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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Classify the following polynomials based on number of terms.
(i) x3 – x2
(ii) 5x
(iii) 4x4 + 2x3 + 1
(iv) 4.x3
(v) x + 2
(vi) 3x2
(vii) y4 + 1
(viii) y20 + y18 + y2
(ix) 6
(x) 2u3 + u2 + 3
(xi) u23 – u4
(xii) y
Solution:
5x, 3x2, 4x3, y and 6 are monomials because they have only one term.
x3 – x2, x + 2, y4 + 1 and u23 – u4 are binomials as they contain only two terms.
4x4 + 2x3 + 1 , y20 + y18 + y2 and 2u3 + u2 + 3 are trinomials as they contain only three terms.

Question 2.
Classify the following polynomials based on their degree.
Solution:
p(x) = 3, p(x) = -7, p(x) = \(\frac{3}{2}\) are constant polynomials
p(x) = x + 3, p(x) = 4x, p(x) = \(\sqrt{3}\)x + 1 are linear polynomials, since the highest degree of the variable x is one.
p(x) = 5x2 – 3x + 2, p(y) = \(\frac{5}{2}\) y2 + 1, p(x) = 3x2 are quadratic polynomials, since the highest degree of the variable is two.
p(x) = 2x3 – x2 + 4x + 1, p(x) = x3 + 1, p(y) = y3 + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.
Find the product of given polynomials p(x) = 3x3+ 2x – x2 + 8 and q (x) = 7x + 2.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 1
Solution:
(7x +2) (3x3 + 2x – x2 + 8) = 7x(3x3 + 2x – x2 + 8) + 2x
(3x3 + 2x – x2 + 8) = 21x4 + 14x2 – 7x3 + 56x + 6x3 + 4x – 2x2+ 16 = 21x4 – x3 + 12x4 + 60x + 16

Question 4.
Let P(x) = 4x4 – 3x + 2x3 + 5 and q(x) = x2 + 2x + 4 find p (x) – q(x).
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 1
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 50

Exercise 3.2

Question 1.
If p(x) = 5x3 – 3x2 + 7x – 9, find
(i) p(-1)
(ii) p(2).
Solution:
Given that p(x) = 5x3 – 3x2 + 7x – 9
(i) p(-1) = 5(-1)3 – 3(-1)2 + 7(-1) – 9 = -5 – 3 – 7 – 9
= -24
(ii) p(2) = 5(2)3 – 3(2)3 + 7(2) – 9 = 40 – 12 + 14 – 9
∴ p(2) = 33

Question 2.
Find the zeros of the following polynomials.
(i) p(x) = 2x – 3
(ii) p(x) = x – 2
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 51

Exercise 3.3

Question 1.
Find the remainder using remainder theorem, when
(i) 4x3 – 5x2 + 6x – 2 is divided by x – 1.
(ii) x3 – 7x2 – x + 6 is divided by x + 2.
Solution:
(i) Let p(x) = 4x3 – 5x2 + 6x – 2. The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p( 1). Now,
p(1) = 4(1)3 – 5(1)2 + 6(1) – 2 = 4 – 5 + 6 – 2 = 3
∴ The remainder is 3.

(ii) Let p(x) = x3 – 7x2 – x + 6. The zero of x + 2 is -2.
When p(x) is divided by x + 2, the remainder is p(-2). Now,
p(-2) = (-2)3 – 7(-2)2 – (-2) +6 = – 8 – 7(4) + 2 + 6
= – 8 – 28 + 2 + 6 = -28.
∴ The remainder is -28.

Question 2.
Find the value of a if 2x3 – 6x2 + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.
Solution:
Let p(x) = 2x3 – 6x2 + 5ax – 9
When p(x) is divided by (x – 2) the remainder is p(2).
Given that p(2) = 13
⇒ 2(2)3 – 6(2)2 + 5a(2) – 9 = 13
⇒ 2(8) – 6(4) + 10a – 9 = 13
⇒ 16 – 24 + 10a – 9 = 13
⇒ 10a – 17 = 13
⇒ 10a = 30
⇒ ∴ a = 3

Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Question 3.
If the polynomials f(x) = ax3+ 4x2 + 3x – 4 and g (x) = x3 – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 52
Solution:
Let f(x) = ax3 + 4x2 + 3x – 4 and g(x) = x3 – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).
Now f(3) = a(3)34(3)2 + 3(3) – 4 = 27a + 36 + 9 – 4
f (3) = 27a + 41 …(1)
When g(x) is divided by (x – 3), the remainder is g(3).
Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 +a
27a – a = 15 – 41
26a = -26
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 53
Sustituting a = -1, in f(3), we get
f(3) = 27(-1) + 41 = -27 + 41
f(3) = 14
∴ The remainder is 14.

Question 4.
Show that x + 4 is a factor of x3 + 6x2 – 7x – 60.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 54

Question 5.
In (5x + 4) a factor of 5x3 + 14x2 – 32x – 32
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 100

Question 6.
Find the value of k, if (x – 3) is a factor of polynomial x3 – 9x2 + 26x + k.
Solution:
Let p (x) = x3 – 9x2 + 26x + k
By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0
p(3) = 0
33 – 9(3)2 + 26 (3) + k = 0
27 – 81 + 78 + k = 0
k = -24
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 60

Question 7.
Show that (x – 3) is a factor of x3 + 9x2 – x – 105.
Solution:
Let p(x) = x3 + 9x2 – x – 105
By factor theorem, x – 3 is a factor of p(x), if p(3) = 0
p (3) = 33 + 9(3)3 – 3 – 105
= 27 + 81 – 3 – 105
= 108 – 108
p(3) = 0
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 60
Therefore, x – 3 is a factor of x3 + 9x2 – x – 105.

Question 8.
Show that (x + 2) is a factor of x3 – 4x2 – 2x + 20.
Solution:
Let p(x) = x3 – 4x2 – 2x + 20
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)3 – 4(-2)2 – 2(-2) + 20 = -8 – 4(4) + 4 + 20
P(- 2) = 0
Therefore, (x + 2) is a factor of x3 – 4x2 – 2x + 20

Exercise 3.4

Question 1.
Expand the following using identities :
(i) (7x + 2y)2
(ii) (4m – 3m)2
(iii) (4a + 3b) (4a – 3b)
(iv) (k + 2)(k – 3)
Solution:
(i) (7x + 2y)2 = (7x)2 + 2 (7x) (2y) + (2y)2 = 49x2 + 28xy + 4y2
(ii) (4m – 3m)2 = (4m)2 – 2(4m) (3m) + (3m)2 = 16m2 – 24mn + 9n2
(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a2 – b2 ]
Put [a = 4a, b = 3b]
(4a + 3b) (4a – 3b) = (4a)2 – (3b)2 = 16a2 – 9b2
(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x2 + (a – b)x – ab]
Put [x = k, a = 2, b = 3]
(k + 2) (k – 3) = k2 + (2 – 3)x – 2 × 3 = k2 – x – 6

Question 2.
Expand : (a + b – c)2
Solution:
Replacing ‘c’ by ‘-c’ in the expansion of
(a + b + c)2 = a2 + b2 + c2 + 2 ab+ 2 bc + 2ca
(a + b + (-c))2 = a2 + b2 + (-c)2 + 2ab + 2b(-c) + 2(-c)a
= a2 + b2 + c2 + 2ab – 2bc – 2ca

Question 3.
Expand : (x + 2y + 3z)2
Solution:
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3 z)2 = x2 + (2y)2 + (3z)2 + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x2 + 4y2 + 9z2 + 4xy + 12y2 + 6zx

Question 4.
Find the area of square whose side length is m + n – q.
Solution:
Area of square = side × side = (m + n – q)2
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
[m + n + (-q)]2 = m2 + n2 + (-q)2 + 2mn + 2n(-q) + 2(-q)m
= m2 + n2 + q2 + 2mn – 2nq – 2qm
Therefore, Area of square = m2 + n2 + q2 + 2mn – 2nq – 2qm = [m2 + n2 + q2 + 2mn – 2nq – 2qm] sq. units.

Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

EXERCISE 3.5

Question 1.
Factorise the following
(i) 25m-2 – 16n2
(ii) x4 – 9x2
Solution:
(i) 25m2 – 16n2 = (5m)2 – (4n)2
= (5m – 4n) (5m + 4n) [∵ a2 – b2 = (a – b)(a + b)]
(ii) x4 – 9x2 = x2(x2 – 9) = x2(x2 – 32) = x2(x – 3)(x + 3)

Question 2.
Factorise the following.
(i) 64m3 + 27n3
(ii) 729x3 – 343y3
Solution:
(i) 64m3 + 27n3 = (4m)3 + (3n)3
= (4m + 3n) ((4m)2 – (4m) (3n) + (3n)2) [∵ a2 + b2 = (a + b) (a2 – ab + b2)]
= (4m + 3n) (16m2 – 12mn + 9n2)
(ii) 729x3 – 343y3 = (9x)3 – (7y)3 = (9x – 7y) ((9x)2 + (9x) (7y) + (7y)2
= (9x – 7y) (81x2 + 63xy + 49y2)
[∵ a3 – b3 = (a – b) (a2 + ab + b2)]

Question 3.
Factorise 81x2 + 90x + 25
Solution:
81x2 + 90x + 25 = (9x)2 + 2 (9x) (5) + 52
= (9x + 5)2 [ ∵ a2 + 2ab + b2 = (a + b)2]

Question 4.
Find a3 + b2 if a + b = 6, ab = 5.
Solution:
Given, a + b = 6, ab = 5
a2 + b2 = (a + b)2 – 3ab(a + b) = (6)3 – 3(5)(6) = 126

Question 5.
Factorise (a – b)2 + 7(a – b) + 10
Solution:
Let a – b = p, we get p2 + 7p + 10,
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 70
p2 + 7p + 10 = p2 + 5p + 2p + 10
= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)
Put p = a – b we get,
(a – b)2 + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.
Factorise 6x2 + 17x + 12
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 71

Exercise 3.6

Question 1.
Factorise x2 + 4x – 96
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 72

Question 2.
Factorise x2 + 7xy – 12y2
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 73

Question 3.
Find the quotient and remainder when 5x3 – 9x2 + 10x + 2 is divided by x + 2 using synthetic division.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 74
Hence the quotient is 5x2 – 19x + 48 and remainder is -94.

Question 4.
Find the quotient and remainder when -7 + 3x – 2x2 + 5x3 is divided by -1 + 4x using synthetic division.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 75
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 76

Question 5.
If the quotient on dividing 5x4 + 4x3 + 2x + 1 by x + 3 is 5x3 + 9x2 + bx – 97 then find the values of a, b and also remainder.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 77
Quotient 5x3 + 11x2 + 33x – 97 is compared with given quotient 5x3 + ax2 + bx – 97
co-efficient of x2 is – 11 = a and co-efficient of x is 33 = b.
Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.
Find the quotient and the remainder when 10 – 4x + 3x2 is divided by x – 2.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending
order). Thus the given problem becomes (3x2 – 4x + 10) ÷ (x – 2)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 80
∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x2 – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form
Dividend = (Divisor × Quotient) + Remainder

Question 2.
If 8x3 – 14x2 – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 101

Exercise 3.8

Question 1.
Factorise 2x3 – x2 – 12x – 9 into linear factors.
Solution:
Let p(x) = 2x3 – x2 – 12x – 9
Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = -1 – 9 = -10
Sum of co-efficients of odd powers = 2 – 12 = -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 102
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 103
Then p (x) = (x + 1)(2x2 – 3x – 9)
Now 2x2 – 3x – 9 = 2x2 – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)
= (x – 3)(2x + 3)
Hence 2x3 – x2 – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.
Factorize x3 + 3x2 – 13x – 15.
Solution:
Let p(x) = x3 + 3x2 – 13x – 15
Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = 3 – 15 = – 12
Sum of coefficients of odd powers = 1 – 13 = – 12
Hence x + 1 is a factor of p (x)
Now use synthetic division to find the other factors.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 104
Hence p(x) = (x + 1) (x2 + 2x – 15)
Now x3 + 3x2 – 13x – 15 = (x + 1)(x2 + 2x – 15)
Now x2 + 2x – 15 = x2 + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)
Hence x3 + 3x2 – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.
Factorize x3 + 13x2 + 32x + 20 into linear factors.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 52
Solution:
Let, p(x) = x3 + 13x2 + 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x – 1) is not a factor.
Sum of coefficients of even powers with constant = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x + 1) is a factor of p(x)
Now we use synthetic division to find the other factors
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 106

Exercise 3.9

Question 1.
Find GCD of 25x3y2 z, 45x2y4z3b
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 107

Question 2.
Find the GCD of (y3 – 1) and (y – 1).
Solution:
y3 – 1 = (y – 1)(y3 + y + 1)
y – 1 = y – 1
Therefore, GCD = y – 1

Question 3.
Find the GCD of 3x2 – 48 and x2 – 7x + 12.
Solution:
3x2 – 48 = 3(x2 – 16) = 3(x2 – 43) = 3(x + 4)(x – 4)
x2 – 7x + 12 = x2 – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)
Therefore, GCD = x – 4

Question 4.
Find the GCD of ax , ax + y, ax + y + z.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 108

Exercise 3.10

Question 5.
Solve graphically. x – y = 3 ; 2x – y = 11
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 109

Exercise 3.11

Question 1.
Solve using the method of substitution. 5x – y = 5, 3x + y = 11
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 110

Exercise 3.12

Question 2.
Solve by the method of elimination. 2x + y = 10; 5x – y = 11
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 111

Exercise 3.13

Question 3.
Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 112

Exercise 3.14

Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be x
Let the sum of the age of his sons be y
At present x = 2y ⇒ x – 2y = 0 …. (1)
After 20 years x + 20 = y + 40
x – y = 40 – 20
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 113

Exercise 3.15

Multiple Choice Questions :

Question 1.
The polynomial 3x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial

Question 2.
The polynomial 4x2 + 2x – 2 is a
(1) linear polynomial
(2) quadratic polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of x is 2
Solution:
(2) quadratic polynomial]

Question 3.
The zero of the polynomial 2x – 5 is
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 114
Hint:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 115
Solution:
(1) \(\frac{5}{2}\)

Question 4.
The root of the polynomial equation 3x – 1 = 0 is
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 116
Hint:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 117
Solution:
(2) x = \(\frac{1}{3}\)

Question 5.
Zero of (7 + 4x) is ___
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 118
Solution:
(2) \(\frac{-7}{4}\)

Question 6.
Which of the following has as a factor?
(1) x2 + 2x
(2) (x – 1)2
(3) (x + 1)2
(4) (x2 – 22)
Solution:
(1) x2 + 2x

Question 7.
If x – 2 is a factor of q (x), then the remainder is _____
(1) q (-2)
(2) x – 2
(3) 0
(4) -2
Solution:
(3) 0

Question 8.
(a – b) (a2 + ab + b2) =
Solution:
(1) a3 + b3 + c3 – 3abc
(2) a2 – b2
(3) a3 + b3
(4) a3 – b3
Solution:
(4) a3 – b3

Question 9.
The polynomial whose factors are (x + 2) (x + 3) is
(1) x2 + 5x + 6
(2) x2 – 4
(3) x2 – 9
(4) x2 + 6x + 5
Solution:
(1) x2 + 5x + 6

Question 10.
(-a – b – c)2 is equal to
(1) (a – b + c)2
(2) (a + b – c)2
(3) (-a + b + c)2
(4) (a + b + c)2
Solution:
(4) (a + b + c)2

Text Book Activities

Activity – 1
Write the Variable, Coefficient and Constant in the given algebraic expression,
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 90

Activity – 2
Write the following polynomials in standard form.
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 91

Activity – 3

Question 1.
Find the value of k for the given system of linear equations satisfying the condition below:
(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution
(ii) kx + 3y = 3; 12x + ky = 6 has no solution
(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 92
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 93

Question 2.
Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 94

Activity – 4
For the given linear equations, find another linear equation satisfying each of the given condition
Solution:
Samacheer Kalvi 9th Maths Chapter 3 Algebra Additional Questions 95

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