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Grade 9 Exam  >  Grade 9 Tests  >  Test: Polynomials - 1 - Grade 9 MCQ

Test: Polynomials - 1 - Grade 9 MCQ


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25 Questions MCQ Test - Test: Polynomials - 1

Test: Polynomials - 1 for Grade 9 2025 is part of Grade 9 preparation. The Test: Polynomials - 1 questions and answers have been prepared according to the Grade 9 exam syllabus.The Test: Polynomials - 1 MCQs are made for Grade 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Polynomials - 1 below.
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Test: Polynomials - 1 - Question 1
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A cubic polynomial is a polynomial with degree

Detailed Solution for Test: Polynomials - 1 - Question 1

A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form f(x) = a3 x3 + a2 x2 + a1 x + a0. An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.

Test: Polynomials - 1 - Question 2
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A polynomial of degree 5 in x has at most

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A polynomial of degree 5 is of the form p(x) =, where a, b, c, d, e, and f are real numbers and a ≠ 0.

Thus, p(x) can have at most 6 terms and at least one term containing. 

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Test: Polynomials - 1 - Question 3
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The coefficient of x3 in the polynomial 5 + 2x + 3x2 – 7x3 is

Detailed Solution for Test: Polynomials - 1 - Question 3
To determine the coefficient of x3 in the polynomial 5 + 2x + 3x2 - 7x3, identify the term containing x3. In this case, the term is -7x3, so the coefficient is -7.
Test: Polynomials - 1 - Question 4
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The quadratic polynomial whose sum of zeroes is 3 and the product of zeroes is –2 is :
Detailed Solution for Test: Polynomials - 1 - Question 4

Sum of zeros = 3/1 
-b/a = 3/1 .....................(1)

Product of zeros = -2/1
c/a = -2/1 ...................(2)

From equation (1) and (2)
a = 1
-b = 3, b = -3
c = -2

The required quadratic equation is 
ax2+bx+c
= x2-3x-2

Test: Polynomials - 1 - Question 5
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A linear polynomial :-
Detailed Solution for Test: Polynomials - 1 - Question 5

A number of zeroes of an n -degree polynomial = n. 

First, a linear polynomial is in the form of ax + b, a≠0, a,b ∈R 

The degree of the polynomial = highest degree of the terms 
So here the highest degree is 1. 

Hence, Linear polynomial has only one zero.

Test: Polynomials - 1 - Question 6
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If x + y = 3, x2 + y2 = 5 then xy is
Detailed Solution for Test: Polynomials - 1 - Question 6

Test: Polynomials - 1 - Question 7
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When the polynomial x3 + 3x2 + 3x + 1 is divided by x + 1, the remainder is :-
Detailed Solution for Test: Polynomials - 1 - Question 7

The zero of x + 1 is –1

                And by remainder theorem, when

                p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, then remainder is p(–1).

                ∴ p(–1) = (–1)3 + 3 (–1)2 + 3(–1) + 1

                = –1 + (3 × 1) + (–3) + 1

                = –1 + 3 – 3 + 1

                = 0

                Thus, the required = 0

Test: Polynomials - 1 - Question 8
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If the polynomial 2x3 – 3x2 + 2x – 4 is divided by x – 2, then the remainder is :-
Detailed Solution for Test: Polynomials - 1 - Question 8
Using the Remainder Theorem, we substitute x = 2 into the polynomial: f(2) = 2(2)3 - 3(2)2 + 2(2) - 4 = 16 - 12 + 4 - 4 = 4. Therefore, the remainder is 4, which corresponds to option B.
Test: Polynomials - 1 - Question 9
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The value of k for which x – 1 is a factor of the polynomial 4x3+ 3x2 – 4x + k is :-
Detailed Solution for Test: Polynomials - 1 - Question 9

X - 1 is a factor of 4x3 + 3x2 -4x +k
then x=1 is one root of 4x3 + 3x2 -4x +k

put x= 1
4x3 +3x2 -4x +k = 0

=> 4 (1)3 +3 (1)2-4 (1) +k =0

=> 4 + 3 - 4 + k = 0

=> k = -3 

Test: Polynomials - 1 - Question 10
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The value of k for which x + 1 is a factor of the polynomial x3 + x2 + x + k is :-
Detailed Solution for Test: Polynomials - 1 - Question 10
To determine the value of k, we use the Factor Theorem, which states that if x + 1 is a factor of the polynomial P(x) = x3 + x2 + x + k, then P(-1) = 0. Substituting x = -1: P(-1) = (-1)3 + (-1)2 + (-1) + k = -1 + 1 - 1 + k = -1 + k. Setting this equal to zero gives: -1 + k = 0 implies k = 1. Thus, the value of k is 1.
Test: Polynomials - 1 - Question 11
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The value of m for which x – 2 is a factor of the polynomial x4 – x3 + 2x2 – mx + 4 is :-
Detailed Solution for Test: Polynomials - 1 - Question 11
To determine the value of m, we use the Factor Theorem, which states that if x - a is a factor of a polynomial P(x), then P(a) = 0. Here, a = 2. Substitute x = 2 into the polynomial: P(2) = (2)4 - (2)3 + 2(2)2 - m(2) + 4. Calculate each term: - (2)4 = 16, - (2)3 = -8, - 2 * (2)2 = 8, - m * 2 = -2m, - Constant term is 4. Combine the terms: 16 - 8 + 8 - 2m + 4 = 20 - 2m. Set equal to zero and solve for m: 20 - 2m = 0, -2m = -20, m = 10. Verification using synthetic division confirms that the remainder is zero when m = 10, ensuring x - 2 is a factor. Thus, the correct answer is A. 10.
Test: Polynomials - 1 - Question 12
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The factors of 2x2 – 3x – 2 are :-
Detailed Solution for Test: Polynomials - 1 - Question 12

2x2 – 3x – 2

2x2 - 4x + x - 2 = 0

2x2(x-2) +1(x-2) = 0

(2x+1) (x-2)

Test: Polynomials - 1 - Question 13
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If x + 2 is a factor of x3 – 2ax2 + 16, then value of a is
Detailed Solution for Test: Polynomials - 1 - Question 13
To determine the value of a when x + 2 is a factor of x3 - 2ax2 + 16, we apply the Factor Theorem, which states that substituting x = -2 into the polynomial should yield zero. Substituting x = -2: -8 - 8a + 16 = 0. Simplify each term: 8 - 8a = 0. Solving for a: -8a = -8 implies a = 1. Thus, the value of a is 1.
Test: Polynomials - 1 - Question 14
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The factors of x3 – 2x2 – 13x – 10 are :-
Detailed Solution for Test: Polynomials - 1 - Question 14

Test: Polynomials - 1 - Question 15
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The expanded form of (2x – 3y – z)2 is :-
Detailed Solution for Test: Polynomials - 1 - Question 15
To expand (2x - 3y - z)2, we use the formula for squaring a trinomial: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc. Here, a = 2x, b = -3y, and c = -z. Applying the formula: 1. Compute the squares of each term: (2x)2 = 4x2, (-3y)2 = 9y2, (-z)2 = z2. 2. Compute the cross terms: 2(2x)(-3y) = -12xy, 2(2x)(-z) = -4xz, 2(-3y)(-z) = 6yz. Combine all these results: 4x2 + 9y2 + z2 - 12xy - 4xz + 6yz. This matches option D.
Test: Polynomials - 1 - Question 16
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The expanded form of (x + y + 2z)2 is :-
Detailed Solution for Test: Polynomials - 1 - Question 16
To expand (x + y + 2z)2, we use the formula for trinomial expansion: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2xz + 2yz. Here, substituting z with 2z: = x2 + y2 + (2z)2 + 2*x*y + 2*x*(2z) + 2*y*(2z) = x2 + y2 + 4z2 + 2xy + 4xz + 4yz. Thus, the expanded form is x2 + y2 + 4z2 + 2xy + 4yz + 4zx, which corresponds to option C.
Test: Polynomials - 1 - Question 17
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The expanded form of (x+1/3)3 is :-  
Detailed Solution for Test: Polynomials - 1 - Question 17
To expand (x + 1/3)3, we use the binomial expansion formula for (a + b)3, which is a3 + 3a2b + 3ab2 + b3. Applying this to (x + 1/3)3: - First term: x3 - Second term: 3 * x2 * (1/3) = x2 - Third term: 3 * x * (1/3)2 = 3 * x * 1/9 = (1/3)x - Fourth term: (1/3)3 = 1/27 Combining these terms, the expanded form is: x3 + x2 + (1/3)x + 1/27 Therefore, the correct option is B.
Test: Polynomials - 1 - Question 18
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x3 + y3 + z3 – 3xyz is :-
Detailed Solution for Test: Polynomials - 1 - Question 18

We know that x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx).
If x + y + z = 0, then x3 + y3 + z3 – 3xyz = 0 or x3 + y3 + z3 = 3xyz.

Test: Polynomials - 1 - Question 19
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(a – b)3 + (b – c)3 + (c – a)3 is equal to :-
Detailed Solution for Test: Polynomials - 1 - Question 19

Let x = (a – b), y = (b – c) and z = (c – a)
Consider, x + y + z = (a – b) + (b – c) + (c – a) = 0
⇒ x3 + y3 + z3 = 3xyz
That is (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b)(b – c)(c – a)

Test: Polynomials - 1 - Question 20
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 is equal to :-
Detailed Solution for Test: Polynomials - 1 - Question 20

Test: Polynomials - 1 - Question 21
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√2 is a polynomial of degree
Detailed Solution for Test: Polynomials - 1 - Question 21

The highest power of the variable is known as the degree of the polynomial.

√2x^0 = √2
hence the degree of the polynomial is zero.

Test: Polynomials - 1 - Question 22
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The degree of the polynomial 4x4+0x3+0x5+5x+74x4+0x3+0x5+5x+7 is
Detailed Solution for Test: Polynomials - 1 - Question 22

The degree of the polynomial 4x4+0x3+0x5+5x+74x4+0x3+0x5+5x+7 is

Test: Polynomials - 1 - Question 23
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The degree of the zero polynomial is
Detailed Solution for Test: Polynomials - 1 - Question 23

The degree of zero polynomial is not defined, because, in zero polynomial, the coefficient of any variable is zero i.e., Ox2 or Ox5, etc. Hence, we cannot exactly determine the degree of the variable.  

Test: Polynomials - 1 - Question 24
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The value of the polynomial 5x−4x2+3, when x = −1 is
Detailed Solution for Test: Polynomials - 1 - Question 24

 Let p (x) = 5x – 4x2 + 3 …(i)
On putting x = -1 in Eq. (i), we get
p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Test: Polynomials - 1 - Question 25
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If p(x) = x + 3, then p(x) + p(-x) is equal to
Detailed Solution for Test: Polynomials - 1 - Question 25

 p(x)=x+3

 p(-x)=-x+3

p(x)+p(-x)=x+3-x+3=6

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