Q: Question 3Verify whether the following are zeroes of the polynomial, indicated against them.(i) p(x) = 3x + 1, x = −13-\dfrac{1}{3}−31(ii) p(x) = 5x - π, x = 45\dfrac{4}{5}54(iii) p(x) = x2 - 1, x = 1, -1(iv) p(x) = (x + 1)(x - 2), x = -1, 2(v) p(x) = x2, x = 0(vi) p(x) = lx + m, x = −ml-\dfrac{m}{l}−lm(vii) p(x) = 3x2 - 1, x = −(13)-\Big(\dfrac{1}{\sqrt{3}}\Big)−(31), (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(32)(viii) p(x) = 2x + 1, x = 12\dfrac{1}{2}21

(i) p(x) = 3x + 1Putting x = −13-\dfrac{1}{3}−31 we get,p(−13)\text{p}\Big(-\dfrac{1}{3}\Big)p(−31) = 3(−13)3\Big(-\dfrac{1}{3}\Big)3(−31) + 1= -1 + 1= 0Hence, x = −13-\dfrac{1}{3}−31 is zeroes of this polynomial.(ii) p(x) = 5x - πPutting x = 45\dfrac{4}{5}54 we get,p(45)=5(45)−π\text{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pip(54)=5(54)−π= 4 - πHence, x = 45\dfrac{4}{5}54 is not a zeroes of this polynomial.(iii) p(x) = x2 - 1Putting x = 1 we get,p(1) = (1)2 - 1= 0Putting x = -1 we getp(-1) = (-1)2 - 1= 1 - 1= 0Hence, x = 1, -1 is zeroes of this polynomial.(iv) p(x) = (x + 1)(x - 2)Putting x = -1 we get,p(1) = (-1 + 1)(-1 - 2)= 0 x (-3)= 0Putting x = 2 we get,p(2) = (2 + 1)(2 - 2)= 3 x 0= 0Hence, x = -1, 2 are the zeroes of this polynomial.(v) p(x) = x2Putting x = 0 we get,p(0) = (0)2= 0Hence, x = 0 is the zero of this polynomial.(vi) p(x) = lx + mPutting x = -ml\dfrac{m}{l}lm we get,p - (ml)=lx−(ml)\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big)(lm)=lx−(lm) + m= -m + m= 0Hence, x = −ml-\dfrac{m}{l}−lm is the zero of this polynomial.(vii) p(x) = 3x2 - 1Putting x = −13-\dfrac{1}{\sqrt{3}}−31 we get,p-(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(31) = 3 x -(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(31)2 - 1= 3 x 13\dfrac{1}{3}31 - 1= 1 - 1= 0Putting x = 23\dfrac{2}{\sqrt{3}}32 we get,p (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(32) = 3 x (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(32)2 - 1= 3 x 43\dfrac{4}{3}34 - 1= 4 - 1= 3Hence, x = −13-\dfrac{1}{\sqrt{3}}−31 is zeroes of polynomial but x = 23\dfrac{2}{\sqrt{3}}32 is not a zeroes of this polynomial.(viii) p(x) = 2x + 1Putting x = 12\dfrac{1}{2}21 we get,p(12)\Big(\dfrac{1}{2}\Big)(21) = 2 x 12\dfrac{1}{2}21 + 1= 1 + 1= 2No, x = 12\dfrac{1}{2}21 is not the zeroes of this polynomial.

Question 1

Question 1Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.(i) 4x2 - 3x + 7(ii) y2 + 2\sqrt{2}2​(iii) 3t+t23\sqrt{t} + t\sqrt{2}3t​+t2​(iv) y+2yy + \dfrac{2}{y}y+y2​(v) x10 + y3 + t50

Accepted Answer

(i) 4x2 - 3x + 7It's a polynomial in one variable i.e., x(ii) y2 + 2\sqrt{2}2​It's a polynomial in one variable i.e., y(iii) 3t3\sqrt{t}3t​ + t2t\sqrt{2}t2​ = 3t1/2 + t 2\sqrt{2}2​It is not a polynomial as degree of t is not a whole number.(iv) y + 2y\dfrac{2}{y}y2​ = y + 2y-1It is in not a polynomial as degree cannot be negative (-ve) in polynomial.(v) x10 + y3 + t50It is polynomial but not in one variable as variables are x, y and t here. Hence, it is a polynomial in three variables.

Question 2

Question 2Write the coefficients of x2 in each of the following:(i) 2 + x2 + x(ii) 2 - x2 + x3(iii) π2\dfrac{π}{2}2π​ x2 + x(iv) 2\sqrt{2}2​x - 1

Accepted Answer

(i) 2 + x2 + xCoefficient of x2 is 1.(ii) 2 - x2 + x3Coefficient of x2 is (-1).(iii) π2\dfrac{π}{2}2π​ x2 + xCoefficient of x2 is = π2\dfrac{π}{2}2π​(iv) 2\sqrt{2}2​x - 1Coefficient of x2 is 0.

Question 3

Question 3Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Accepted Answer

One example of a binomial of degree 35 is:3y35 + 5One example of a monomial of degree 100 is:2x100

Question 4

Question 4Write the degree of each of the following polynomials:(i) 5x3 + 4x2 +7x(ii) 4 - y2(iii) 5t - 7\sqrt{7}7​(iv) 3

Accepted Answer

(i) 5x3 + 4x2 +7xDegree of polynomial is 3.(ii) 4 - y2Degree of polynomial is 2.(iii) 5t - 7\sqrt{7}7​Degree of polynomial is 1.(iv) 3Degree of polynomial is 0. Degree of constant term is always zero.

Question 5

Question 5Classify the following as linear, quadratic and cubic polynomials:(i) x2 + x(ii) x - x3(iii) y + y2 + 4(iv) 1 + x(v) 3t(vi) r2(vii) 7x3

Accepted Answer

(i) x2 + x — Quadratic polynomial [∵ Degree is 2](ii) x - x3 — Cubic polynomial [∵ Degree is 3](iii) y + y2 + 4 — Quadratic polynomial [∵ Degree is 2](iv) 1 + x — Linear polynomial [∵ Degree is 1](v) 3t — Linear polynomial [∵ Degree is 1](vi) r2 — Quadratic polynomial [∵ Degree is 2](vii) 7x3 — Cubic polynomial [∵ Degree is 3]

Question 6

Question 1Find the value of the polynomial 5x - 4x2 + 3 at(i) x = 0(ii) x = -1(iii) x = 2

Accepted Answer

p(x) = 5x - 4x2 + 3(i) x = 0We put x = 0 in the place of x in p(x)p(0) = 5(0) - 4(0)2 +3= 0 + 0 + 3p(0) = 3Hence, at x = 0, 5x - 4x2 + 3 = 3(ii) x = -1We put x = -1 in the place of x in p(x)p(-1) = 5(-1) - 4(-1)2 + 3= (-5) (-4) + 3= (-9) + 3= -6Hence, at x = -1, 5x - 4x2 + 3 = -6(iii) x = 2We put x = 2 in the place of x in p(x)p(2) = 5(2) - 4(2)2 +3= 10 - 4 x 4 + 3= 10 - 16 + 3= 13 - 16= -3Hence, at x = 2, 5x - 4x2 + 3 = -3

Question 7

Question 2Find p(0), p(1) and p(2) for each of the following polynomials:(i) p(y) = y2 - y + 1(ii) p(t) = 2 + t + 2t2 - t3(iii) p(x) = x3(iv) p(x) = (x - 1) (x + 1)

Accepted Answer

(i) p(y) = y2 - y + 1Putting y = 0 we get,p(0) = (0)2 - 0 + 1= 0 - 0 + 1= 1∴ p(0) = 1Putting y = 1 we get,p(1) = (1)2 - 1 + 1= 1 - 1 + 1= 2 - 1= 1∴ p(1) = 1Putting y = 2 we get,p(2) = (2)2 - 2 + 1= 4 - 2 + 1= 5 - 2= 3∴ p(2) = 1Hence, for p(y) = y2 - y + 1, p(0) = 1, p(1) = 1 and p(2) = 1(ii) p(t) = 2 + t + 2t2 - t3Putting t = 0 we get,p(0) = 2 + 0 + 2(02) - 03= 2 + 0 + 0 - 0= 2∴ p(0) = 2Putting t = 1 we get,p(1) = 2 + 1 + 2(12) - 13= 2 + 1 + 2 - 1= 4∴ p(1) = 4Putting t = 2 we get,p(2) = 2 + 2 + 2(22) - 23= 4 + 2(4) - 8= 4 + 8 - 8= 4∴ p(2) = 4Hence, for p(t) = 2 + t + 2t2 - t3, p(0) = 2, p(1) = 4 and p(2) = 4(iii) p(x) = x3Putting x = 0 we get,p(0) = (0)3= 0∴ p(0) = 0Putting x = 1 we get,p(1) = (1)3= 1∴ p(1) = 1Putting x = 2 we get,p(2) = (2)3= 8∴ p(2) = 8Hence, for p(x) = x3, p(0) = 0, p(1) = 1 and p(2) = 8(iv) p(x) = (x - 1)(x + 1)Putting x = 0 we get,p(0) = (0 - 1)(0 + 1)= (-1) x 1= (-1)∴ p(0) = -1Putting x = 1 we get,p(1) = (1 - 1)(1 + 1)= 0 x 2= 0∴ p(1) = 0Putting x = 2 we getp(2) = (2 - 1) (2 + 1)= 1 x 3= 3∴ p(2) = 3Hence, for p(x) = (x - 1) (x + 1), p(0) = -1, p(1) = 0 and p(2) = 3

Question 8

Question 3Verify whether the following are zeroes of the polynomial, indicated against them.(i) p(x) = 3x + 1, x = −13-\dfrac{1}{3}−31​(ii) p(x) = 5x - π, x = 45\dfrac{4}{5}54​(iii) p(x) = x2 - 1, x = 1, -1(iv) p(x) = (x + 1)(x - 2), x = -1, 2(v) p(x) = x2, x = 0(vi) p(x) = lx + m, x = −ml-\dfrac{m}{l}−lm​(vii) p(x) = 3x2 - 1, x = −(13)-\Big(\dfrac{1}{\sqrt{3}}\Big)−(3​1​), (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​)(viii) p(x) = 2x + 1, x = 12\dfrac{1}{2}21​

Accepted Answer

(i) p(x) = 3x + 1Putting x = −13-\dfrac{1}{3}−31​ we get,p(−13)	ext{p}\Big(-\dfrac{1}{3}\Big)p(−31​) = 3(−13)3\Big(-\dfrac{1}{3}\Big)3(−31​) + 1= -1 + 1= 0Hence, x = −13-\dfrac{1}{3}−31​ is zeroes of this polynomial.(ii) p(x) = 5x - πPutting x = 45\dfrac{4}{5}54​ we get,p(45)=5(45)−π	ext{p}\Big(\dfrac{4}{5}\Big) = 5\Big(\dfrac{4}{5}\Big) - \pip(54​)=5(54​)−π= 4 - πHence, x = 45\dfrac{4}{5}54​ is not a zeroes of this polynomial.(iii) p(x) = x2 - 1Putting x = 1 we get,p(1) = (1)2 - 1= 0Putting x = -1 we getp(-1) = (-1)2 - 1= 1 - 1= 0Hence, x = 1, -1 is zeroes of this polynomial.(iv) p(x) = (x + 1)(x - 2)Putting x = -1 we get,p(1) = (-1 + 1)(-1 - 2)= 0 x (-3)= 0Putting x = 2 we get,p(2) = (2 + 1)(2 - 2)= 3 x 0= 0Hence, x = -1, 2 are the zeroes of this polynomial.(v) p(x) = x2Putting x = 0 we get,p(0) = (0)2= 0Hence, x = 0 is the zero of this polynomial.(vi) p(x) = lx + mPutting x = -ml\dfrac{m}{l}lm​ we get,p - (ml)=lx−(ml)\Big(\dfrac{m}{l}\Big) = l x -\Big(\dfrac{m}{l}\Big)(lm​)=lx−(lm​) + m= -m + m= 0Hence, x = −ml-\dfrac{m}{l}−lm​ is the zero of this polynomial.(vii) p(x) = 3x2 - 1Putting x = −13-\dfrac{1}{\sqrt{3}}−3​1​ we get,p-(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(3​1​) = 3 x -(13)\Big(\dfrac{1}{\sqrt{3}}\Big)(3​1​)2 - 1= 3 x 13\dfrac{1}{3}31​ - 1= 1 - 1= 0Putting x = 23\dfrac{2}{\sqrt{3}}3​2​ we get,p (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​) = 3 x (23)\Big(\dfrac{2}{\sqrt{3}}\Big)(3​2​)2 - 1= 3 x 43\dfrac{4}{3}34​ - 1= 4 - 1= 3Hence, x = −13-\dfrac{1}{\sqrt{3}}−3​1​ is zeroes of polynomial but x = 23\dfrac{2}{\sqrt{3}}3​2​ is not a zeroes of this polynomial.(viii) p(x) = 2x + 1Putting x = 12\dfrac{1}{2}21​ we get,p(12)\Big(\dfrac{1}{2}\Big)(21​) = 2 x 12\dfrac{1}{2}21​ + 1= 1 + 1= 2No, x = 12\dfrac{1}{2}21​ is not the zeroes of this polynomial.

Question 9

Question 4Find the zero of the polynomial in each of the following cases:(i) p(x) = x + 5(ii) p(x) = x - 5(iii) p(x) = 2x + 5(iv) p(x) = 3x - 2(v) p(x) = 3x(vi) p(x) = ax, a ≠ 0(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Accepted Answer

(i) p(x) = x + 5⇒ p(x) = 0⇒ x + 5 = 0⇒ x = -5Hence, -5 is a zero of polynomial x + 5(ii) p(x) = x - 5⇒ p(x) = 0⇒ x - 5 = 0⇒ x = 5Hence, 5 is a zero of polynomial x - 5(iii) p(x) = 2x + 5⇒ p(x) = 0⇒ 2x + 5 = 0⇒ 2x = -5⇒ x = −52-\dfrac{5}{2}−25​Hence, −52-\dfrac{5}{2}−25​ is a zero of polynomial 2x + 5(iv) p(x) = 3x - 2⇒ p(x) = 0⇒ 3x - 2 = 0⇒ 3x = 2⇒ x = 23\dfrac{2}{3}32​Hence, 23\dfrac{2}{3}32​ is a zero of polynomial 3x - 2(v) p(x) = 3x⇒ p(x) = 0⇒ 3x = 0⇒ x = 03\dfrac{0}{3}30​⇒ x = 0Hence, 0 is a zero of polynomial 3x(vi) p(x) = ax⇒ p(x) = 0⇒ ax = 0⇒ x = 0a\dfrac{0}{a}a0​⇒ x = 0Hence, 0 is a zero of polynomial ax(vii) p(x) = cx + d⇒ p(x) = 0⇒ cx + d = 0⇒ cx = -dx = −dc\dfrac{-d}{c}c−d​Hence, −dc-\dfrac{d}{c}−cd​ is a zero of polynomial cx + d

Question 10

Question 1Determine which of the following polynomials has (x + 1) a factor :(i) x3 + x2 + x + 1(ii) x4 + x3 + x2 + x + 1(iii) x4 + 3x3 + 3x2 + x + 1(iv) x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​

Accepted Answer

(i) x3 + x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x3 + x2 + x + 1p(-1) = (-1)3 + (-1)2 + (-1) + 1= -1 + 1 -1 + 1= 0Remainder is zero (0), so (x + 1) is factor of this polynomial.(ii) x4 + x3 + x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x4 + x3 + x2 + x + 1p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) + 1= 1 -1 + 1 -1 + 1= 1Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.(iii) x4 + 3x3 + 3x2 + x + 1⇒ x + 1 = 0⇒ x = -1p(x) = x4 + 3x3 + 3x2 + x + 1p(-1) = (-1)4 + 3 x (-1)3 + 3 x (-12) + (-1) + 1= 1 -3 + 3 -1 + 1= 1Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.(iv) x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​⇒ x + 1 = 0⇒ x = -1p(x) = x3 - x2 - (2 + 2\sqrt{2}2​)x + 2\sqrt{2}2​p(-1) = (-1)3 - (-1)2 - (2 + 2\sqrt{2}2​)(-1) + 2\sqrt{2}2​= -1 - 1 + 2 + 2\sqrt{2}2​ + 2\sqrt{2}2​= -2 + 2 + 2 2\sqrt{2}2​= 2 2\sqrt{2}2​Remainder is not zero (0), so (x + 1) is not a factor of this polynomial.

Question 11

Question 2Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:(i) p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2(iii) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3

Accepted Answer

(i) p(x) = 2x3 + x2 - 2x - 1g(x) = x + 1⇒ x + 1 = 0⇒ x = -1Putting x = -1 we get,p(-1) = 2(-1)3 + (-1)2 - 2(-1) - 1= -2 + 1 + 2 -1= 0Remainder is zero (0), so g(x) is factor of p(x).(ii) p(x) = x3 + 3x2 + 3x + 1g(x) = x + 2⇒ x + 2 = 0⇒ x = -2Putting x = -2 we get,p(-2) = (-2)3 + 3(-2)2 + 3(-2) + 1= -8 + 3(4) - 6 + 1= -8 + 12 - 6 + 1= 13 - 14= -1Remainder is not zero (0), so g(x) is not a factor of p(x).(iii) p(x) = x3 - 4x2 + x + 6g(x) = x - 3⇒ x - 3 = 0⇒ x = 3Putting x = -3 we get,p(3) = (3)3 - 4(3)2 + 3 + 6= 27 - 4(9) + 9= 27 - 36 + 9= 36 - 36= 0Remainder is zero (0), so g(x) is a factor of p(x).

Question 12

Question 3Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:(i) p(x) = x2 + x + k(ii) p(x) = 2x2 + kx + 2\sqrt{2}2​(iii) p(x) = kx2 - 2\sqrt{2}2​x + 1(iv) p(x) = kx2 - 3x + k

Accepted Answer

(i) p(x) = x2 + x + k⇒ x - 1 = 0⇒ x = 1Putting x = 1 we get,p(1) = (1)2 + 1 + k= 1 + 1 + k⇒ 2 + k = 0⇒ k = -2So the value of k = -2(ii) p(x) = 2x2 + kx + 2\sqrt{2}2​⇒ x - 1 = 0⇒ x = 1Putting x = 1 we get,p(1) = 2(1)2 + k(1) + 2\sqrt{2}2​⇒ 2 + k + 2\sqrt{2}2​⇒ k = -2 -2\sqrt{2}2​⇒ k = -(2 + 2\sqrt{2}2​)So the value of k = -(2 + 2\sqrt{2}2​)(iii) p(x) = kx2 - 2\sqrt{2}2​x + 1⇒ x - 1 = 0⇒ x = 1Putting x = 1 we get,p(1) = k(1)2 - 2\sqrt{2}2​(1) + 1⇒ k - 2\sqrt{2}2​ + 1 = 0⇒ k = 2\sqrt{2}2​ - 1So the value of k = 2\sqrt{2}2​ - 1(iv) p(x) = kx2 - 3x + k⇒ x - 1 = 0⇒ x = 1Putting x = 1 we get,p(1) = k(1)2 -3(1) + k= k - 3 + k⇒ 2k - 3 = 0⇒ 2k = 3⇒ k = 32\dfrac{3}{2}23​So the value of k = 32\dfrac{3}{2}23​

Question 13

Question 4Factorise :(i) 12x2 - 7x + 1(ii) 2x2 + 7x + 3(iii) 6x2 + 5x - 6(iv) 3x2 - x - 4

Accepted Answer

(i) 12x2 - 7x + 1= 12x2 - (4 + 3)x + 1= 12x2 -4x -3x + 1= 4x(3x - 1) -1(3x - 1)= (3x - 1)(4x - 1)Hence, 12x2 - 7x + 1 = (3x - 1)(4x - 1)(ii) 2x2 + 7x + 3= 2x2 + (6 + 1)x + 3= 2x2 + 6x + x + 3= 2x(x + 3) + 1(x + 3)= (2x + 1)(x + 3)Hence, 2x2 + 7x + 3 = (2x + 1)(x + 3)(iii) 6x2 + 5x - 6= 6x2 + (9 - 4)x - 6= 6x2 +9x - 4x - 6= 3x(2x + 3) -2(2x + 3)= (2x + 3)(3x - 2)Hence, 6x2 + 5x - 6 = (2x + 3)(3x - 2)(iv) 3x2 - x - 4= 3x2 - (4 - 3)x - 4= 3x2 - 4x + 3x - 4= x(3x - 4) + 1(3x - 4)= (3x - 4)(x + 1)Hence, 3x2 - x - 4 = (3x - 4)(x + 1)

Question 14

Question 5Factorise :(i) x3 - 2x2 - x + 2(ii) x3 - 3x2 - 9x - 5(iii) x3 + 13x2 + 32x + 20(iv) 2y3 + y2 - 2y - 1

Accepted Answer

(i) x3 - 2x2 - x + 2Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = 1x - 1 = 0x = 1p(x) = x3 - 2x2 - x + 2p(1) = (1)3 - 2 x (1)2 - 1 + 2= 1 - 2 - 1 + 2= 0Hence, (x - 1) is the factor of x3 - 2x2 - x + 2On dividing, x3 - 2x2 - x + 2 by (x - 1)x−1)x2−x−2x−1)x3−2x2−x+2‾x−1)2+−x3−+x2‾x−1x3−2−x2−x+2x−1)x3−2−+x2+−x‾x−1)x3−2x2(3)−2x+2x−1)x3−2x2(31)−+2x+−2‾x−1)x3−2x2(31)−2x×\begin{array}{l} \phantom{x - 1)}{\quad x^2 -x - 2} \ x - 1\overline{\smash{\big)}\quad x^3 - 2x^2 - x + 2} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{+}{-}x^2} \ \phantom{{x - 1}x^3-2}-x^2 - x + 2 \ \phantom{{x - 1)}x^3-2}\underline{\underset{+}{-}x^2 \underset{-}{+} x} \ \phantom{{x - 1)}{x^3-2x^{2}(3)}}-2x + 2 \ \phantom{{x - 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}2x \underset{-}{+} 2} \ \phantom{{x - 1)}{x^3-2x^{2}(31)}{-2x}}	imes \end{array}x−1)x2−x−2x−1)x3−2x2−x+2​x−1)2−+​x3+−​x2​x−1x3−2−x2−x+2x−1)x3−2+−​x2−+​x​x−1)x3−2x2(3)−2x+2x−1)x3−2x2(31)+−​2x−+​2​x−1)x3−2x2(31)−2x×​we get quotient = x2 - x - 2Factorising x2 - x - 2,= x2 - x - 2= x2 -(2 - 1)x - 2= x2 - 2x + x -2= x( x - 2) + 1(x + 2)= (x + 1)(x - 2)∴ x2 - x - 2 = (x + 1)(x - 2)∴ p(x) = x3 - 2x2 - x + 2 = (x - 1) (x + 1) (x - 2)So, factors are (x - 1) (x + 1) (x - 2)(ii) x3 - 3x2 - 9x - 5Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = -1x + 1 = 0x = -1p(x) = x3 - 3x2 - 9x - 5p(-1) = (-1)3 - 3 x (-1)2 - 9 x (-1) - 5= -1 - 3 + 9 - 5= -9 + 9= 0Hence, (x + 1) is a factor of x3 - 3x2 - 9x - 5On dividing, x3 - 3x2 - 9x - 5 by (x + 1)x+1)x2−4x−5x+1)x3−3x2−9x−5‾x−1)2+−x3+−x2‾x+1x3−2−4x2−9xx+1)x3−2−+4x2−+4x‾x+1)x3−2x2(3)−5x−5x+1)x3−2x2(31)−+5x−+5‾x+1)x3−2x2(31)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 -4x - 5} \ x + 1\overline{\smash{\big)}\quad x^3 - 3x^2 - 9x - 5} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}-4x^2 - 9x \ \phantom{{x + 1)}x^3-2}\underline{\underset{+}{-}4x^2 \underset{+}{-} 4x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}-5x - 5 \ \phantom{{x + 1)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}5x \underset{+}{-} 5} \ \phantom{{x + 1)}{x^3-2x^{2}(31)}{-2}}	imes \end{array}x+1)x2−4x−5x+1)x3−3x2−9x−5​x−1)2−+​x3−+​x2​x+1x3−2−4x2−9xx+1)x3−2+−​4x2+−​4x​x+1)x3−2x2(3)−5x−5x+1)x3−2x2(31)+−​5x+−​5​x+1)x3−2x2(31)−2×​We get quotient = x2 - 4x - 5Factorising x2 - 4x - 5x2 -(5 - 1)x - 5x2 -5x + x - 5x(x - 5) + 1(x - 5)(x + 1) (x - 5)∴ x2 - 4x - 5 = (x + 1) (x - 5)∴ p(x) = x3 - 3x2 -9x - 5 = (x + 1) (x + 1) (x - 5)So, factors are (x + 1) (x + 1) (x - 5)(iii) x3 + 13x2 + 32x + 20Let (x - a) is a factor, a = 1, -1, 2, -2.......Putting a = -1x + 1 = 0x = -1p(x) = x3 + 13x2 + 32x + 20p(-1) = (-1)3 + 13 x (-1)2 + 32 x (-1) + 20= -1 + 13 - 32 + 20= -33 + 33= 0Hence, (x + 1) is a factor of x3 + 13x2 + 32x + 20On dividing, x3 + 13x2 + 32x + 20 by (x + 1)x+1)x2+12x+20x+1)x3+13x2+32x+20‾x−1)2+−x3+−x2‾x+1x3−212x2+32x+20x+1)x31+−12x2+−12x‾x+1)x3−2x2(3)20x+20x+1)x3−2x23+−20x+−20‾x+1)x3−2x2(3)−2×\begin{array}{l} \phantom{x + 1)}{\quad x^2 +12x + 20} \ x + 1\overline{\smash{\big)}\quad x^3 + 13x^2 + 32x + 20} \ \phantom{x - 1)}\phantom{2}\underline{\underset{-}{+}x^3 \underset{-}{+}x^2} \ \phantom{{x + 1}x^3-2}12x^2 + 32x + 20 \ \phantom{{x + 1)}x^31}\underline{\underset{-}{+}12x^2 \underset{-}{+} 12x} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}}20x + 20 \ \phantom{{x + 1)}{x^3-2x^{2}3}}\underline{\underset{-}{+}20x \underset{-}{+} 20} \ \phantom{{x + 1)}{x^3-2x^{2}(3)}{-2}}	imes \end{array}x+1)x2+12x+20x+1)x3+13x2+32x+20​x−1)2−+​x3−+​x2​x+1x3−212x2+32x+20x+1)x31−+​12x2−+​12x​x+1)x3−2x2(3)20x+20x+1)x3−2x23−+​20x−+​20​x+1)x3−2x2(3)−2×​We get quotient = x2 + 12x + 20Factorising x2 + 12x + 20x2 + 10x + 2x + 20x(x + 10) + 2(x + 10)(x + 10)(x + 2)∴ x2 + 12x + 20 = (x + 10)(x + 2)∴ p(x) = x3 + 13x2 + 32x + 20 = (x + 1) (x + 10) (x + 2)So, factors are (x + 1) (x + 10) (x + 2)(iv) 2y3 + y2 - 2y - 1Let (y - a) is a factor, a = 1, -1, 2, -2.......Putting a = 1y - 1 = 0y = 1p(y) = 2y3 + y2 - 2y - 1p(1) = 2 x (1)3 + (1)2 - 2 x 1 - 1= 2 + 1 - 2 - 1= 0Hence, (y - 1) is a factor of 2y3 + y2 - 2y - 1On dividing, 2y3 + y2 - 2y - 1 by (y - 1)y−1)2y2+3y+1y−1)2y3+y2−2y−1‾y−1)+−2y3−+2y2‾y−1x3−2333y2−2yy−1)x3−2+−3y2−+3y‾y−1)x3−2x2(3333)y−1y−1)x3−2x23333+−y−+1‾y−1)x3−2x2(333)−2×\begin{array}{l} \phantom{y - 1)}{\quad 2y^2 + 3y + 1} \ y - 1\overline{\smash{\big)}\quad 2y^3 + y^2 - 2y - 1} \ \phantom{y - 1)}\underline{\underset{-}{+}2y^3 \underset{+}{-}2y^2} \ \phantom{{y - 1}x^3-233}3y^2 - 2y \ \phantom{{y - 1)}x^3-2}\underline{\underset{-}{+}3y^2 \underset{+}{-} 3y} \ \phantom{{y - 1)}{x^3-2x^{2}(3333)}}y - 1 \ \phantom{{y - 1)}{x^3-2x^{2}3333}}\underline{\underset{-}{+}y \underset{+}{-} 1} \ \phantom{{y - 1)}{x^3-2x^{2}(333)}{-2}}	imes \end{array}y−1)2y2+3y+1y−1)2y3+y2−2y−1​y−1)−+​2y3+−​2y2​y−1x3−2333y2−2yy−1)x3−2−+​3y2+−​3y​y−1)x3−2x2(3333)y−1y−1)x3−2x23333−+​y+−​1​y−1)x3−2x2(333)−2×​We get quotient = 2y2 + 3y + 1Factorising 2y2 + 3y + 12y2 + 2y + y +12y(y + 1) + 1(y + 1)(y + 1)(2y + 1)∴ 2y2 + 3y + 1 = (y + 1)(2y + 1)∴ p(y) = 2y3 + y2 - 2y - 1 = (y - 1)(y + 1)(2y + 1)So, factors are (y - 1)(y + 1)(2y + 1)

Question 15

Question 1Use suitable identities to find the following products:(i) (x + 4)(x + 10)(ii) (x + 8)(x - 10)(iii) (3x + 4)(3x - 5)(iv) (y2 + 32\dfrac{3}{2}23​) (y2 - 32\dfrac{3}{2}23​)(v) (3 - 2x)(3 + 2x)

Accepted Answer

(i) (x + 4)(x + 10)[∵ (x + a)(x + b) = x2 + (a + b)x + ab]Putting a = 4, b = 10= x2 + (4 + 10)x + 4 x 10= x2 + 14x + 40Hence, (x + 4)(x + 10) = x2 + 14x + 40(ii) (x + 8)(x - 10)[∵ (x + a)(x + b) = x2 + (a + b)x + ab]Putting a = 8 , b = -10= x2 + [8 + (-10)]x + 8 x (-10)= x2 - 2x - 80Hence, (x + 8)(x - 10) = x2 - 2x - 80(iii) (3x + 4)(3x - 5)[∵ (x + a)(x + b) = x2 + (a + b)x + ab]Putting a = 4, b = -5, x = 3x= (3x)2 + [4 + (-5)](3x) + (4)(-5)= 9x2 + (-1)(3x) - 20= 9x2 - 3x - 20Hence, (3x + 4) (3x - 5) = 9x2 - 3x - 20(iv) (y2 + 32\dfrac{3}{2}23​)(y2 - 32\dfrac{3}{2}23​)[∵ (a + b)(a - b) = a2 - b2]Putting a = y2, b = 32\dfrac{3}{2}23​= (y2)2 - (32\dfrac{3}{2}23​)2= y4 - 94\dfrac{9}{4}49​Hence, (y2 + 32\dfrac{3}{2}23​) (y2 - 32\dfrac{3}{2}23​) = y4 - 94\dfrac{9}{4}49​(v) (3 - 2x)(3 + 2x)[∵ (a - b)(a + b) = a2 - b2]Putting a = 3, b = 2x= (3)2 - (2x)2= 9 - 4x2Hence, (3 - 2x)(3 + 2x) = 9 - 4x2

Question 16

Question 2Evaluate the following products without multiplying directly:(i) 103 × 107(ii) 95 × 96(iii) 104 × 96

Accepted Answer

(i) 103 × 107We can write it as,= (100 + 3)(100 + 7)[∵ (x + a)(x + b) = x2 + (a + b)x + ab]Putting x = 100, a = 3, b = 7= (100)2 + (3 + 7)(100) + (3)(7)= 10000 + (10)(100) + 21= 10000 + 1000 + 21= 11021Hence, 103 × 107 = 11021(ii) 95 × 96We can write it as,[100 + (-5)][100 + (-4)][∵ (x + a)(x + b) = x2 + (a + b)x + ab]Putting x = 100, a = -5 , b = -4= (100)2 - [(-5) + (-4)](100) + (-5) x (-4)= 10000 - (9)(100) + 20= 10000 - 900 + 20= 10020 - 900= 9120Hence, 95 × 96 = 9120(iii) 104 × 96We can write it as,(100 + 4)(100 - 4)[∵ (a + b)(a - b) = a2 - b2]Putting a = 100, b = 4= (100)2 - (4)2= 10000 - 16= 9984Hence, 104 × 96 = 9984

Question 17

Question 3Factorise the following using appropriate identities:(i) 9x2 + 6xy + y2(ii) 4y2 - 4y + 1(iii) x2−(y2100)x^2 - \Big(\dfrac{y^2}{100}\Big)x2−(100y2​)

Accepted Answer

(i) 9x2 + 6xy + y2[∵ a2 + 2ab + b2 = (a + b)2]= (3x)2 + 2(3x)(y) + (y)2= (3x + y)2Hence, 9x2 + 6xy + y2 = (3x + y)(3x + y)(ii) 4y2 - 4y + 1[∵ a2 - 2ab + b2 = (a - b)2]= (2y)2 - 2(2y)(1) + (1)2= (2y - 1)2Hence, 4y2 - 4y + 1 = (2y - 1)(2y - 1)(iii) x2−(y2100)x^2 - \Big(\dfrac{y^2}{100}\Big)x2−(100y2​)[∵ a2 - b2 = (a - b)(a + b)]x2−(y2100)=x2−(y10)2=(x−y10)(x+y10)x^2 - \Big(\dfrac{y^2}{100}\Big) \[1em] = x^2 - \Big(\dfrac{y}{10}\Big)^2 \[1em] = \Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)x2−(100y2​)=x2−(10y​)2=(x−10y​)(x+10y​)Hence, x2−(y2100)x^2 - \Big(\dfrac{y^2}{100}\Big)x2−(100y2​) = (x−y10)(x+y10)\Big(x - \dfrac{y}{10}\Big) \Big(x + \dfrac{y}{10}\Big)(x−10y​)(x+10y​)

Question 18

Question 4Expand each of the following, using suitable identities:(i) (x + 2y + 4z)2(ii) (2x - y + z)2(iii) (-2x + 3y + 2z)2(iv) (3a - 7b - c)2(v) (-2x + 5y - 3z)2(vi) [14a−12b+1]2\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2[41​a−21​b+1]2

Accepted Answer

(i) (x + 2y + 4z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = x, y = 2y, z = 4z= (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx(ii) (2x - y + z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = 2x, y = (-y), z = z= (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)= 4x2 + y2 + z2 - 4xy - 2yz + 4zx(iii) (-2x + 3y + 2z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = -2x, y = 3y, z = 2z= (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8zx(iv) (3a - 7b - c)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = 3a, y = (-7b), z = (-c)= (3a)2 + (-7b)2 + (-c)2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)= 9a2 + 49b2 + c2 - 42ab + 14bc - 6ac(v) (-2x + 5y - 3z)2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = -2x, y = 5y, z = (-3z)= (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12zx(vi) [14a−12b+1]2\Big[\dfrac{1}{4}a - \dfrac{1}{2}b + 1\Big]^2[41​a−21​b+1]2[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Putting x = a4,y=−b2\dfrac{a}{4}, y = -\dfrac{b}{2}4a​,y=−2b​, z = 1=(a4)2+(−b2)2+(1)2+2(a4)(−b2)+2(−b2)(1)+2(1)(a4)=a216+b24+1−ab4−b+a2= \Big(\dfrac{a}{4}\Big)^2 + \Big(-\dfrac{b}{2}\Big)^2 + (1)^2 + 2\Big(\dfrac{a}{4}\Big) \Big(-\dfrac{b}{2}\Big) + 2\Big(-\dfrac{b}{2}\Big)(1) + 2(1) \Big(\dfrac{a}{4}\Big) \[1em] = \dfrac{a^2}{16} + \dfrac{b^2}{4} + 1 - \dfrac{ab}{4} -b + \dfrac{a}{2}=(4a​)2+(−2b​)2+(1)2+2(4a​)(−2b​)+2(−2b​)(1)+2(1)(4a​)=16a2​+4b2​+1−4ab​−b+2a​

Question 19

Question 5Factorise:(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz(ii) 2x2 + y2 + 8z2 - 222\sqrt{2}22​xy + 424\sqrt{2}42​yz - 8xz

Accepted Answer

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)= (2x + 3y - 4z)2 [∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]Hence, 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz = (2x + 3y - 4z)(2x + 3y - 4z)(ii) 2x2 + y2 + 8z2 - 222\sqrt{2}22​xy + 424\sqrt{2}42​yz - 8xz[∵ (a + b + c)2 = (a)2 + (b)2 + (c)2 + 2ab + 2bc + 2ca]=(−2x)2+(y)2+(22z)2+2(−2x)(y)+2(y)(22z)+2(22z)(−2x)=(−2x+y+22z)2= (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x) \[1em] = (-\sqrt{2}x + y + 2\sqrt{2}z)^2=(−2​x)2+(y)2+(22​z)2+2(−2​x)(y)+2(y)(22​z)+2(22​z)(−2​x)=(−2​x+y+22​z)2Hence, 2x2 + y2 + 8z2 -222\sqrt{2}22​xy + 424\sqrt{2}42​yz - 8zx = (-2\sqrt{2}2​x + y + 222\sqrt{2}22​z)(-2\sqrt{2}2​x + y + 222\sqrt{2}22​z)

Question 20

Question 6Write the following cubes in expanded form:(i) (2x + 1)3(ii) (2a - 3b)3(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3[23x​+1]3(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3[x−32​y]3

Accepted Answer

(i) (2x + 1)3We know that:(a + b)3= (a)3 + (b)3 + 3a2b + 3ab2Putting a = 2x, b = 1= (2x)3 + (1)3 + 3(2x)2(1) + 3(2x)(1)2= 8x3 + 12x2 + 6x + 1Hence, (2x + 1)3 = 8x3 + 12x2 + 6x + 1(ii) (2a - 3b)3We know that:(a - b)3= (a)3 - (b)3 - 3a2b + 3ab2Putting a = 2a, b = 3b= (2a)3 - (3b)3 - 3(2a)2(3b) + 3(2a)(3b)2= 8a3 - 27b3 - 3(4a2)(3b) + 3(2a)(9b2)= 8a3 - 27b3 - 36a2b + 54ab2Hence, (2a - 3b)3 = 8a3 - 27b3 - 36a2b + 54ab2(iii) [3x2+1]3\Big[\dfrac{3x}{2} + 1\Big]^3[23x​+1]3We know that:(a + b)3 = (a)3 + (b)3 + 3ab(a + b)Putting a = 32x\dfrac{3}{2}x23​x, b = 1=(32)3+(1)3+3(3x2)(1)(3x2+1)=27x38+1+(9x2)(3x2+1)=27x38+1+(9x2)(3x2)+9x2(1)=27x38+27x24+9x2+1= \Big(\dfrac{3}{2}\Big)^3 + (1)^3 + 3 \Big(\dfrac{3x}{2}\Big)(1) \Big(\dfrac{3x}{2} + 1\Big) \[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2} + 1\Big) \[1em] = \dfrac{27x^3}{8} + 1 + \Big(\dfrac{9x}{2}\Big) \Big(\dfrac{3x}{2}\Big) + \dfrac{9x}{2}(1)\[1em] = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1=(23​)3+(1)3+3(23x​)(1)(23x​+1)=827x3​+1+(29x​)(23x​+1)=827x3​+1+(29x​)(23x​)+29x​(1)=827x3​+427x2​+29x​+1Hence, [3x2+1]3=27x38+27x24+9x2+1\Big[\dfrac{3x}{2} + 1\Big]^3 = \dfrac{27x^3}{8} + \dfrac{27x^2}{4} + \dfrac{9x}{2} + 1[23x​+1]3=827x3​+427x2​+29x​+1(iv) [x−23y]3\Big[x - \dfrac{2}{3}y\Big]^3[x−32​y]3We know that:(a - b)3 = (a)3 - (b)3 - 3ab(a - b)Putting a = x, b = −23y-\dfrac{2}{3}y−32​y=(x)3−[−23y]3−3(x)[−23y][x−23y]=x3−827y3−2xy[x−23y]=x3−827y3−2x2y+43xy2= (x)^3 - \Big[-\dfrac{2}{3}y\Big]^3 - 3(x)\Big[-\dfrac{2}{3}y\Big] \Big[x - \dfrac{2}{3}y\Big]\[1em] = x^3 - \dfrac{8}{27}y^3 - 2xy\Big[x - \dfrac{2}{3}y\Big]\[1em] = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2=(x)3−[−32​y]3−3(x)[−32​y][x−32​y]=x3−278​y3−2xy[x−32​y]=x3−278​y3−2x2y+34​xy2Hence, [x−23y]3=x3−827y3−2x2y+43xy2\Big[x - \dfrac{2}{3}y\Big]^3 = x^3 - \dfrac{8}{27}y^3 - 2x^2y + \dfrac{4}{3}xy^2[x−32​y]3=x3−278​y3−2x2y+34​xy2

Polynomials

Exercise 21

Exercise 22

Exercise 23

Exercise 24