + )2 – 3] 4. 3 + 3 = ( + )3 – 3 ( + ) 5. 3 – 3 = ( – )3 + 3 ( – ) = b a   b  2 3c      a   a  = b a  b 2 3c   b  2  = a  a  a 6. 4 + 4 =[( + )2 – 2]2 –2()2 7. 4 – 4 = (2 + 2) (2 – 2) then use (1) and (3)  EXAMPLES  Ex.16 If  and  are the zeroes of the polynomial ax2 + bx + c. Find the value of (i)  –  (ii) 2 + 2. Since  and  are the zeroes of the polynomial ax2 + bx + c. Sol.  b  +=– ; a c  = a (i) ( – )2 = ( + )2 – 4  b  a 2 =   – 4c b2 4c b 2  4ac = 2 – = a a a a2 b 2  4ac a  – =  b  a  TO FORM A QUADRATIC POLYNOMIAL  WITH THE GIVEN ZEROES Let zeroes of a quadratic polynomial be  and .  x = , x= x –  = 0, x – = 0 The obviously the quadratic polynomial is (x – ) (x – ) i.e., x2 – ( + ) x +  x 2  (Sum of the zeroes) x  Pr oduct of the zeroes Sol. c a 2 b  2ac Hence the polynomial formed (ii) 3 + 3. Since  and  are the zeroes of ax2 + bx + c Sol.   += b , a = c a =– =– b a = x2 – (sum of zeroes) x + Product of zeroes = x2 – 10x + 24 Ex.19 Form the quadratic polynomial whose zeroes are –3, 5. Sol. Here, zeroes are – 3 and 5. Sum of the zeroes = – 3 + 5 = 2 (i) 2 – 2 = ( + ) (– ) b =– a Sum of the zeroes = 4 + 6 = 10 Product of the zeroes = 4 × 6 = 24 a2 Ex.17 If  and  are the zeroes of the quadratic polynomial ax2 + bx + c. Find the value of (i) 2 –  2 a3 Ex.18 Form the quadratic polynomial whose zeroes are 4 and 6. = ( + )2 – 2 =   – 2  =  b 3  3abc  EXAMPLES  (ii) 2 +  2 = 2 + 2 + 2 – 2 2 =  b 2  3ac     a2    Product of the zeroes = (–3) × 5 = – 15 Hence the polynomial formed 2 (  )  4 = x2 – (sum of zeroes) x + Product of zeroes 2 c b b   4 = – a a  a  b 2  4ac b b 2  4ac a2 (ii) 3 + 3 = ( + ) (2 + 2 – ) a 2 = x2 – 2x – 15 Ex.20 Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively(i) 1 ,–1 4 (ii) 2, 1 3 (iii) 0, 5 Edubull Sol. Let the polynomial be ax2 + bx + c and its zeroes be  and . (i) Here,  +  = ++=2=– 1 and  .  = – 1 4  +  +  = – 7 = Thus the polynomial formed  = – 14 = – = x2 – (Sum of zeroes) x + Product of zeroes 1 4 = x2 –   x – 1 = x2 – x –1 4  The other polynomial are k  x 2   x3 + (–2) x2 + (–7)x + 14  = x2 – (Sum of zeroes) x + Product of zeroes Sol.  1 3  x3 + b 2 c d x + x+ a a a Other polynomial are k  x 2  2 x   and its zeroes are , , . Then If k = 3, then the polynomial is ++=0=– 3x2 – 3 2 x + 1 (iii) Here, +=0 . = and 5  = – 6 = = x2 – (Sum of zeroes) x + Product of zeroes = x2 – (0) x + 5 = x2 + Ex.21 Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2, – 7 and –14, respectively. Let the cubic polynomial be b 2 c d x + x+ a a a x3 – (0) x2 + (–7) x + (–6) or x3 – 7x + 6 Ex.23 If  and  are the zeroes of the polynomials ax2 + bx + c then form the polynomial whose and its zeroes are ,  and , then zeroes are ....(1) b c d , and in (1), a a a we get ax3 + bx2 + cx + d  x3 + c a d a Putting the values of 5 ....(1) b a  +  +  = – 7 = Thus the polynomial formed  Let the cubic polynomial be ax3 + bx2 + cx + d 1 1 or x2 – 2 x + 3 3  Sol.  x3 – 2x2 – 7x + 14 Ex.22 Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively. 1 3  b c d , and in (1), a a a we get x   1 4  Thus the polynomial formed = x2 – ( 2 ) x + c a d a Putting the values of If k = 4, then the polynomial is 4x2 – x – 4. (ii) Here,  + = 2 ,  = b a Sol. 1 1 and .   Since  and  are the zeroes of ax2 + bx + c So  +  = b c , = a a Edubull Sum of the zeroes = x2 – (sum of zeroes) x + product of zeroes 1 1  + =    or x2 – b a = b c c a  =  or 3  x 2   1 1 1 a . = = c   c a But required polynomial is x2  b a  x+   c   c  b c a c or c  x 2  x     Sol. First arrange the term of dividend and the divisor in the decreasing order of their degrees. Step 2 : To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor. Step 3 : To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor. Step 4 : Continue this process till the degree of remainder is less than the degree of divisor. Since  and  are the zeroes of the polynomial x2 + 4x + 3. Then,  +  = – 4,  = 3  Division Algorithm for Polynomial If p(x) and g(x) are any two polynomials with Sum of the zeroes =1+ = g(x)  0, then we can find polynomials q(x) and r(x) such that     2     2  +1+ =    p(x) = q(x) × g(x) + r(x)  2   2  2 (  ) 2 (4) 2 16 = = =   3 3 where r(x) = 0 or degree of r(x) < degree of g(x). The result is called Division Algorithm for polynomials. Product of the zeroes      1   = 1 + + +       = 1     =2+  2  2 2   2   2 =     (   ) 2 (4) 2 16 = = =  3 3 But required polynomial is (if k = 3) Step 1:  cx2 + bx + a Ex.24 If  and  are the zeroes of the polynomial x2 + 4x + 3, form the polynomial whose   zeroes are 1 + and 1 + .   16 16  x  3 3  b a x+ c c or x2 +   WORKING RULE TO DIVIDE A POLYNOMIAL  BY ANOTHER POLYNOMIAL – (sum of zeroes) x + Product of zeroes  x2 –  16 16  x  3 3  or k  x 2   3x2 – 16x + 16 Product of the zeroes = 16 16 x+ 3 3 Dividend  Quotient  Divisor  Remainder  EXAMPLES  Ex.25 Divide 3x3 + 16x2 + 21x + 20 by x + 4. Sol. Edubull x2 + x – 3 x – x + 1 x4 – 3x2 + 4x + 5 x4 – x3 + x2 – + – x3 – 4x2 + 4x + 5 x3 – x2 + x – + – –3x2 + 3x + 5 – 3x2 + 3x – 3 + – + 8 3x2 + 4x + 5 3x 3 x+4 3x + 16x + 21x + 20 First term of q(x) = = 3x2 3 2 x 3x + 12x – – 4x 2 Second term of q(x) = = 4x 2 4x + 21x + 20 x 2 4x + 16x – – 5x Third term of q(x) = =5 5x + 20 x 5x + 20 – – 0 3 2 2 Quotient = 3x2 + 4x + 5 We stop here since Remainder = 0 degree of (8) < degree of (x2 – x + 1). Ex.26 Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below : So, quotient = x2 + x – 3, remainder = 8 Therefore, p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 Sol. Quotient × Divisor + Remainder We have, p(x) = x3 = (x2 + x – 3) (x2 – x + 1) + 8 – 3x2 + 5x – 3 and g(x) = x–3 x2 –2 = x4 – x3 + x2 + x3 – x2 + x – 3x2 + 3x – 3 + 8 3 x x2–2 x3 – 3x2 + 5x – 3 First term of quotient is 2 = x x 3 x – 2x  3x 2 – + – 3x2 + 7x – 3 Second term of quotient is x 2 = –3 – 3x2 +6 + – 7x – 9 = x4 – 3x2 + 4x + 5 = Dividend Therefore the Division Algorithm is verified. Ex.28 Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.t2 – 3; 2t4 + 3t3 – 2t2 – 9t – 12. Sol. We divide 2t4 + 3t3 – 2t2 – 9t – 12 by t2 – 3 We stop here since degree of (7x – 9) < degree of (x2 2t2 + 3t + 4 4 3 2 t – 3 2t + 3t – 2t – 9t – 12 4 2t – 6t2 – + 3 3t + 4t2 + 9t – 12 3t3 – 9t – + 4t2 – 12 4t2 – 12 – + 0 2 – 2) So, quotient = x – 3, remainder = 7x – 9 Therefore, Quotient × Divisor + Remainder = (x – 3) (x2 – 2) + 7x – 9 = x3 – 2x – 3x2 + 6 + 7x – 9 = x3 – 3x2 + 5x – 3 = Dividend Therefore, the division algorithm is verified. Ex.27 Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given below p(x) = x4 – 3x2 + 4x + 5, g (x) = x2 + 1 – x Sol. We have, p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x Here, remainder is 0, so t2 – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12. 2t4 + 3t3 – 2t2 – 9t – 12 = (2t2 + 3t + 4) (t2 – 3) Ex.29 Obtain all the zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are 5 and – 3 5 . 3 Edubull Sol. 5 5 and – , 3 3 Since two zeroes are 5 5 ,x=– 3 3 x=   5   5  5  x  x = x2 – or 3x2 – 5   3   3  3 is a factor of the given polynomial. x2 – x + 1 x3 x–2 x3 – 3x2 + 3x – 2 First term of quotient is =x x x3 – 2x2 – +  x2 = –x –x2 + 3x – 2 Second term of quotient is x –x2 + 2x + – x =1 x – 2 Third term of quotient is x x –2 – + 0 Now, we apply the division algorithm to the given polynomial and 3x2 – 5. x2 + 2x + 1 4 3 2 3x – 5 3x + 6x – 2x – 10x – 5 4 3x – 5x2 – + 6x3 + 3x2 – 10x – 5 6x3 – 10x – + 3x2 –5 3x2 –5 – + 0 Hence, g(x) = x2 – x + 1. Ex.31 Give examples of polynomials p(x), q(x) and r(x), which satisfy the division algorithm and 2 (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg q(x) = 0 Sol. (i) Let q(x) = 3x2 + 2x + 6, degree of q(x) = 2 p(x) = 12x2 + 8x + 24, degree of p(x) = 2 Here, deg p(x) = deg q(x) (ii) p(x) = x5 + 2x4 + 3x3 + 5x2 + 2 So, 3x4 + 6x3 – 2x2 – 10x – 5 q(x) = x2 + x + 1, = (3x2 – 5) (x2 + 2x + 1) + 0 Quotient = x2 + 2x + 1 = (x + 1)2 g(x) = x3 + x2 + x + 1 Zeroes of (x + 1)2 are –1, –1. r(x) = 2x2 – 2x + 1, degree of r(x) = 2 Hence, all its zeroes are 5 ,– 3 Here, deg q(x) = deg r(x) 5 , –1, –1. 3 (iii) Let p(x) = 2x4 + 8x3 + 6x2 + 4x + 12 q(x) = 2, degree of q(x) = 0 Ex.30 On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x). Sol. degree of q(x) = 2 g(x) = x4 + 4x3 + 3x2 + 2x + 6 r(x) = 0 p(x) = x3 – 3x2 + x + 2 q(x) = x – 2 and r (x) = –2x + 4 By Division Algorithm, we know that p(x) = q(x) × g(x) + r(x) Therefore, x3 – 3x2 + x + 2 = (x – 2) × g(x) + (–2x + 4)   x3 – 3x2 + x + 2 + 2x – 4 = (x – 2) × g(x)  x 3  3x 2  3x  2  g(x) = x2 On dividing x3 – 3x2 + 3x – 2 by x – 2, we get g(x) Here, deg q(x) = 0 Ex.32 If the zeroes of polynomial x3 – 3x2 + x + 1 are a – b, a , a + b. Find a and b. Sol.  a – b, a, a + b are zeros  product (a – b) a(a + b) = –1  (a2 – b2) a = –1 …(1) and sum of zeroes is (a – b) + a + (a + b) = 3  3a = 3  a = 1 by (1) and (2) (1 – b2)1 = –1 …(2) Edubull  2 = b2  b = ± 2  a = –1 & b = ± 2 Ans. Ex.33 If two zeroes of the polynomial x4 – 6x3 –26x2 + 138x – 35 are 2 ± find other zeroes. Sol. 3,  2 ± 3 are zeroes. x [4k – 25 + 16 – 2k] + 10 – 8k + k2  x=2± 3  x – 2 = ± 3 (squaring both sides)  (x – 2)2 = 3  x2 + 4 – 4x – 3 = 0 2  x – 4x + 1 = 0 , is a factor of given polynomial  other factors × 2  other factors = x – 2x – 35 = x2 – 7x + 5x – 35 = x(x – 7) + 5(x – 7) = (x – 7) (x + 5)  other zeroes are (x – 7) = 0  x = 7 x+5=0x=–5 Ans. 3  coefficient of x = 1  2k – 9 = 1  k = (10/2) = 5 Also constant term = a  a = 25 – 40 + 10 x2 – 2x – 35 4 3 2 x2 – 4x + 1 x 4 – 6x 3 – 26x2 + 138x – 35 x – 4x + x – – + – 2x3 – 27x2 + 138x – 35 – 2x3 + 8x2 – 2x – + + – 35x2 + 140x – 35 – 35x2 + 140x – 35 + – + 4 According to questions, remainder is x + a  k2 – 8k + 10 = a  (5)2 – 8(5) + 10 = a x 4  6x 3  26x 2  138x  35 = x 2  4x  1 2 Ex.34 If the polynomial x – 6x + 16x –25x + 10 is divided by another polynomial x2 –2x + k, the remainder comes out to be x + a, find k & a. Sol. x2 – 4x + (8 – k) 4 3 2 x2 – 2x + k x – 6x + 16x – 25x + 10 4 3 2 x – 2x + x k – – + – 4x3 + x2 (16 – k)– 25x + 10 – 4x3 + x2 (8) – 4xk – + + x2 [8 – k] + x[4k– 25] + 10 x2 [8 – k] – 2x[8 – k] +k(8 – k) – + – a=–5  k = 5, a = –5 Ans.