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Grade 9 Exam > Grade 9 Tests > Test: Polynomials - 2 - Grade 9 MCQ

Test: Polynomials - 2 - Grade 9 MCQ

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25 Questions MCQ Test - Test: Polynomials - 2

Test: Polynomials - 2 for Grade 9 2024 is part of Grade 9 preparation. The Test: Polynomials - 2 questions and answers have been prepared according to the Grade 9 exam syllabus.The Test: Polynomials - 2 MCQs are made for Grade 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Polynomials - 2 below.

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Test: Polynomials - 2 - Question 1

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√2 is a polynomial of degree

A.
√2
B.
2
C.
0
D.
1

Detailed Solution for Test: Polynomials - 2 - Question 1

The degree of a polynomial is the highest power that is present in it. The degree of the constant polynomial is Zero.

Test: Polynomials - 2 - Question 2

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If a+b+c = 0, then a³+b³+c³is equal to

A.
1
B.
3abc
C.
abc
D.
2abc

Detailed Solution for Test: Polynomials - 2 - Question 2

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Test: Polynomials - 2 - Question 3

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A polynomial containing two nonzero terms is called a ________.

A.
binomial
B.
trinomial
C.
monomial
D.
none of these

Detailed Solution for Test: Polynomials - 2 - Question 3

A binomial is a mathematical expression with two terms.

Examples of binomials.

binomial

All of these examples are binomials. Study them for a bit, and see if you can spot a pattern. The following is a list of what binomials must have:

They must have two terms.

If the variables are the same, then the exponents must be different.

Exponents must be whole positive integers. They cannot be negatives or fractions.

A term is a combination of numbers and variables. In the example 3x + 5, our first term is 3x, and our second term is 5. Terms are separated by either addition or subtraction. In our first example, notice how the 3x and 5 are separated by addition. In the last example, we have a binomial whose two terms both have the same variable s. Notice how each term has its variable to a different exponent. The first term has an exponent of 5, and the second term has an exponent of 4. While we can have fractions for our numbers, we cannot have fractional exponents.

Here are some examples of expressions that are not binomials.

Test: Polynomials - 2 - Question 4

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The expanded form of (3x−5)³ is

A.
27x³−135x²+225x−125
B.
27x³+135x²−225x−125
C.
27x³+135x²+225x−125
D.
none of these

Detailed Solution for Test: Polynomials - 2 - Question 4

Let a = 3x and b = 5 (3x)³ - (5)³ - 3×3x×5(3x- 5) = 27x³ - 125 -45x ( 3x -5) = 27x³ -125 - 135x² + 225x = 27x³ - 135x² + 225x - 125

Test: Polynomials - 2 - Question 5

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A.
231
B.
320
C.
336
D.
322

Detailed Solution for Test: Polynomials - 2 - Question 5

X+ 1/x = 7 then,
cubing both side
= (x +1/x)³ = (7)³
= x³ + 1/x³ + 3×x×1/x×(x+1/x) = 343
= x³ +1/x³ +3(7) = 343
= x³ +1/x³ +21 =343
= x³ +1/x³ =343 - 21
= x³ +1/x³ = 322

Test: Polynomials - 2 - Question 6

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Degree of the polynomial 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is:

A.
5
B.
4
C.
6
D.
7

Detailed Solution for Test: Polynomials - 2 - Question 6

Degree is the highest exponent of any value in an equation. Here, this highest exponent is 4. Therefore, 4 is the degree of the given equation.

Test: Polynomials - 2 - Question 7

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(x + 1) is a factor of the polynomial

A.
x⁴+3x³+3x²+x+1
B.
x³+x²−x+1
C.
x⁴+x³+x²+1
D.
x³+x²+x+1

Detailed Solution for Test: Polynomials - 2 - Question 7

Option a
f(x) = x⁴ + 3x³ + 3x² + x + 1
Substitute x = -1:
( -1 )⁴ + 3( -1 )³ + 3( -1 )² + ( -1 ) + 1 = 1 - 3 + 3 - 1 + 1 = 1 ≠ 0
So, (x + 1) is not a factor of this polynomial.

Option b
f(x) = x³ + x² - x + 1
Substitute x = -1:
( -1 )³ + ( -1 )² - ( -1 ) + 1 = -1 + 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.

Option c
f(x) = x⁴ + x³ + x² + 1
Substitute x = -1:
( -1 )⁴ + ( -1 )³ + ( -1 )² + 1 = 1 - 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.

Option d
f(x) = x³ + x² + x + 1
Substitute x = -1:
( -1 )³ + ( -1 )² + ( -1 ) + 1 = -1 + 1 - 1 + 1 = 0
Since the result is zero, (x + 1) is a factor of this polynomial.

Correct answer: d) x³ + x² + x + 1

Test: Polynomials - 2 - Question 8

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A polynomial containing three nonzero terms is called a ________.

A.
monomial
B.
binomial
C.
trinomial
D.
none of these

Test: Polynomials - 2 - Question 9

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If x + 2 is a factor of x³ – 2ax² + 16, then value of a is

A.
3
B.
1
C.
4
D.
2

Detailed Solution for Test: Polynomials - 2 - Question 9

use factor theorem as x+2 is factor of
x³-2ax²+16 so put x = -2 and equate the equation to 0
so putting x = -2
(-2)³-2a(-2)²+16 =0
-8-8a+16=0
-8a = -8
a = 8/8= 1
So,
a = 1

Test: Polynomials - 2 - Question 10

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A.
2
B.
1
C.
-1
D.
0

Detailed Solution for Test: Polynomials - 2 - Question 10

(0.75 * 0.75 * 0.75 + 0.25 * 0.25 * 0.25)/(0.75 * 0.75 - 0.75 * 0.25 + 0.25 * 0.25)

= (0.421875 + 0.015625)/(0.5625 - 0.1875 + 0.0625)

= (0.4375)/(0.4375)

= 1

Test: Polynomials - 2 - Question 11

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If one of the factor of x² + x – 20 is (x + 5). Find the other

A.
x – 4
B.
x + 2
C.
x + 4
D.
x – 5

Test: Polynomials - 2 - Question 12

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If P(x) = 10x−4x²−3, then the value of p(0)+p(1) is

A.
1
B.
3
C.
-3
D.
0

Detailed Solution for Test: Polynomials - 2 - Question 12

Test: Polynomials - 2 - Question 13

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The coefficient of x³ in 2x+x²−5x³+x⁴ is

A.
-5
B.
2
C.
1
D.
-1

Test: Polynomials - 2 - Question 14

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If both x - 2 and are the factors of px² + 5x + r, then

A.
2p = r
B.
p = 2r
C.
p = r
D.
none of these

Detailed Solution for Test: Polynomials - 2 - Question 14

Let f(x) = px² + 5 x + r
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)² + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...(1)

If is a factor of f (x), then by factor theorem,

Subtracting (2) from (1), we get
3p - 3r = 0
⇒ p = r

Test: Polynomials - 2 - Question 15

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If x + 2 is a factor of x³ – 2ax² + 16, then value of a is

A.
3
B.
1
C.
4
D.
2

Test: Polynomials - 2 - Question 16

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The value of the polynomial 5x−4x²+3, when x = −1 is

A.
- 6
B.
1
C.
9
D.
-1

Detailed Solution for Test: Polynomials - 2 - Question 16

It is given that

p(x) = 5x - 4x² + 3

We have to find the value when x = -1

p(-1) = 5(-1) - 4(-1)² + 3

By further calculation

p(-1) = -5 - 4 + 3

So we get

p(-1) = -9 + 3

p(-1) = -6

Therefore, the value is -6.

Test: Polynomials - 2 - Question 17

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If x²+kx+6 = (x+2)(x+3), then the value of ‘k’ is

A.
3
B.
1
C.
5
D.
0

Test: Polynomials - 2 - Question 18

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The remainder when the polynomial x⁴+2x³−3x²+x−1 is divided by (x−2) is

A.
21
B.
19
C.
20
D.
-21

Test: Polynomials - 2 - Question 19

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The remainder obtained when the polynomial p(x) is divided by (b – ax) is

A.
p(-b/a)
B.
p(b/a)
C.
p(-a/b)
D.
p(a/b)

Detailed Solution for Test: Polynomials - 2 - Question 19

b-ax=0

b=ax

b/a=x

i.e. remainder is p(b/a)

Test: Polynomials - 2 - Question 20

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The value of x³+y³+15xy−125 when x+y = 5 is

A.
0
B.
3
C.
2
D.
1

Detailed Solution for Test: Polynomials - 2 - Question 20

x³+y³+15xy-125
=x³ + y³ +3 xy ×5 - 125
=x³ + y³ +3xy(x+y) - (5)³
= (x+y) ³ - (5)³
=(5)³ - (5)³
=0

Test: Polynomials - 2 - Question 21

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If p(x) = x + 3, then p(x) + p(-x) is equal to

A.
0
B.
3
C.
6
D.
2x

Detailed Solution for Test: Polynomials - 2 - Question 21

Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3

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

Test: Polynomials - 2 - Question 22

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One of the factors of (16y²−1)+(1−4y)² is

A.
(4 – y)
B.
8y+2
C.
(4y - 1)
D.
(4 + y)

Detailed Solution for Test: Polynomials - 2 - Question 22

Test: Polynomials - 2 - Question 23

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If the polynomial x³−6x²+ax+3 leaves a remainder 7 when divided by (x−1), then the value of ‘a’ is

A.
8
B.
0
C.
7
D.
9

Test: Polynomials - 2 - Question 24

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Test: Polynomials - 2 - Question 25

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The value of (a²−b²)³+(b²−c²)3+(c²−a²)³ is

A.
3(a−b)(b−c)(c−a)
B.
3(a+b)(b+c)(c+a)
C.
3(a+b)(b+c)(c+a)(a−b)(b−c)(c−a)
D.
none of these

Detailed Solution for Test: Polynomials - 2 - Question 25

Let (a²-b²) =x , (b²-c²) =y , (c²-a²) = z we know that a+b+ c=0 and, a³ + b³ + c³ = 3abc so, 3(a+b)(a-b)(b+c) (b-c)(c+a)(c-a) = 3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)

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