Page 1 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, , Exercise 2.1, 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes, of p(x), in each case., , Solutions:, Graphical method to find zeroes:Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis., , (i), (ii), (iii), (iv), (v), (vi), , In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis, does not cut it at any point., In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at, only one point., In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at, any three points., In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at, two points., In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at, four points., In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at, three points., , πππ«π€ππ‘π₯ππ―'π°
Page 3 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2), Product of zeroes = 0Γ-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 ), (v) t2β15, β t2 = 15 or t = Β±β15, Therefore, zeroes of polynomial equation t2 β15 are (β15, -β15), Sum of zeroes =β15+(-β15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2), Product of zeroes = β15Γ(-β15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 ), (vi) 3x2βxβ4, β 3x2β4x+3xβ4 = x(3x-4)+1(3x-4) = (3x - 4)(x + 1), Therefore, zeroes of polynomial equation3x2 β x β 4 are (4/3, -1), Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2), Product of zeroes=(4/3)Γ(-1) = (-4/3) = (Constant term) /(Coefficient of x2 ), , 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively. (i) 1/4 , -1, Solution:, From the formulas of sum and product of zeroes, we know,, Sum of zeroes = Ξ±+Ξ², Product of zeroes = Ξ± Ξ², , Sum of zeroes = Ξ±+Ξ² = 1/4, Product of zeroes = Ξ± Ξ² = -1, β΄ If Ξ± and Ξ² are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written, directly as:x2β(Ξ±+Ξ²)x +Ξ±Ξ² = 0 x2β(1/4)x, +(-1) = 0, 4x2βx-4 = 0, Thus,4x2βxβ4 is the quadratic polynomial., (ii)β2, 1/3, , πππ«π€ππ‘π₯ππ―'π°
Page 9 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2. (iii), x3-3x+1, x5-4x3+x2+3x+1, Solutions:, Given,, First polynomial = x3-3x+1, Second polynomial = x5-4x3+x2+3x+1, , As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 ., 3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are β(5/3) and - β(5/3)., , Solutions:, Since this is a polynomial equation of degree 4, hence there will be total 4 roots., β(5/3) and - β(5/3) are zeroes of polynomial f(x)., , β΄ (x -β(5/3)) (x+β(5/3) = x2-(5/3) = 0, (3x2β5)=0, is a factor of given polynomial f(x)., Now, when we will divide f(x) by (3x2β5) the quotient obtained will also be a factor of f(x) and the remainder, will be 0., , πππ«π€ππ‘π₯ππ―'π°
Page 10 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, , Therefore, 3x4 +6x3 β2x2 β10xβ5 = (3x2 β5)(x2+2x+1), Now, on further factorizing (x2+2x+1) we get,, x2+2x+1 = x2+x+x+1 = 0 x(x+1)+1(x+1) = 0, (x+1)(x+1) = 0, So, its zeroes are given by: x= β1 and x = β1., Therefore, all four zeroes of given polynomial equation are:, β(5/3),- β(5/3) , β1 and β1., , Hence, is the answer., 4. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were xβ2 and β2x+4,, respectively. Find g(x)., Solutions:, Given,, Dividend, p(x) = x3-3x2+x+2, Quotient = x-2, Remainder = β2x+4, We have to find the value of Divisor, g(x) =?, , πππ«π€ππ‘π₯ππ―'π°
Page 11 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, As we know,, Dividend = Divisor Γ Quotient + Remainder, x3-3x2+x+2 = g(x)Γ(x-2) + (-2x+4), x3-3x2+x+2-(-2x+4) = g(x)Γ(x-2) Therefore,, g(x) Γ (x-2) = x3-3x2+3x-2, Now, for finding g(x) we will divide x3-3x2+3x-2 with (x-2), , Therefore, g(x) = (x2βx+1), 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and, (i) deg p(x) = deg q(x), (ii) deg q(x) = deg r(x), (iii) deg r(x) = 0, Solutions:, According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)β 0. Then we, can find the value of quotient q(x) and remainder r(x), with the help of below given formula;, Dividend = Divisor Γ Quotient + Remainder, β΄ p(x) = g(x)Γq(x)+r(x), Where r(x) = 0 or degree of r(x)< degree of g(x)., Now let us proof the three given cases as per division algorithm by taking examples for each., (i) deg p(x) = deg q(x), Degree of dividend is equal to degree of quotient, only when the divisor is a constant term., Let us take an example, p(x) = 3x2 + 3x + 3 is a polynomial to be divided by g(x) = 3., So, (3x2 + 3x + 3)/3 = x2 + x + 1 = q(x), Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend, p(x)., Hence, division algorithm is satisfied here., , πππ«π€ππ‘π₯ππ―'π°
Page 12 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, (ii) deg q(x) = deg r(x), Let us take an example, p(x) = x2 + 3 is a polynomial to be divided by g(x) = x - 1., So, x2 + 3 = (x - 1)Γ(x) + (x + 3), Hence, quotient q(x) = x, Also, remainder r(x) = x + 3, Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of, remainder r(x)., Hence, division algorithm is satisfied here., (iii) deg r(x) = 0, The degree of remainder is 0 only when the remainder left after division algorithm is constant., Let us take an example, p(x) = x2 + 1 is a polynomial to be divided by g(x) = x. So, x2 + 1 =, (x)Γ(x) + 1, Hence, quotient q(x) = x, And, remainder r(x) = 1, Clearly, the degree of remainder here is 0., Hence, division algorithm is satisfied here., , Exercise 2., 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the, relationship between the zeroes and the coefficients in each case:, (i) 2x3+x2-5x+2; -1/2, 1, -2, Solution:, Given, p(x) = 2x3+x2-5x+2, And zeroes for p(x) are = 1/2, 1, -2, 5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0, p(1) = 2(1)3+(1)2-5(1)+2 = 0 p(-2), = 2(-2)3+(-2)2-5(-2)+2 = 0, Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2., Now, comparing the given polynomial with general expression, we get;, +cx+d = 2x3+x2-5x+2, a=2, b=1, c= -5 and d = 2, As we know, if Ξ±, Ξ², Ξ³ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;, Ξ± +Ξ²+Ξ³ = βb/a Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ± = c/a Ξ± Ξ²Ξ³ = β d/a., Therefore, putting the values of zeroes of the polynomial,, Ξ±+Ξ²+Ξ³ = Β½+1+(-2) = -1/2 = βb/a Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ± = (1/2Γ1)+(1, , πππ«π€ππ‘π₯ππ―'π°
Page 13 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, Γ-2)+(-2Γ1/2) = -5/2 = c/a Ξ± Ξ² Ξ³ = Β½Γ1Γ(-2) = -2/2, = -d/a, Hence, the relationship between the zeroes and the coefficients are satisfied., (ii) x3-4x2+5x-2 ;2, 1, 1, Solution:, Given, p(x) = x3-4x2+5x-2, And zeroes for p(x) are 2,1,1., , p(1) = 13-(4Γ12 )+(5Γ1)-2 = 0, Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2, Now, comparing the given polynomial with general expression, we get;, +cx+d =, a = 1, b = -4, c = 5 and d = -2, As we know, if Ξ±, Ξ², Ξ³ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;, Ξ± + Ξ² + Ξ³ = βb/a Ξ±Ξ² + Ξ²Ξ³ + Ξ³Ξ± = c/a Ξ± Ξ² Ξ³ = β d/a., Therefore, putting the values of zeroes of the polynomial,, Ξ± +Ξ²+Ξ³ = 2+1+1 = 4 = -(-4)/1 = βb/a Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ± =, 2Γ1+1Γ1+1Γ2 = 5 = 5/1= c/a Ξ±Ξ²Ξ³ = 2Γ1Γ1 = 2 = -(2)/1 = -d/a, Hence, the relationship between the zeroes and the coefficients are satisfied., , 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the, product of its zeroes as 2, β7, β14 respectively., Solution:, Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be Ξ±, Ξ², Ξ³., As per the given question, Ξ±+Ξ²+Ξ³ = -b/a = 2/1 Ξ±Ξ² +Ξ²Ξ³+Ξ³Ξ± = c/a = -7/1 Ξ± Ξ²Ξ³ = -d/a = -14/1, Thus, from above three expressions we get the values of coefficient of polynomial. a, = 1, b = -2, c = -7, d = 14, Hence, the cubic polynomial is x3-2x2-7x+14, , πππ«π€ππ‘π₯ππ―'π°
Page 14 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, 3. If the zeroes of the polynomial x3-3x2+x+1 are a β b, a, a + b, find a and b., Solution:, We are given with the polynomial here, p(x), = x3-3x2+x+1, And zeroes are given as a β b, a, a + b, Now, comparing the given polynomial with general expression, we get;, +rx+s = x3-3x2+x+1, p = 1, q = -3, r = 1 and s = 1, Sum of zeroes = a β b + a + a + b, -q/p = 3a, Putting the values q and p., -(-3)/1 = 3a a=1, Thus, the zeroes are 1-b, 1, 1+b., Now, product of zeroes = 1(1-b)(1+b), -s/p = 1-b2 1/1 = 1-b2 b2, = 1+1 = 2 b, = Β±β2, Hence,1-β2, 1 ,1+β2 are the zeroes of x3-3x2+x+1., , 4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 Β±β3, find other zeroes., Solution:, Since this is a polynomial equation of degree 4, hence there will be total 4 roots., Let f(x) = x4-6x3-26x2+138x-35, , Since 2 +β3 and 2-β3 are zeroes of given polynomial f(x)., β΄ [xβ(2+β3)] [xβ(2-β3)], , πππ«π€ππ‘π₯ππ―'π°, , =0
Page 15 :
NCERT Solution For Class 10 Maths Chapter 2- Polynomials, (xβ2ββ3)(xβ2+β3) = 0, On multiplying the above equation we get, x2-4x+1,, this is a factor of a given polynomial f(x)., Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0., , So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 β2xβ35), Now, on further factorizing (x2β2xβ35) we get,, x2β(7β5)x β35 = x2β 7x+5x+35 = 0 x(x, β7)+5(xβ7) = 0, (x+5)(xβ7) = 0, So, its zeroes are given by: x=, β5 and x = 7., Therefore, all four zeroes of given polynomial equation are: 2+β3 , 2-β3, β5 and 7., , πππ«π€ππ‘π₯ππ―'π°