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

Multiple Choice Questions.

Question 1.

If x^{3} + 6x^{2} + 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) = x^{3} + 6x^{2} + 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 – 3x^{3} is ……..

(a) constant polynomial

(b) linear polynomial

(c) quadratic polynomial

(d) cubic polynomial

Solution:

(d) cubic polynomial

Question 4.

If x^{51} + 51 is divided by x + 1, then the remainder is …….

(a) 0

(b) 1

(c) 49

(d) 50

Solution:

(d) 50

Hint:

p(x) = x^{51} + 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) = x^{3} – x^{2} – 2, q(x) = x^{2} – 3x + 1

(a) x^{3} – 3x – 1

(b) x^{3} + 2x^{2} – 1

(c) x^{3} – 2x^{2} – 3x

(d) x^{3} – 2x^{2} + 3x – 1

Solution:

(a) x^{3} – 3x – 1

Hint:

p(x) + q(x) = (x^{3} – x^{2} – 2) + (x^{2} – 3x + 1) = x^{3} – x^{2} – 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) = y^{6} + y³ – 2y³ – 2

= y^{6} – y³ – 2

Question 8.

Let the polynomials be

(A) -13q^{5} + 4q² + 12q

(B) (x² + 4)(x² + 9)

(C) 4q^{8} – q^{6} + q²

(D) –\(\frac{5}{7}\) y^{12} + y³ + y^{5}.

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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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 x^{4} – y^{4} and x² – y² is ……..

(a) x^{4} – y^{4}

(b) x² – y²

(c) (x + y)²

(d) (x + y)^{4}

Solution:

(b) x² – y²

Hint:

x^{4} – y^{4} = (x²)² – (y²)²

= (x² + y²)(x² – y²)

x² – y² = (x² – y²)

G.C.D. = x² – y²