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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.15

Multiple Choice Questions.

Question 1.
If x3 + 6x2 + kx + 6 is exactly divisible by x + 2, then k = 2
(a) 6
(b) -7
(c) -8
(d) 11
Solution:
(d) 11
Hint:
p(x) = x3 + 6x2 + kx + 6
Given p(-2) = 0
(-2)3 + 6(-2)2 + k(-2) + 6 = 0
-8 + 24 – 2k + 6 = 0
22 – 2k = 0
k = $$\frac{22}{2}$$
= 11

Question 2.
The root of the polynomial equation 2x + 3 = 0 is…….
(a) $$\frac{1}{3}$$
(b) –$$\frac{1}{3}$$
(c) –$$\frac{3}{2}$$
(d) –$$\frac{2}{3}$$
Solution:
(c) –$$\frac{3}{2}$$
Hint:
2x + 3 = 0
2x = – 3 ⇒ x = –$$\frac{3}{2}$$

Question 3.
The type of the polynomial 4 – 3x3 is ……..
(a) constant polynomial
(b) linear polynomial
(d) cubic polynomial
Solution:
(d) cubic polynomial

Question 4.
If x51 + 51 is divided by x + 1, then the remainder is …….
(a) 0
(b) 1
(c) 49
(d) 50
Solution:
(d) 50
Hint:
p(x) = x51 + 51
p(-1)= (-1)51 + 51
= -1 + 51
= 50

Question 5.
The zero of the polynomial 2x + 5 is ……..
(a) $$\frac{5}{2}$$
(b) –$$\frac{5}{2}$$
(c) $$\frac{2}{5}$$
(d) –$$\frac{2}{5}$$
Solution:
(b) –$$\frac{5}{2}$$
Hint:
2x + 5 = 0 ⇒ 2x = -5 ⇒ x = –$$\frac{5}{2}$$

Question 6.
The sum of the polynomials p(x) = x3 – x2 – 2, q(x) = x2 – 3x + 1
(a) x3 – 3x – 1
(b) x3 + 2x2 – 1
(c) x3 – 2x2 – 3x
(d) x3 – 2x2 + 3x – 1
Solution:
(a) x3 – 3x – 1
Hint:
p(x) + q(x) = (x3 – x2 – 2) + (x2 – 3x + 1) = x3 – x2 – 2 + x² – 3x + 1
= x³ – 3x – 1

Question 7.
Degree of the polynomial (y³ – 2) (y³ + 1) is
(a) 9
(b) 2
(c) 3
(d) 6
Solution:
(d) 6
(y³ – 2) (y³ + 1) = y6 + y³ – 2y³ – 2
= y6 – y³ – 2

Question 8.
Let the polynomials be
(A) -13q5 + 4q² + 12q
(B) (x² + 4)(x² + 9)
(C) 4q8 – q6 + q²
(D) –$$\frac{5}{7}$$ y12 + y³ + y5.
Then ascending order of their degree is
(a) A, B, D, C
(b) A, B, C, D
(c) B, C, D, A
(d) B, A, C, D
Solution:
(d) B, A, C, D

Question 9.
If p(a) = 0 then (x – a) is a …….. of p(x)
(a) divisor
(b) quotient
(c) remainder
(d) factor
Solution:
(d) factor

Question 10.
Zeros of (2 – 3x) is ……..
(a) 3
(b) 2
(c) $$\frac{2}{3}$$
(d) $$\frac{3}{2}$$
Solution:
(c) $$\frac{2}{3}$$

Question 11.
Which of the following has x -1 as a factor?
(a) 2x – 1
(b) 3x – 3
(c) 4x – 3
(d) 3x – 4
Solution:
(b) 3x – 3

Question 12.
If x – 3 is a factor of p(x), then the remainder is ……..
(a) 3
(b) -3
(c) p(3)
(d) p(-3)
Solution:
(c) p(3)

Question 13.
(x +y)(x² – xy + y²) is equal to ……..
(a) (x + y)³
(b) (x – y)³
(c) x³ + y³
(d) x³ – y³
Solution:
(c) x³ + y³

Question 14.
(a + b – c)² is equal to ……..
(a) (a – b + c)²
(b) (-a – b + c)²
(c) (a + b + c)²
(d) (a – b – c)²
Solution:
(b) (-a – b + c)²
Hint:
(a + b – c)² = a² + b² + c² + 2ab – 2bc – 2ac
(- a – b + c)² = a² + b² + c² + 2ab – 2bc – 2ac
(OR)
(- a – b + c)² = (-1)² (a + b + c)² (taking – 1 as common)
= (a + b – c)²

Question 15.
In an expression ax² + bx + c the sum and product of the factors respectively ……..
(a) a, bc
(b) b, ac
(c) ac, b
(d) bc, a
Solution:
(b) b, ac

Question 16.
If (x + 5) and (x – 3) are the factors of ax² + bx + c, then values of a, b and c are ………
(a) 1, 2, 3
(b) 1, 2, 15
(c) 1, 2, -15
(d) 1, -2, 15
Solution:
(c) 1, 2, -15
Hint:
(x + 5) (x – 3) = x² + (5 – 3) x + (5) (-3)
= x² + 2x – 15
compare with ax² + bx + c
a = 1, b = 2 and c = -15

Question 17.
Cubic polynomial may have maximum of ……… linear factors.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 18.
Degree of the constant polynomial is ……..
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 19.
Find the value of m from the equation 2x + 3y = m. If its one solution is x = 2 and y = -2.
(a) 2
(b) -2
(c) 10
(d) 0
Solution:
(b) – 2
Hint:
The equation is 2x + 3y = m
Substitute x – 2 and y = -2 we get
2(2) + 3(-2) = m ⇒ 4 – 6 = m ⇒ -2 = m

Question 20.
Which of the following is a linear equation?
(a) x + $$\frac{1}{2}$$ = 2
(b) x (x – 1) = 2
(c) 3x + 5 = $$\frac{2}{3}$$
(d) x³ – x = 5
Solution:
(c) 3x + 5 = $$\frac{2}{3}$$

Question 21.
Which of the following is a solution of the equation 2x – y = 6?
(a) (2, 4)
(b) (4, 2)
(c) (3, -1)
(d) (0, 6)
Solution:
(b) (4, 2)
Hint:
2x – y = 6
Substitute x – 4 and y = 2 we get
2(4) – 2 = 6 ⇒ 8 – 2 = 6 ⇒ 6 = 6
∴ (4, 2) is the solution

Question 22.
If (2, 3) is a solution of linear equation 2x + 3y = k then, the value of k is ……..
(a) 12
(b) 6
(c) 0
(d) 13
Solution:
(d) 13
Hint:
The equation is 2x + 3y = k
Substitute x = 2 and y = 3 we get,
2(2) + 3(3) = k ⇒ 4 + 9 = k ⇒ 13 = k

Question 23.
Which condition does not satisfy the linear equation ax + by + c = 0 ……..
(a) a ≠ 0, b = 0
(b) a = 0, b ≠ 0
(c) a = 0, b = 0, c ≠ 0
(d) a ≠ 0, b ≠ 0
Solution:
(c) a = 0, b = 0, c ≠ 0

Question 24.
Which of the following is not a linear equation in two variable?
(a) ax + by + c = 0
(b) 0x + 0y + c = 0
(c) 0x + by + c = 0
(d) ax + 0y + c = 0
Solution:
(b) 0x + 0y + c = 0
Hint:
0x + 0y + c = 0
0 + 0 + c = 0 ⇒ c = 0
There is no variable.
∴ It is not a linear equation

Question 25.
The value of k for which the pair of linear equations 4x + 6y – 1 = 0 and 2x + ky – 1 = 0 represents parallel lines is ……..
(a) k = 3
(b) k = 2
(c) k = 4
(d) k = -3
Solution:
(a) k = 3
Hint:
Slope of 4x + 6y – 1 = 0 is
6y = -4x + 1 ⇒ y = $$\frac{-4}{6}$$ x + $$\frac{1}{6}$$
Slope = $$\frac{-4}{6}$$ = $$\frac{-2}{3}$$
Slope of 2x + ky – 7 = 0
ky = -2x + 7
y = $$\frac{-2}{k}$$x + $$\frac{7}{k}$$
Slope of a line = $$\frac{-2}{k}$$
Since the lines are parallel
$$\frac{-2}{3}$$ = $$\frac{-2}{k}$$
-2k = – 6
k = $$\frac{6}{2}$$
= 3

Question 26.
A pair of linear equations has no solution then the graphical representation is ……..

Solution:

Hint:
Since there is no solution the two lines are parallel. (l11m)

Question 27.
If $$\frac{a_1}{a_2}$$ ≠ $$\frac{b_1}{b_2}$$ where a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) unique
(d) infinite
Solution:
(c) unique
Hint:
Since it has unique solution
$$\frac{a_1}{a_2}$$ ≠ $$\frac{b_1}{b_2}$$

Question 28.
$$\frac{a_1}{a_2}$$ = $$\frac{b_1}{b_2}$$ ≠ $$\frac{c_1}{c_2}$$ where a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then the given pair of linear equation has …….. solution(s).
(a) no solution
(b) two solutions
(c) infinite
(d) unique
Solution:
(a) no solution
Hint:
$$\frac{a_1}{a_2}$$ = $$\frac{b_1}{b_2}$$ ≠ $$\frac{c_1}{c_2}$$ the linear equation has no solution.

Question 29.
GCD of any two prime numbers is …….
(a) -1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Question 30.
The GCD of x4 – y4 and x² – y² is ……..
(a) x4 – y4
(b) x² – y²
(c) (x + y)²
(d) (x + y)4
Solution:
(b) x² – y²
Hint:
x4 – y4 = (x²)² – (y²)²
= (x² + y²)(x² – y²)
x² – y² = (x² – y²)
G.C.D. = x² – y²