Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.15

Multiple Choice Questions.

 

Question 1. If \( x^3 + 6x^2 + kx + 6 \) is exactly divisible by \( x + 2 \), then \( k = \)
(a) 6
(b) -7
(c) 8
(d) 11
Answer: (d) 11
In simple words: When a polynomial can be divided evenly by \( x + 2 \), it means that if you put \( x = -2 \) into the polynomial, the answer should be zero. This helps us find the value of \( k \).

🎯 Exam Tip: Remember the Remainder Theorem: if a polynomial \( p(x) \) is divided by \( (x - a) \), the remainder is \( p(a) \). If it's *exactly* divisible, the remainder is 0.

 

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) \( -\frac{2}{3} \)
Answer: (c) \( -\frac{3}{2} \)
In simple words: To find the root of an equation, you need to find the value of \( x \) that makes the equation true. Here, we move the constant term and then divide to solve for \( x \).

🎯 Exam Tip: The root of a polynomial is also called its zero. It's the value of \( x \) that makes the polynomial equal to zero.

 

Question 3. The type of the polynomial \( 4 – 3x^3 \) is ........
(a) constant polynomial
(b) linear polynomial
(c) quadratic polynomial
(d) cubic polynomial
Answer: (d) cubic polynomial
In simple words: A polynomial's type is decided by its highest power. Since \( x \) has a power of 3, this is a cubic polynomial.

🎯 Exam Tip: The degree of a polynomial is the highest power of the variable in the expression. This determines its classification (linear, quadratic, cubic, etc.).

 

Question 4. If \( x^{51} + 51 \) is divided by \( x + 1 \), then the remainder is .......
(a) 0
(b) 1
(c) 49
(d) 50
Answer: (d) 50
In simple words: When you divide \( x^{51} + 51 \) by \( x + 1 \), the leftover part is 50. This is found by plugging \( x = -1 \) into the expression.

🎯 Exam Tip: Apply the Remainder Theorem: if \( p(x) \) is divided by \( (x+1) \), the remainder is \( p(-1) \). Be careful with negative numbers raised to odd or even powers.

 

Question 5. The zero of the polynomial \( 2x + 5 \) is ........
(a) \( \frac{5}{2} \)
(b) \( -\frac{5}{2} \)
(c) \( \frac{2}{5} \)
(d) \( -\frac{2}{5} \)
Answer: (b) \( -\frac{5}{2} \)
In simple words: To find the zero of this polynomial, we set the expression equal to zero and solve for \( x \). This value of \( x \) makes the polynomial zero.

🎯 Exam Tip: To find the zero of a linear polynomial \( ax + b \), set it to zero: \( ax + b = 0 \), then solve for \( x \), which gives \( x = -\frac{b}{a} \).

 

Question 6. The sum of the polynomials \( p(x) = x^3 – x^2 – 2 \), \( q(x) = x^2 – 3x + 1 \) is
(a) \( x^3 – 3x – 1 \)
(b) \( x^3 + 2x^2 - 1 \)
(c) \( x^3 – 2x^2 – 3x \)
(d) \( x^3 – 2x^2 + 3x – 1 \)
Answer: (a) \( x^3 – 3x – 1 \)
In simple words: To add these two polynomials, we combine the terms that have the same power of \( x \). This makes the expression simpler.

🎯 Exam Tip: When adding or subtracting polynomials, always combine like terms. Arrange them in descending order of powers for clarity.

 

Question 7. Degree of the polynomial \( (y^3 – 2) (y^3 + 1) \) is
(a) 9
(b) 2
(c) 3
(d) 6
Answer: (d) 6
In simple words: To find the degree of this product, we first multiply the terms. The highest power of \( y \) in the result will be the degree. Here, \( y^3 \) multiplied by \( y^3 \) gives \( y^6 \).

🎯 Exam Tip: When multiplying polynomials, the degree of the product is the sum of the degrees of the individual polynomials. Here, \( 3 + 3 = 6 \).

 

Question 8. Let the polynomials be
(A) \( -13q^5 + 4q^2 + 12q \)
(B) \( (x^2 + 4)(x^2 + 9) \)
(C) \( 4q^8 – q^6 + q^2 \)
(D) \( -\frac{5}{7} y^{12} + y^3 + y^5 \)
**Arrange them in descending order of their degree.**
(a) A, B, D, C
(b) A, B, C, D
(c) B, C, D, A
(d) B, A, C, D
Answer: (d) B, A, C, D
In simple words: We need to find the highest power in each polynomial and then list them from the polynomial with the biggest power to the one with the smallest power. This helps us arrange them by their degree.

🎯 Exam Tip: For polynomials in product form (like B), first expand or simply add the highest powers of the variables to find the degree. For example, \( (x^2+4)(x^2+9) \) has a degree of \( 2+2=4 \).

 

Question 9. If \( p(a) = 0 \) then \( (x – a) \) is a ........ of \( p(x) \)
(a) divisor
(b) quotient
(c) remainder
(d) factor
Answer: (d) factor
In simple words: If plugging in 'a' makes the polynomial equal to zero, it means \( (x - a) \) can divide the polynomial perfectly without any remainder. So, \( (x - a) \) is a factor.

🎯 Exam Tip: This is a direct application of the Factor Theorem. If \( p(a) = 0 \), then \( (x-a) \) is a factor of \( p(x) \).

 

Question 10. Zeros of \( (2 – 3x) \) is ........
(a) 3
(b) 2
(c) \( \frac{2}{3} \)
(d) \( \frac{3}{2} \)
Answer: (c) \( \frac{2}{3} \)
In simple words: To find the zero, we set \( 2 - 3x \) equal to zero. Then we solve for \( x \) by moving terms around.

🎯 Exam Tip: Always remember that the zero of a polynomial is the value of the variable that makes the entire expression equal to zero.

 

Question 11. Which of the following has \( x - 1 \) as a factor?
(a) \( 4x - 1 \)
(b) \( 3x - 3 \)
(c) \( 4x - 3 \)
(d) \( 3x – 4 \)
Answer: (b) \( 3x - 3 \)
In simple words: If \( (x - 1) \) is a factor, then plugging in \( x = 1 \) into the polynomial should give zero. We check each option to see which one works.

🎯 Exam Tip: To quickly check if \( (x-1) \) is a factor, substitute \( x=1 \) into each option. The one that results in 0 is the correct answer.

 

Question 12. If \( x – 3 \) is a factor of \( p(x) \), then the remainder is ........
(a) 3
(b) -3
(c) \( p(3) \)
(d) \( p(-3) \)
Answer: (c) \( p(3) \)
In simple words: The Remainder Theorem tells us that if we divide a polynomial \( p(x) \) by \( (x - 3) \), the leftover amount is exactly what we get if we put \( 3 \) in place of \( x \).

🎯 Exam Tip: This question tests your understanding of the Remainder Theorem. If \( (x-a) \) is a factor, the remainder is \( p(a) \).

 

Question 13. \( (x + y)(x^2 – xy + y^2) \) is equal to ........
(a) \( (x + y)^3 \)
(b) \( (x - y)^3 \)
(c) \( x^3 + y^3 \)
(d) \( x^3 – y^3 \)
Answer: (c) \( x^3 + y^3 \)
In simple words: This is a special algebraic identity. When you multiply these two expressions, the terms in the middle cancel out, leaving just \( x^3 + y^3 \). It's a key formula for summing cubes.

🎯 Exam Tip: Memorize common algebraic identities like sum and difference of cubes. This will save you time in calculations.

 

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) \( (a – b – c)^2 \)
Answer: (b) \( (-a – b + c)^2 \)
In simple words: When squaring an expression, changing the sign of all terms inside the bracket does not change the final squared value. So, \( (a + b - c)^2 \) is the same as \( (-(a + b - c))^2 \), which simplifies to \( (-a - b + c)^2 \).

🎯 Exam Tip: Remember that \( (-x)^2 = x^2 \). This property means that changing all the signs inside a parenthesis before squaring will not change the result.

 

Question 15. In an expression \( ax^2 + bx + c \) the sum and product of the factors respectively are ........
(a) a, bc
(b) b, ac
(c) ac, b
(d) bc, a
Answer: (b) b, ac
In simple words: For a quadratic expression, when we look for two factors, their sum should be the coefficient of the middle term (b), and their product should be the multiplication of the first and last coefficients (ac).

🎯 Exam Tip: For a quadratic \( ax^2 + bx + c \), to factorize, you look for two numbers whose sum is \( b \) and whose product is \( ac \).

 

Question 16. If \( (x + 5) \) and \( (x – 3) \) are the factors of \( ax^2 + bx + c \), then values of a, b and c are ........
(a) 1, 2, 3
(b) 1, 2, 15
(c) 1, 2, -15
(d) 1, -2, 15
Answer: (c) 1, 2, -15
In simple words: Since \( (x + 5) \) and \( (x - 3) \) are factors, we multiply them to get the original quadratic expression. Then we match the terms to \( ax^2 + bx + c \) to find the values of \( a \), \( b \), and \( c \).

🎯 Exam Tip: When given factors like \( (x+p) \) and \( (x+q) \), the quadratic polynomial is \( (x+p)(x+q) = x^2 + (p+q)x + pq \). Use this to find \( a, b, c \).

 

Question 17. Cubic polynomial may have maximum of ......... linear factors.
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: A cubic polynomial is one where the highest power of the variable is 3. The highest number of straight-line factors it can have is equal to its degree.

🎯 Exam Tip: The Fundamental Theorem of Algebra states that a polynomial of degree \( n \) has exactly \( n \) roots (counting multiplicity) in the complex numbers, which means it can have at most \( n \) linear factors.

 

Question 18. Degree of the constant polynomial is ........
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (d) 0
In simple words: A constant polynomial is just a number, like 5. We can write this as \( 5x^0 \), so the power of \( x \) is 0.

🎯 Exam Tip: A non-zero constant (like 7 or -2) has a degree of 0. The zero polynomial (just 0) is usually said to have an undefined degree or a degree of \( -\infty \).

 

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
Answer: (b) -2
In simple words: We are given the values of \( x \) and \( y \) that make the equation true. We simply put these values into the equation to find the value of \( m \).

🎯 Exam Tip: If a pair of values \( (x, y) \) is a solution to an equation, substituting these values into the equation must satisfy it.

 

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) \( x^3 – x = 5 \)
Answer: (c) \( 3x + 5 = \frac{2}{3} \)
In simple words: A linear equation is one where the highest power of the variable is 1. We look for an equation that fits this rule.

🎯 Exam Tip: A linear equation in one variable can be written in the form \( ax + b = 0 \), where \( a \neq 0 \). A linear equation in two variables is \( ax + by + c = 0 \).

 

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)
Answer: (b) (4, 2)
In simple words: A solution means that if we put the \( x \) and \( y \) values into the equation, both sides will be equal. We test each option to find the correct pair.

🎯 Exam Tip: To verify if a pair \( (x, y) \) is a solution, substitute the \( x \) and \( y \) values into the equation and check if the left-hand side equals the right-hand side.

 

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
Answer: (d) 13
In simple words: Since (2, 3) is a solution, we can replace \( x \) with 2 and \( y \) with 3 in the equation. This will allow us to calculate the value of \( k \).

🎯 Exam Tip: A solution to an equation means that the coordinate pair satisfies the equation when substituted. This is a direct substitution problem.

 

Question 23. Which condition does not satisfy the linear equation \( ax + by + c = 0 \) ........
(a) \( a \neq 0 \), \( b = 0 \)
(b) \( a = 0 \), \( b \neq 0 \)
(c) \( a = 0 \), \( b = 0 \), \( c \neq 0 \)
(d) \( a \neq 0 \), \( b \neq 0 \)
Answer: (c) \( a = 0 \), \( b = 0 \), \( c \neq 0 \)
In simple words: For an equation to be a linear equation, at least one of \( a \) or \( b \) must not be zero. If both \( a \) and \( b \) are zero, there are no variables, so it's not a linear equation.

🎯 Exam Tip: A linear equation in two variables must have at least one variable term. If \( a=0 \) and \( b=0 \), the equation becomes \( c=0 \), which is not a linear equation if \( c \neq 0 \).

 

Question 24. Which of the following is not a linear equation in two variables?
(a) \( 0x + 5y + 6 = 0 \)
(b) \( 0x + 0y + c = 0 \)
(c) \( 0x + by + c = 0 \)
(d) \( ax + 0y + c = 0 \)
Answer: (b) \( 0x + 0y + c = 0 \)
In simple words: A linear equation in two variables must have at least one variable, either \( x \) or \( y \), with a coefficient that is not zero. If both \( x \) and \( y \) terms vanish, it's no longer a two-variable equation.

🎯 Exam Tip: For an equation to be a linear equation in two variables, at least one of the coefficients of \( x \) or \( y \) must be non-zero. If both are zero, it becomes a constant equation.

 

Question 25. The value of k for which the pair of linear equations \( 4x + 6y – 1 = 0 \) and \( 2x + ky – 7 = 0 \) represents parallel lines is ........
(a) \( k = 3 \)
(b) \( k = 2 \)
(c) \( k = 4 \)
(d) \( k = -3 \)
Answer: (a) \( k = 3 \)
In simple words: For two lines to be parallel, their slopes must be equal. We calculate the slope of each line and set them equal to each other to find \( k \).

🎯 Exam Tip: For lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) to be parallel, the condition is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Applying this directly can save time.

 

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

(b)

(c)

(d)

Answer: (b) (The graphical representation of two parallel lines.)
In simple words: If two lines are parallel and never meet, there is no point where they cross each other. This means there is no common solution that satisfies both equations.

🎯 Exam Tip: Remember the three graphical interpretations of linear equations: intersecting lines (unique solution), parallel lines (no solution), and coincident lines (infinite solutions).

 

Question 27. If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) where \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) then the given pair of linear equation has ........ solution(s).
(a) no solution
(b) two solutions
(c) unique
(d) infinite
Answer: (c) unique
In simple words: When the ratios of the coefficients of \( x \) and \( y \) are not equal, it means the two lines will cross at exactly one point. This point is the unique solution.

🎯 Exam Tip: This condition \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) is the test for intersecting lines, which always yield a unique solution. Always match the conditions with the type of solution.

 

Question 28. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) where \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) then the given pair of linear equation has ........ solution(s).
(a) no solution
(b) two solutions
(c) infinite
(d) unique
Answer: (a) no solution
In simple words: This condition means the two lines are parallel but not the same. They will never meet, so there is no point that satisfies both equations.

🎯 Exam Tip: This condition means the lines are parallel and distinct. Remember to check all three ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \) to determine the nature of the solutions.

 

Question 29. GCD of any two prime numbers is ........
(a) -1
(b) 0
(c) 1
(d) 2
Answer: (c) 1
In simple words: Prime numbers only have two factors: 1 and themselves. Since they don't share any other common factors besides 1, their greatest common divisor is always 1.

🎯 Exam Tip: By definition, prime numbers have no common factors other than 1. This makes their Greatest Common Divisor (GCD) always 1.

 

Question 30. The GCD of \( x^4 - y^4 \) and \( x^2 - y^2 \) is ........
(a) \( x^4 – y^4 \)
(b) \( x^2 – y^2 \)
(c) \( (x + y)^2 \)
(d) \( (x + y)^4 \)
Answer: (b) \( x^2 – y^2 \)
In simple words: We can factor \( x^4 - y^4 \) into \( (x^2 - y^2)(x^2 + y^2) \). Since \( x^2 - y^2 \) is a factor of both expressions, it is their greatest common divisor. Factoring expressions simplifies finding common terms.

🎯 Exam Tip: Always factorize expressions completely to find their GCD. Recognize and use algebraic identities like \( a^2 - b^2 = (a-b)(a+b) \) for efficient factorization.

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