2

Q1. Which of the following polynomials has -3 as a zero?

• 1) x² + 3
• 2) (x - 3)
• 3) x² - 9
• 4) x² - 3x

Solution

Let p(x) = x² - 9 Therefore, p(-3) = 9 - 9 = 0. So -3 is the zero of p(x).

Q2. Factorize :

Solution

Q3. Simplify: .

Solution

Q4. Zero of the polynomial p(x) where p (x) = ax, a 0 is :

• 1) 1
• 2) a
• 3)
• 4) 0

Solution

Since, p(0) = a x 0 = 0 Therefore, x = 0 is the zero.

Q5. Factorise : (x² - 2x)² - 2 (x² - 2x) - 3.

Solution

(x² - 2x)² - 2(x² - 2x) - 3 Put y = (x² - 2x) (x² - 2x)² - 2(x² - 2x) -3 = y² - 2y - 3 = y² - 3y + y - 3 = y(y - 3) + 1(y - 3) = (y + 1) (y - 3) = (x² - 2x+ 1) (x² - 2x - 3) = (x - 1)² (x² - 3x + x - 3) = (x - 1)² (x (x- 3) + 1 (x - 3)) = (x - 1)² (x - 3) (x + 1)

Q6.

Solution

Q7. P(x) = q(x) × g(x) + r(x);  

• 1) deg[q(x)] > deg[p(x)]
• 2) None of these
• 3) deg[q(x)] = deg[p(x)]
• 4) deg[q(x)] < deg[p(x)]

Solution

P(x) = q(x) × g(x) + r(x) r(x) is either zero, a constant or a polynomial whose degree is less than q(x). When a polynomial f(x) is divided by a linear polynomial (x - a), then the remainder is f(a).

Q8. In y⁶ − 10y⁴ − 12y³ + 1, the coefficient of y³ is 

• 1) −12
• 2) −1
• 3) −10
• 4) 1

Solution

In y⁶ − 10y⁴ − 12y³ + 1, the coefficient of y³ is −12. 

Q9. Which of the following is a trinomial in x?

• 1) x³ + 1
• 2) x³ + x² + x
• 3) x³ + 2x
• 4)

Solution

is trinomial in x.

Q10. Divide p(x) = x⁴ − 3x² + 2x + 5 by (x − 1) using the long division method. Find the remainder. 

• 1) 5
• 2) 7
• 3) 3
• 4) 6

Solution

f(x) = x⁴ + 0x³ − 3x² + 2x + 5 Long division method:   

Q11. In p(x) = 0, r is ___ of a polynomial p(x) if p(r) = 0. 

• 1) a root
• 2) the solution
• 3) zero
• 4) All of these

Solution

In p(x) = 0, r is a root, zero or the solution of a polynomial p(x) if p(r) = 0. 

Q12. Prove that x + 3 is a multiple of q(x) = x³ + 6x² + 11x + 6 using Remainder theorem.

Solution

Q13. If p(x) = 7 - 3x + 2x² then value of p(-2) is:

• 1) 22
• 2) 21
• 3) 31
• 4) 12

Solution

p(x) = 7 - 3x + 2x² p(-2) = 7 + 6 + 8 = 21. 21

Q14. On dividing f(x) = 2x⁴ − 9x³ − 21x² + 88x + 48 by (x − 2), we get the remainder 

• 1) 100
• 2) 150
• 3) 50
• 4) 75

Solution

f(x) = 2x⁴ − 9x³ − 21x² + 88x + 48 We have (x − 2) = 0 ∴ x = 2 By the remainder theorem, ∴ f(2) = 2(2)⁴ - 9(2)³ − 21(2)² + 88(2) + 48  = 32 - 72 - 84 + 176 + 48 = 100

Q15. Factorise : (i) (8b³ - 1) (ii) (p³q³ + 343)

Solution

(i) (8b³ - 1) = [(2b)³ - 1] = (2b - 1) [(2b)² + 2b 1 + 1²) = (2b - 1) (4b² + 2b + 1) (ii) (p³q³ + 343) = [(pq)³ + 7³] = (pq + 7) ((pq)² - pq7 + 7²) = (pq + 7) (p²q² - 7pq + 49)

Q16. q(x) is a multiple of (x + a); if on dividing q(x) by (x + a), the remainder is 

• 1) variable
• 2) one
• 3) zero
• 4) constant

Solution

q(x) is a multiple of (x + a); if on dividing q(x) by (x + a), the remainder is zero. 

Q17. Degree of which of the following polynomial is zero?

• 1) 15
• 2) _{x + y}
• 3) x
• 4) y

Solution

Degree of the polynomial 15 is zero.

Q18. The factors of (2a - b)³ + (b - 2c)³ + 8(c - a)³ are

• 1) 2a x b x 2c
• 2) (2a - b)(b - 2c)(c - a)
• 3) 3(2a - b)(b - 2c)(c - a)
• 4) 6(2a - b)(b - 2c)(c - a)

Solution

(2a - b)³ + (b - 2c)³ + 8(c - a)³ = (2a - b)³ + (b - 2c)³ + (2c - 2a)³ Since (2a - b) + (b - 2c) + (2c - 2a) = 0 Therefore, (2a - b)³ + (b - 2c)³ + (2c - 2a)³ = 3(2a - b)(b - 2c)(2c - 2a) = 6(2a - b)(b - 2c)(c - a)

Q19. Which of the following is a binomial in y?

• 1)
• 2) y² +
• 3) y +
• 4)

Solution

y² +  _{is a binomial in y.}

Q20. In −1x + 2y = z, what are the variables? 

• 1) −1, 2, −1
• 2) x, y, z
• 3) −1, 2, 1
• 4) −1, x, y

Solution

A variable is a symbol which takes on numerical values. ∴ x, y, z are variables.

Q21. If x + y + z = 1, xy + yz + zx = -1 and xyz = -1 then find the value of x³ + y³ + z³.

Solution

Since (x + y + z)² = x² + y² + z² - 2xy - 2yz - 2zx Or, 1 = x² + y² + z² - 2 x -1 Or, 1 = x² + y² + z² + 2 Or, x² + y² + z² = -1 Now, x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx) x³ + y³ + z³ = 3xyz + (x + y + z) (x² + y² + z² - xy - yz - zx) = -3 + 1(-1 + 1) x³ + y³ + z³ = -3.

Q22. A coefficient is a constant in the term of a 

• 1) constant
• 2) alphabet
• 3) None of these
• 4) Variable

Solution

A coefficient is a constant in the term of a polynomial.

Q23.

Solution

Q24. What is the remainder when x⁵ + 5x⁴ + 9x³ + 11x² + 12x + 13 is divided by x + 2? (Use the remainder theorem) 

• 1) 6
• 2) 9
• 3) 12
• 4) 3

Solution

P(x) = x⁵ + 5x⁴ + 9x³ + 11x² + 12x + 13 x + 2 = 0 x = −2 P(−2) = (−2)⁵ + 5(−2)⁴ + 9(−2)³ + 11(−2)² + 12(−2) + 13  = −32 + 80 - 72 + 44 - 24 + 13 = 9

Q25. Zero of the polynomial p(x) = cx + d is:

• 1)
• 2) -c
• 3)
• 4) -d

Solution

Q26. Find the remainder when p(x) = 2x³ - 13x² + 17x + 12 is divided by (x - 2). 

• 1) 5
• 2) -10
• 3) 15
• 4) 10

Solution

f(x) = 2x³ - 13x² + 17x + 12 We have (x − 2) = 0 ∴ x = 2 By the remainder theorem, ∴ f(2) = 2(2)³ − 13(2)² + 17(2) + 12  = 10

Q27. Which of the following is a polynomial in one variable?

• 1)
• 2) 3 - x² + x
• 3)
• 4) x³ + y³ + 7

Solution

3 - x² + x is a polynomial in one variable

Q28. (a) if p(y) = 4 + 3y - y² + 5y³, find p(2).(b) Check whether (y + 2) is a factor of p(y).(c) Find the remainder when p(y) is divided by .

Solution

(a) Since p(y) = 4 + 3y - y² + 5y³ Therefore, p(2) = 4 + 3(2) - (2)² + 5(2)³ = 46 (b) p(-2) = 4 + 3(-2) - (-2)² + 5(-2)³ = -46 0 (y + 2) is not a factor of p(y) (c) Required remainder = p  

Q29. A monomial in a is a single term of the form axⁿ, a is a real number and n is 

• 1) a whole number
• 2) a natural number
• 3) an integer
• 4) a real number

Solution

A monomial in a is a single term of the form axⁿ, a is a real number and n is a whole number.

Q30. If, then the remainder is 

• 1)
• 2)
• 3)
• 4)

Solution

If, then ax − b = 0, ∴ Remainder =  

Q31. If 1875 is divided by 12, then what is the remainder? 

• 1) 4
• 2) 3
• 3) 1
• 4) 2

Solution

1875 = (12 × 156) + 3 …Dividend = (Divisor × Quotient) + Remainder 

Q32. If a - b = 7, a² + b² = 85 find a³ - b³.

Solution

a - b = 7 a² + b² = (a - b)² + 2ab 85 = 49 + 2ab 2ab = 36 ab = 18 a³ - b³ = (a - b)³ + 3ab (a - b) = 343 + 54 7 = 343 + 378 = 721

Q33. What is the remainder when 2x² − 5x − 1 is divided by x −3? (Use the remainder theorem) 

• 1) 1
• 2) 4
• 3) 3
• 4) 2

Solution

P(x) = 2x² − 5x − 1 x − 3 = 0 x = 3 P(3) = 2(3)² - 5(3) − 1 = 18 - 15 − 1 = 2

Q34. If, the value of x³ - y³ is:

• 1) 1
• 2) -1
• 3) 0
• 4)

Solution

Q35. Prove that = 1.

Solution

= a + b = 0.75 + 0.25 = 1

Q36. Prove that: 2x³ + 2y³ + 2z³ - 6xyz = (x + y + z) [(x - y)² + (y - z)² + (z - x)² hence evaluate 2(7)³ + 2(9)³ + 2(13)³ - 6(7) (9) (13).

Solution

Now, put x = 7, y = 9 and z = 13, we get

Q37.   In the following problem use the factor theorem to find if g(x) is a factor of p(x): p(x) = x³ + x² + 3x + 175 and g(x) = x + 5

Solution

Q38. In a polynomial, none of the variables has a _____power. 

• 1) None of these
• 2) positive
• 3) zero
• 4) negative

Solution

In a polynomial, none of the variables has a negative power.

Q39. Factorize : 3x² - x - 4

Solution

3x² - x - 4 = 3x² + 3x - 4x - 4 = 3x (x + 1) -4 (x + 1) = (x + 1) (3x - 4)

Q40. Evaluate using suitable identity (999)³.

Solution

Q41. In 3z³ - 9x⁶ - 6y⁴ + z, the degree of the polynomial is 

• 1) 6
• 2) 4
• 3) 1
• 4) 3

Solution

The degree of a polynomial is the degree of the leading term. In 3z³ - 9x⁶ - 6y⁴ + z, the leading term is -9x⁶. ∴ The degree of the polynomial is 6.

Q42. Factorise:

Solution

= (3p)³ - =

Q43. Find the remainder when 3x³ - 4x² + 7x - 5 is divided by (x - 3) and (x + 3).

Solution

p(x) = 3x³ - 4x² + 7x - 5 Remainder when (x - 3) divides p(x) is = p(3) = 3 3³ - 4 3² + 7 3 - 5 = 3 27 - 4 9 + 21 - 5 = 61 Remainder when (x + 3) divides p(x) is = p(-3) = 3 (-3)³ - 4(-3)² + 7 (-3) - 5 = 3 (-27) - 4 9 - 21 - 5 = -143 _{Therefore, the required remainders are 61 and -143.}

Q44. Which of the following is a polynomial? 

• 1)
• 2) 2x + 3y + 5z
• 3) x⁻² + 2
• 4)

Solution

An algebraic expression in which the variable has non-negative integral exponents only is a polynomial. ∴ 2x + 3y + 5z is a polynomial.

Q45. Degree of the polynomial p(x) = 4x⁴ + 2x³ + x⁵ + 2x + 7 is:

• 1) 4
• 2) 5
• 3) 7
• 4) 3

Solution

The exponent of the highest degree term in a polynomial is known as its degree. Here, highest degree of a term = 5.Therefore,degree ofpolynomial=5

Q46. If (x + y + z) = 0, then prove that (x³ + y³ + z³) = 3xyz.

Solution

x+y+z = 0 x+y = -z (x+y)³= (-z)³ x³+y³+3xy(x+y)= -z³ x³+y³+3xy(-z)= -z³ [x+y= -z] x³+y³-3xyz= -z³ x³+y³+z³= 3xyz Hence, x+y+z= 0 x³+y³+z³= 3xyz

Q47. Factorize : 8 - 27x³ - 36x + 54x²

Solution

8 - 27x³ - 36x + 54x² = (2)³ + (-3x)³ + 3(2)²(-3x) + 3(2)( - 3x)² = (2 - 3x)³ = (2 - 3x) (2 - 3x) (2 - 3x)

Q48. Factorize: x³ + 6x² + 11x + 6.

Solution

Q49. Factorize: .

Solution

Q50. What is the remainder when 5x³ - x² + 6 is divided by x − 4?

• 1) 562
• 2) 236
• 3) 310
• 4) 378

Solution

P(x) = 5x³ - x² + 6 x − 4 = 0 x = 4 P(4) = 5(4)³ - 4² + 6 = 310

Q51. If a polynomial f(x) is divided by x - a then remainder is

• 1) f (-a)
• 2) f(a)-f(0)
• 3) f (0)
• 4) f (a)

Solution

By remainder theorem the remainder is f(a).

Q52. The remainder obtained when the polynomial p(x) is divided by (b - ax) is:

• 1)
• 2)
• 3)
• 4)

Solution

Taking, b - ax = 0 we get x= b/a, therefore remainder is:

Q53. Factorise: 2x³ + 7x² - 3x - 18.

Solution

p(x) = 2x³ + 7x² - 3x - 18 consider, p(1) = 2 + 7 - 3 - 18 0 p(-1) = -2 + 7 + 3 - 18 0 p(2) = 16 + 28 - 6 - 18 0 p(-2) = -16 + 28 + 6 - 18 = 0 (x + 2) is a factor of p(x). p(x) = (x + 2) (2x² + 3x - 9) = (x + 2) (2x² + 6x - 3x - 9) = (x + 2) (2x (x + 3) - 3 (x + 3) = (x + 2) (x + 3) (2x - 3)

Q54. A polynomial with 3 terms is called a 

• 1) Binomial
• 2) All of these
• 3) Monomial
• 4) Trinomial

Solution

A polynomial with 3 terms is called a trinomial. 

Q55. Factorise x² -1 -2a - a²

Solution

Q56. If the polynomial p(x) = x³ - 3x² + ax - b is such that when divided by (x - 2 ) and (x + 2), the remainders are 21 and 1 respectively. Find the values of a and b.

Solution

Q57. One of the factors of (16y² - 1) + (1 - 4y)² is

• 1) 8y
• 2) (4y + 1)
• 3) (4 + y)
• 4) (4 - y)

Solution

Q58. What is remainder when x³ - 2x² + x + 1 is divided by (x -1)?

• 1) 1
• 2) 2
• 3) 0
• 4) -1

Solution

Remainder = 1 - 2 + 1 + 1 = 1.

Q59. If (x - 2) is a factor of the polynomial x³ - 2ax² + ax - 1, find the value of 'a'.

Solution

Q60. If x⁵¹ + 51 is divided by (x + 1) the remainder is:

• 1) 50
• 2) 49
• 3) 1
• 4) 0

Solution

If f(x) = x⁵¹ + 51 is divided by (x + 1) the remainder is given by: f(-1) = (-1)⁵¹ + 51 = -1 + 51 = 50.

Q61. The coefficient of x² in (3x + x³) is:

• 1) 1
• 2) 4
• 3) 3
• 4) 2

Solution

Hence the coefficient of x² is 4.

Q62. The degree of a non-zero constant polynomial is always 

• 1) 2
• 2) 0
• 3) −1
• 4) 1

Solution

The degree of a non-zero constant polynomial is always zero.

Q63. Degree of zero polynomial is:

• 1) 0
• 2) 1
• 3) Any natural number
• 4) Not defined

Solution

Degree of zero polynomial is not defined.

Q64. Factorise: (x² - 2x)² - 11 (x² - 2x) + 24

Solution

Put x² - 2x = y So, (x² - 2x)² - 11 (x² - 2x) + 24 = y² - 11y + 24 = y² - 8y - 3y + 24 = y(y - 8) - 3(y - 8) = (y - 8) (y - 3) = (x² - 2x - 8)(x² - 2x - 3) = (x² - 4x + 2x - 8)(x² - 3x + x - 3) = (x + 2) (x - 4) (x - 3) (x + 1)

Q65. Factorize : (x - y)³ + (y - z)³ + (z - x)³

Solution

Let a = x - y, b = y - z, c = z - x Here, a + b + c = x - y + y - z + z - x = 0 Now if a + b + c = 0 then x³ + y³ + z³ = 3xyz Hence, (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y) (y - z) (z - x).

Q66. Find the remainder when f(x) = 2x² - 11x − 20 is divided by (2x + 3). 

• 1) 2
• 2) 0
• 3) 1
• 4) -4

Solution

2x + 3 = 0 ∴ x = We have f(x) = 2x² - 11x − 20  

Q67. Factorise : a⁷b +ab⁷

Solution

Q68. Evaluate (999)³.

Solution

(999)³ = (1000 - 1)³ (a - b)³ = a³ - b³ - 3a²b + 3ab² (999)³ = (1000)³ - (1)³ - 3 (1000)² 1 + 3 1000 1² = 1000000000 - 1 - 3000000 + 3000 = 997000000 + 3000 - 1 = 997002999

Q69. If p(x) = then p (-1) is:

• 1)
• 2)
• 3)
• 4)

Solution

Q70. The value of p for which x + p is a factor of x² + px + 3 - p is:

• 1) 1
• 2) 3
• 3) -1
• 4) -3

Solution

(x + p) is a factor if (-p)² + p(-p) +3 - p =0

Q71. On dividing f(x) = x³ − 4x² + 2x − 3 by (x + 2), we get the remainder 

• 1) 14
• 2) 29
• 3) -17
• 4) -31

Solution

f(x) = x³ − 4x² + 2x - 3 We have (x + 2) = 0 ∴ x = −2 By the remainder theorem, ∴ f(−2) = (−2)³ - 4(−2)² + 2(−2) - 5 = −8 - 16 - 4 − 3= −31

Q72. If a + b = 10 and a² + b² = 58, find the value of a³ + b³.

Solution

Q73. Prove that x² + 6x + 15 has no zero.

Solution

Q74. Without actually calculating the cubes, find the value of 75³ - 25³ - 50³.

Solution

Let x = 75, y = -25, z = -50 x + y + z = 75 - 25 - 50 = 0 We know, if x + y + z = 0 then x³ + y³ + z³ = 3xyz 75³ - 25³ - 50³ = 3(75) (-25) (-50) = 281250

Q75.

Solution

Q76. Which of the following is a multiple of x³ + 7x² + 7x − 6? 

• 1) x + 3
• 2) x + 2
• 3) x - 3
• 4) x - 2

Solution

q(x) = x³ + 7x² + 7x − 6 x + 2 = 0 ∴ x = −2 q(−2) = (−2)³ + 7(−2)² + 7(−2) − 6 = −8 + 8 - 14 - 6 = 0

Q77. If x and y be two positive real number such that 8x³ + 27y³ = 730 and 2x²y + 3xy² = 15 then show that 2x + 3y = 10.

Solution

8x³ + 27y³ = 730 2x²y + 3xy² = 15 (2x + 3y)³ = (2x)³ + (3y)³ + 3 (2x) (3y) (2x + 3y) = 8x³ + 27y³ + 18xy (2x + 3y) (2x + 3y)³ = 8x³ + 27y³ + 18 (2x²y + 3xy²) (2x + 3y)³ = 730 + 18 (15) = 730 + 270 (2x + 3y)³ = 1000 2x + 3y = 10

Q78. If a² + b² + c² = 30 and a + b + c = 10 then find the value of ab + bc + ca.

Solution

We have (a + b + c)² = a² + +b² + c² + 2ab + 2bc + 2ca (a + b+ c)² = a² + b² + c² + 2(ab + bc + c + ca) (10)² = 30 + 2(ab + bc + ca) 100 - 30 = 2(ab + bc + ca) 70 = 2(ab + bc + ca) 35 = ab + bc + ca

Q79. If a = 1 - , find

Solution

a = 1 - ,

Q80. If (x-2) is a factor of the polynomial x³ + ax² + b x + 6 and when divided by (x-3), it leaves remainder 3. Find the values of a and b.

Solution

Let p(x) = x³ + ax² + bx +6 (x-2) is a factor of the polynomial x³ + ax² + b x +6 p(2) = 0 p(2) = 2³ + ax 2² + b x 2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0 7 +2 a +b = 0 b = - 7 -2a…(i) x³ + ax² + bx +6 when divided by (x-3) leaves remainder 3. p(3) = 3 p(3) = 3³ + ax 3² + b x 3 +6 = 27+9a +3b +6 =33+9a+3b = 3 11+3a +b =1 3a+b =-10 b=-10-3a….(ii) Equating the value of b from (ii) and (i) , we have (- 7 - 2a) = (-10 - 3a) a = -3 Substituting a = -3 in (i), we get b = - 7 -2(-3) = -7 + 6 = -1 Thus the values of a and b are -3 and -1 respectively.

Q81. Prove that (x+y)³ - (x - y)³ - 6y (x² - y²) = 8y³.

Solution

(x+y)³ -(x-y)³-6y(x² - y²) i.e., where a =x+y b = x-y But = (a-b)³ Therefore, (a-b)³ = {x+y-(x-y)}³ =

Q82. In y⁶ − 10y⁴ − 12y³ + 1, the degree of the polynomial is 

• 1) 3
• 2) 4
• 3) 6
• 4) 1

Solution

The highest power of the variable in the polynomial is the degree of the polynomial. In y⁶ − 10y⁴ − 12y³ + 1, the degree of the polynomial is 6.

Q83. Factorize: p(x) = 2x³ - 13x² + 12x + 10

Solution

Q84. Simplify: .

Solution

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

Q85. For what value of a if the polynomial 2x³ + ax² + 11x + a + 3 is exactly divisible by 2x - 1.

Solution

Let p(x) = If p(x) is exactly divisible by (2x-1) then (2x-1) is a factor of p(x) Thus p = 0 Thus p(x) is exactly divisible by (2x-1) at a = -7

Q86. Expand: (a) (b)

Solution

(a) = = = (b) = =

Q87. What is the remainder when x³ − 12x² − 42 is divided by x − 3? 

• 1) -123
• 2) 245
• 3) 344
• 4) 123

Solution

P(x) = x³ − 12x² − 42 x − 3 = 0 x = 3 P(3) = (3)³ - 12(3)² − 42 = 27 - 108 - 42 = −123

Q88. If f(x) = x⁴ - 2x³ + 3x² - ax + b is divided by(x - 1) and (x + 1), it leaves the remainder 5 and 19 respectively. Find 'a' and 'b'.

Solution

Given f(x) = x⁴ - 2x³ + 3x² - ax + b When f(x) is divided by (x-1), it leaves a remainder 5 f(1) = 5 1 - 2(1)³ + 3(1)² - a(1) + b = 5 1 - 2 + 3 - a +b = 5 -a + b = 3 … (i) When f(x) is divided by (x+1), it leaves a remainder 19 f(-1) = 19 (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + b = 19 1 + 2 + 3 + a + b = 19 a +b = 13 … (ii) Adding (i) and (ii), 2b = 16 _{b = 8 (i) a = b - 3 = 8 - 3 = 5}

Q89. If both (x - 2) and (2x - 1) are factors of ax² + 5x + b, show that a - b = 0.

Solution

Let p(x) = ax² + 5x + b Since both (x - 2) and (2x - 1) are factors of ax² + 5x + b. x = 2 or x = 1/2 Therefore, substituting x = 2 in the equation, 4a + 10 + b = 0 _{Or, 4a + b = -10… (1) Also, x = 1/2, therefore, substituting x = 1/2 in the equation,}    On solving (1) and (2), a = -2 and b = -2 Hence, a - b = -2 - (-2) = 0.

Q90. Verify x³ + y³ + z³ - 3xyz = (x + y + z) [(x - y)² + (y - z)² + (z - x)²] and factorise 64x³ - 125y³ - 64z³ + 240 xyz.

Solution

(x + y + z) [ (x - y)² + (y - z)² + (z - x)²] (x + y + z) (x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2xz) = (x + y + z) (2x² + 2y² + 2z² - 2xy - 2yz - 2xz) = 2 (x + y + z) (x² + y² + z² - xy - yz - xz) = (x + y + z) (x² + y² + z² - xy - yz - xz) = x³ + y³ + z³ - 3xyz = LHS. Consider, 64x³ + 125y³ - 64z³ + 240 xyz = (4x)³ + (5y)³ + (-4z)³ - 3 (4x) (5y) (-4z) = (4x + 5y - 4z) (16x² + 25y² + 16z² - 20xy + 20yz + 16xz)

Q91. Find for p(t) = t² - t + 2

• 1)
• 2)
• 3)
• 4)

Solution

Q92. Without calculating the square of any number , calculate the value of the following: 497 503

Solution

Q93. Factorize: a⁷ + ab⁶.

Solution

Q94. Find the zeroes of the polynomial

Solution

Q95. Without actually evaluating the cubes, find the value of: 15³ - 6³ - 9³

Solution

Q96. Factorise: x⁶ - 729.

Solution

x⁶ - 729 = (x²)³ - (9)³ [(x²) - (9)] [(x²)² + (x²) (9) + (9)²] = (x² - 9) (x⁴ + 9x² + 81) = (x - 3)(x + 3) (x⁴ + 9x² + 81)

Q97. The coefficient of

• 1) 17
• 2) -10
• 3) -3
• 4) -17

Solution

So the coef. of x² is 17.

Q98. Factorise the polynomial x⁴ + 5x² - 6

Solution

x⁴ + 5x² - 6 Putting x² = y , the polynomial can be written as y² + 5y - 6 Factorizing it further, here a = 1, c = -6 ,    ac = -6 b = 5 = 6 - 1 y² + 5y - 6 = y² + 6y - y - 6 = y(y + 6) - 1(y + 6) = (y - 1)( y + 6) = ( x² - 1) (x² + 6)           [ putting back x² = y ] = ( x - 1) (x + 1) (x² + 6)

Q99. If x + = 7 then find the value of x³ + .

Solution

We know … (i)

Q100. Factorise : a³(b - c)³ + b³ (c - a)³ + c³ (a - b)³

Solution

[a (b - c)]³ + [b (c - a)]³ + [c (a - b)]³ Here, a (b - c) + b (c - a) + c (a - b) = ab - ac + bc - ab + ca - bc = 0 [a (b - c)]³ + [b (c - a)]³ + [c (a - b)]³ = 3 a (b - c) b (c - a) c (a - b) = 3 abc (a - b) (b -c) (c - a)

Q101. Without actual division prove that x⁴ + 2x³ - 2x² + 2x -3 is exactly divisible by x² + 2x - 3.

Solution

Let f(x) = and g(x) = = = (x + 3)(x - 1) of g(x) f(x) is divisible by (x+3) and (x-1). Hence f(x) is exactly divisible by g(x).

Q102. Using suitable identity evaluate (-32)³ + (18)³ + (14)³.

Solution

We know x³ + y³ + z³ - 3xyz = 0, if x + +y + z = 0 Let x = -32, y = 18, z = 14 x + y + z = (-32) + (18) + (14) = 0 Therefore, (-32)³ + (18)³ + (14)³ = 3(-32) (18) (14) = -24192

Q103. If x + y + z = 1, xy + yz + zx = -1 and xyz = -1 find the value of x³+y³+z³.

Solution

Identity:x³ +y³ +z³ -3xyz = (x+y+z)( x² +y² +z² -xy-yz-zx) Since x + y + z = 1, xy + yz + zx = -1and xyz= -1, Putting values, we get: So Now (x+y+z)² = x² +y² +z² +2xy +2yz +2zx 1² = x² +y² +z² +2(-1) x² +y² +z² = 3 Put in eq (i) x³ +y³ +z³ =3-2 = 1

Q104. Find the values of a and b so that (x + 1) and (x - 2) are factors of (x³ + ax² + 2x + b).

Solution

If (x + 1) is a factor of x³ + ax² + 2x + b then (-1)³ + a(-1)² + 2(-1) + b = 0 -1 + a - 2 + b = 0 a + b = 3 ...(1) If (x - 2) is a factor of x³ + ax² + 2x+ b then (2)³ + a(2)² + 2(2) + b = 0 4a + b = -12 ...(2) Subtracting (1) from (2), we get 3a = -15 Or, a = -5 Using in (1), we get -5 + b = 3 Or, b = 8 Hence, a = -5, b = 8.

Q105. The volume of a cube is given by the polynomial p(x) = 8x³ - 36x² + 54x - 27. Find the possible expression for the sides of cube.

Solution

8x³ - 36x² + 54x - 27 = (2x)³ + 3(2x)²(-3) + 3(2x)(-3)² + (-3)³ = (2x - 3)³ Since volume of a cube is give by (side)³. Therefore the expression for side is: 2x - 3.

Q106. The area of a square field is 9x² - 24x + 16 sq units. Find the side of the field if x = 8 units.

Solution

Q107.  

Solution

Q108. Which of the following is a binomial? 

• 1) x + y
• 2) x - 3x
• 3) y + 2x + z
• 4) 2y + x - 1

Solution

A binomial has two terms. x + y is a binomial.

Q109. Factorize:

Solution

Let p(x) = Then p(1) = Thus (x -1) is a factor of p(x). Now by long division, Thus,

Q110. Check whether the polynomial t + 1 is a factor of 4t³ + 4t² - t - 1.

Solution

p(x) = 4t³ + 4t² - t - 1 p(-1) = 4(-1)³ + 4(-1)² - (-1) - 1 = -4 + 4 + 1 - 1 = 0 Thus, (t + 1) is a factor of p(x).

Q111. Factorize : 4x⁴ + 7x² - 2

Solution

Let x² = y then equation is 4y² + 7y - 2 = 4y² + 7y - 2 = 4y² + 8y - y - 2 = 4y(y + 2) - (y + 2) = (y + 2)(4y - 1) = (x² + 2) (4x² - 1) = (x² + 2) ((2x)² - 1) = (x² + 2) (2x + 1) (2x - 1)

Q112. Factorize: a¹²y⁴ - a⁴y¹².

Solution

a¹²y⁴ -a⁴y¹² a⁴y⁴ (a⁸ - y⁸) (Taking common terms) a⁴y⁴ [(a⁴)² - (y⁴)²] a⁴y⁴ [(a⁴ + y⁴) (a⁴ -y⁴)] a⁴y⁴ [(a⁴ + y⁴) (a² + y²) (a² - y²)] a⁴y⁴ [(a⁴ + y⁴) (a² + y²) (a + y) (a - y)]

Q113.

Solution

(i)  No                  has negative power of   ( ii ) Yes             (iii ) Yes                         ( iv ) No                 has fractional power of   ( v ) No                                                     has negative power of  

Q114. In 8m - 3m³ + m⁸ - 13m², the leading term is

• 1) 8m
• 2) -13m²
• 3) -3m³
• 4) m⁸

Solution

A leading term is the term of the highest degree in the polynomial. In 8m - 3m³ + m⁸ - 13m², the leading term is m⁸

Q115. If a² + b² + c² = 90 and a + b + c = 20 then find the value of ab + bc + ca.

Solution

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca (a+ b + c)² = a² + b² + c²+ 2 (ab + bc + ca) (20)² = 90 + 2 (ab + bc + ca) 400 - 90 = 2 (ab + bc + ca) ab + bc + ca = 155

Q116. Find the value of 'a' if (x - a) is a factor of x⁵ - a²x³ + 2x + a + 3. Hence, factorise x² - 2ax - 3.

Solution

Let p(x) = x⁵-a²x³+2x+a+3 Since, (x-a) is a factor of p(x), so p(a) = 0

Q117. Use suitable identities to Factorise ³  - 8y ³ - 6²y+ 12y²

Solution

 ³  - 8y ³ - 6²y+ 12y²          =³  - (2y) ³ - 6y(- 2y)                                      =³  - (2y) ³ – 3XX 2y(- 2y)        { [(a - b) ³= a³ - b³- 3 a b(a - b)] where a = , b = 2y}         = (- 2y) ³            = (- 2y) (- 2y) (- 2y)

Q118. Simplify by factorization method: .

Solution

Q119. Factorise: x³ + 13x² + 32x + 20.

Solution

Let p(x) = x³ + 13x² + 32x + 20 p(-1) = -1 + 13 - 32 + 20 = -33 + 33 = 0 Therefore (x + 1) is a factor of p(x). On dividing p(x) by (x + 1) we get p(x) (x + 1) = x² + 12x + 20 Thus, x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20) = (x + 1) (x² + 10x + 2x + 20) = (x + 1)[x(x + 10) + 2(x + 10)] = (x + 1) (x +2) (x + 10) Hence, x³ + 13x² + 32x + 20 = (x + 1) (x +2) (x + 10).

Q120. Determine whether the indicated numbers are zeroes of the given polynomial? (i) g(x) = 3x² -2; x = . (ii) f(x) = x³ -6x² + 11x -6; x = 1, 3.

Solution

(i) g(x) = 3x²-2 At x = (ii) f(x) = x³-6x²+11x-6 f(1) and f(3) are equal to zero x= 1,3 are the zeroes

Q121. Find the remainder when p(x) = x³ - ax² + 6x - a is divided by (x - a). 

• 1) a
• 2) 5a
• 3) 5
• 4) -5

Solution

f(x) = x³ - ax² + 6x - a We have (x − a) = 0 ∴ x = a By the remainder theorem, ∴ f(a) = (a)³ − a(a)² + 6(a) − a  = 5a

Q122. Find the product of .

Solution

Q123. Using suitable identity evaluate (42)³ - (18)³ - (24)³.

Solution

(42)³ - (18)³ - (24)³ Let x = 42, y = -18, z = -24 Now x + y + z = 42 - 18 - 24 = 0 Therefore, x³ + y³ + z³ = 3xyz = 3 (42) (-18) (-24) = 54432

Q124. Find the value of the polynomial p(z) = 3z² - 4z + when z = 3.

Solution

P (z) = P (3) = = (39) - 12 + = 27 - 12 + = 15 +

Q125. Without actual division, prove that a⁴ + 2a³ - 2a² + 2a - 3 is an exact multiple of a² + 2a - 3.

Solution

Q126. a² + b² + c² - ab - bc - ca equals:

• 1) (a + b + c)²
• 2) (a – b – c)²
• 3) (a – b + c)²
• 4)

Solution

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

Q127. (x + 2) is a factor of 2x³ + 5x² - x - k. The value of k is:

• 1) 6
• 2) -24
• 3) -6
• 4) 24

Solution

Here, 2(-2)³ + 5(-2)² - (-2) - k =0 Or, -16 + 20 + 2 - k =0 Or, k = 6.

Q128. If p = 4 - q, prove that p³ + q³ + 12pq = 64.

Solution

We have p = 4 - q Or, p + q - 4 =0 p³ + q³ + (-4)³ = 3 p q (-4) P³ + q³ + 12pq = 64.

Q129. If x³ + ax² + bx + 6 has x - 2 as a factor and leaves a remainder 3 when divided by x - 3, find the values of a and b.

Solution

f(x) = x³ + ax² + bx + 6 Since (x - 2) is a factor therefore f(2) = 8 + 4a + 2b + 6 = 0 2a + b = -7 ...(i) Again, On dividing by (x - 3) it gives remainder 3 therefore f(3) = 27 + 9a + 3b + 6 = 3 3a + b = -30 ... (ii) Subtracting (i) from (ii), we get a = -23 From (i), b = -7 + 46 = 39 Hence, a = -23, b = 39.

Q130. Factorise : x⁶ +8y⁶ - z⁶ + 6x²y²z²

Solution

Q131. Factorize : x³ - 8y³ - 27z³ - 18xyz.

Solution

x³ - 8y³ - 27z³ - 18xyz = (x)³ + (-2y)³ + (-3z)³ - 3(x)(-2y) (-3z) Using the identity: a³ + b³ + c³ - 3abc - (a+ b + c) (a² + b² + c² - ab - bc - ca) = (x - 2y - 3z) (x² + 4y² + 9z² - 2xy + 6yz - 3xz)

Q132. Which of the following options is a quadratic polynomial in one variable?

• 1)
• 2)
• 3)
• 4)

Solution

_{is a quadratic polynomial in one variable.}

Q133. Find the volume of a cubical aquarium of edge 84 m. 

• 1) 512000
• 2) 458963
• 3) 592704
• 4) 80640

Solution

84 = 80 + 4 (80 + 4)³ = 80³ + 4³ + 3 × 80 × 4 (80 + 4)            … using (x + y)³ = x³ + y³ + 3xy(x + y) = 512000 + 64 + 80640 = 592704 

Q134. If x = 2 + , then find the value of .

Solution

Q135. Factorize: x² + x + 6.

Solution

x²+3√3x +6 = x²+2√3x +√3x+6 = x(x+2√3)+√3(x+2√3) = (x+√3) (x+2√3)

Q136. Find the value of a and b so that the polynomial x³ + 10x² + ax + b has (x - 1) and (x + 2) as factors.

Solution

Let f(x) = x³ + 10x² + ax + b Since (x - 1) is a factor therefore f(1) = 0 1 + 10 + a + b = 0 a + b = -11 … (1) Also, (x + 2) is a factor, therefore f(-2) = 0 (-2)³ + 10(-2)² + a(-2) + b = 0 _{-8 + 40 -2a + b = 0} 2a - b = 32 … (2) Adding (1) and (2), we get 3a = 21 Or, a = 7 From (1), we get b = -11 - 7 = -18. Hence, a = 7, b = -18.

Q137. The term mxⁿ… m is 

• 1) a real number
• 2) a natural number
• 3) a whole number
• 4) an integer

Solution

The term mxⁿ… m is a real number. 

Q138. Find the reminder when 3x³ - 4x² + 7x - 5 is divided by (x - 3) and (x + 3).

Solution

Let, p(x) = 3x³ - 4x² + 7x - 5 Remainder when p(x) is divided by (x - 3) is p(3) = 3(3)³ - 4(3)² + 7 3 - 5 = 81 - 36 + 21 - 5 = 102 - 41 = 61 Remainder when p(x) is divided by (x + 3) is p(-3) = 3(-3)³ - 4(-3)² + 7 (-3) - 5 = -81 - 36 - 21 - 5 = -143

Q139. Using suitable identity evaluate (-28)³ + (9)³ + (19)³

Solution

x³ + y³ + z³ = 3xyz if x + y + z = 0 Let x = -28 y = 9 z = 19 x + y + z = (-28) + (9) + (19) = 0 (-28)³ + (9)³ + (19)³ = 3(-28) (9) (19) = -14364

Q140. Factorize :.

Solution

Q141. Given a polynomial p(x) = x² - 5x + 4. (A) Find the value of the polynomial p(x) at x = 2 (B) Check whether x is a factor of p(x). (C) Factorise p(x).

Solution

p(x) = x² - 5x + 4 (A) p(2) = 2² - 5(2) + 4 = 4 - 10 + 4 = -2 (B) p(0) = 0² - 5 0 + 4 = 4 x - 0 = x is not a factor of p(x). (C)      x² - 5x + 4      = x² - 4x - x + 4      = x(x - 4) - (x - 4)      = (x - 1)(x - 4)

Q142. Factorize 9x² + y² + z² - 6xy + 2yz - 6zx. Hence find its value if x = 1, y = 2 and z = -1.

Solution

9x² + y² + z² - 6xy + 2yz - 6zx = (3x)² + y² + z² - 2 3x y + 2yz - 2 z 3x = (3x - y - z)² {Using (a-b-c)²=a²+b²+c²-2ab+2bc-2ca} When x=1,y=2 and z=-1, 9x² + y² + z² - 6xy + 2yz - 6zx = (3x - y - z)² = [3(1) - 2 - (-1)]² = (3 - 2 + 1)² = 2² = 4

Q143. What should be added from 16x² + 40x + 1 to make it a perfect square?

Solution

Q144. If p(x) = x³ - ax² + bx + 3 leaves a remainder - 19 when divided by (x + 2) and a remainder 17 when divided by (x - 2), prove that a + b = 6.

Solution

Here, p(-2) = -8 - 4a - 2b + 3 = -19 4a + 2b = 14 … (1) And, p(2) = 8 - 4a + 2b + 3 = 17 -4a + 2b = 6 … (2) Adding (1) and (2), we get 4b = 20 b = 5 _{From (1), 4a + 10 = 14} Or, 4a = 4 Or, a = 1 Hence, a + b = 6

Q145. Without finding the cubes, factorize and find the value of:

Solution

So a³+ b³ +c³-3abc = 0 a³+ b³ +c³=3abc ₌

Q146. If a + b = 8 and a² + b² = 40 find the value of a³ + b³=^.

Solution

a + b= 8 Now, a² + b² = (a + b)² - 2ab 40 = 8² - 2ab 2ab = 64 - 40 = 24 ab = 12 Therefore, a³ + b³ = (a + b)³ - 3ab (a + b) = 8³ - 3 x 12 x 8 = 512 - 288 = 224

Q147. Check whether (p + 1) is a factor of p¹⁰⁰ - 1 and p¹⁰¹ - 1.

Solution

Q (p) = p¹⁰⁰ - 1 Q(-1) = (-1)¹⁰⁰ - 1 = 1 - 1 = 0 (p + 1) is a factor of Q(p) R(p) = p¹⁰¹ - 1 R(-1) = (-1)¹⁰¹ - 1 = -1 - 1 = -2 (p + 1) is not a factor of R(p)

Q148. The polynomials p(x) = ax³ + 3x² - 3 and q(x) = 2x³ - 5 + a , when divided by (x - 4) leave the remainders R₁ and R₂. Find 'a' if R₁ + R₂ = 0.

Solution

Q149. Factorize:(a) 4a² -9b² -2a -3b (b) a² + b² -2(ab -ac + bc)

Solution

(a) 4a²-9b²-2a-3b =[(2a)²-(3b)²]-(2a+3b) =(2a-3b) (2a+3b)-(2a+3b) =(2a+3b) [2a-3b-1] =(2a+3b) (2a-3b-1) (b) a²+b²-2(ab-ac+bc) =a²+b²-2ab+2ac-2bc =(a-b)²+2c(a-b) =(a-b)(a-b)+2c(a-b) =(a-b)(a-b+2c)

Q150. Factorize: .

Solution

Q151. What are the possible expressions for the dimensions of the cuboid whose volume is given below? Volume = 12ky² + 8ky - 20k.

Solution

Volume = 12ky² + 8ky - 20k = 4k (3y² + 2y - 5) = 4k (3y² + 5y -3y - 5) = 4k (y (3y+5) - 1(3y+5)) = 4k (3y +5) (y-1) Dimensions are l = 4k, b = 3y+5, h = y-1

Q152. Simplify by factorization method:

Solution

Q153. Find the values of p and q if the polynomial is divisible by the polynomial

Solution

Q154. In 5y² + 8y - 3y³, 8 is the coefficient of 

• 1) y²
• 2) y³
• 3) y + 1
• 4) y

Solution

A coefficient is the constant in the term of a polynomial. In 5y² + 8y - 3y³, 8 is the coefficient of y.

Q155. (a) Find the value of 34 x 36 using suitable identify. (b) Simplify using identify:

Solution

(a) 34 36 = (30 + 4) (30 + 6) Using identity (x+ a) (x + b) = x² + (a + b) x + ab Let x = 30, a = 4, b = 6 34 36 = (30 + 4) (30 + 6) = 30² + 30 (4 + 6) + 4 6 = 900 + 300 + 24 = 1224 (b)

Q156. Factorize : a³ + 8b³ - 27c² + 18 abc.

Solution

a³ + 8b³ - 27c² + 18 abc = (a)³ + (2b)³ + (-3c)³ + 18 abc =(a)³ + (2b)³ + (-3c)³ - 3(a) (2b) (-3c) We know that x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx) Therefore, (a)³ + (2b)³ + (-3c)³ - 3(a) (2b) (-3c) = (a + 2b - 3c) (2a² + 4b² + 9c² - 2ab + 6bc + 3 ac)

Q157. Without actually evaluating the cubes, find the value of: (8)³ + (5)³ + (-13)³

Solution

Q158. Factorise: b² + c² + 2 (ab + bc + ca)

Solution

Q159. If the polynomial ax³ + 3x² - 13 and 2x³ - 5x + a, when divided by (x + 2) leaves the same remainder, find the value of 'a'.

Solution

Q160. Using factor theorem, show that (x + 1) is a factor of x¹⁹ + 1.

Solution

Let f(x) = x¹⁹ + 1 (x + 1) is a factor if f(-1) = 0 Now, f(-1) = (-1)¹⁹ + 1 = -1 + 1 = 0 Hence (x + 1) is a factor of x¹⁹ + 1.

Q161.

Solution

Q162. The polynomials kx³ + 3x² - 8 and 3x³ - 5x + k are divided by x + 2. If the remainder in each case is the same, find the value of k.

Solution

Let p(x) = kx³ +3x² -8 and q(x) = 3x³ -5x +k When divided by (x+2), the remainders are p(-2) and q(-2). p(-2) = k(-2)³+3(-2)²-8=-8k + 4 q(-2) = 3(-2)³-5(-2)+k= -14 + k Given that, p(-2) = q(-2) -8k + 4 = -14 + k -9k = -18 k = 2

Q163. Factorise: a³ - b³ + 1 + 3ab.

Solution

a³ - b³ +1 + 3ab

Q164. If a+b+c=9 and ab+bc+ca=40. Find the value

Solution

Q165. Factorise:

Solution

 

Q166. Find the value of a if (x + a) is a factor of x⁴ -a²x² + 3x -a.

Solution

Let f(x) = x⁴-a²x²+3x-a (x+a) is a factor of f (x) f (-a) = 0 (-a)⁴-a²(-a)²+3(-a)-a = 0 a⁴-a⁴-3a-a = 0 -4a = 0 a = 0

Q167. If a = and b = , find a² -b².

Solution

Q168.

Solution

Q169. If a+b = 12 and ab= 27, find the value of a³+b³.

Solution

We know the identity: Given a+b=12 and ab=27, substituting values in the identity above, we get: = = 1728 - 972 = 756

Q170. Expand the following: (i) (x-2y-3z)². (ii) (y-)².

Solution

(i) (ii) Using identity:

Q171. Factorize : 8 a³ - b³ - 12a²b + 6ab²

Solution

8a³ - b³ - 12a²b + 6ab² = (2a)³ - (b)³ - 3 (2a)² b + 3 (2a) b² = (2a - b)³ = (2a - b) (2a - b) (2a - b)

Q172. If x = 1 - , find value of .

Solution

x= 1-√2 = 2² = 4

Q173. Show that (x - 2) is a factor of the Polynomial f(x) = 2x³ - 3x² - 17x + 30 and hence factorize f(x).

Solution

So (x-2) is a factor of f(x) Factorisation of f(x):

Q174. The polynomials x³ + 2x² - 5ax - 8 and x³ + ax² - 12x - 6 when divided by (x - 2) and (x - 3) leave remainder p and q respectively. If q - p = 10, find the value of a.

Solution

Let f(x) = x³ + 2x² -5ax -8 and g(x) = x³ + ax² -12x -6 When divided by (x-2) and (x-3),f(x) and g(x) leave remainder p and q respectively g(x) = x³ + ax² -12x -6

Q175. What is the remainder when h(y) = 4y³ - 12y² + 14y - 3 is divided by (2x - 1)

Solution

Q176. Using factor theorem, show that x² + 5x + 6 is a factor of x⁴ + 5x³ + 9x² + 15x + 18.

Solution

We have, x² + 5x + 6 = x² + 3x + 2x + 6 = x (x + 3) +2 (x + 3) = (x + 2) (x + 3) Let p(x) = x⁴ + 5x³ + 9x² + 15x + 18 Put x = -2 p(-2) = (-2)⁴ + 5(-2)³ + 9(-2)² + 15 (-2) + 18 = 16 - 40 + 36 - 30 + 18 = 70 - 70 = 0 (x + 2) is a factor Put x = -3 p(-3) = (-3)⁴ + 5(-3)³ + 9(-3)² + 15 (-3) + 18 = 81 - 135 + 81 - 45 + 18 = 180 - 180 = 0 (x + 3) is a factor Hence x² + 5x + 6 Is a factor of p(x).

Q177. Using factor theorem, show that (2x + 1) is a factor of 2x³ + 3x² - 11x - 6.

Solution

2x + 1 is a factor of polynomial p(x) _{only if} = 0 _Now,

Q178. (x + 2) is one of the factors of the polynomial x³ + 13x² + 32x + 20. Find its remaining factors.

Solution

Q179. Simplify (x + y + z)² - (x + y - z)².

Solution

(x + y + z)² - (x + y - z)² =(x² + y² + z² + 2xy + 2yz + 2zx) - (x² + y² + z² + 2xy - 2yz - 2zx) = = 4yz + 4zx = 4z(y + x)

Q180. Factorise the following:px² + (4p² - 3q)x - 12pq

Solution

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