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Test: Polynomials - 2


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25 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Polynomials - 2

Test: Polynomials - 2 for Class 9 2022 is part of Mathematics (Maths) Class 9 preparation. The Test: Polynomials - 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Polynomials - 2 MCQs are made for Class 9 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Polynomials - 2 below.
Solutions of Test: Polynomials - 2 questions in English are available as part of our Mathematics (Maths) Class 9 for Class 9 & Test: Polynomials - 2 solutions in Hindi for Mathematics (Maths) Class 9 course. Download more important topics, notes, lectures and mock test series for Class 9 Exam by signing up for free. Attempt Test: Polynomials - 2 | 25 questions in 25 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study Mathematics (Maths) Class 9 for Class 9 Exam | Download free PDF with solutions
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Test: Polynomials - 2 - Question 1
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√2 is a polynomial of degree

Test: Polynomials - 2 - Question 2
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If a+b+c = 0, then a3+b3+c3 is equal to

Test: Polynomials - 2 - Question 3
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A polynomial containing two nonzero terms is called a ________.

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
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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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:
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
Test: Polynomials - 2 - Question 8
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A polynomial containing three nonzero terms is called a ________.
Test: Polynomials - 2 - Question 9
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If x + 2 is a factor of x3 – 2ax2 + 16, then value of a is
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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Detailed Solution for Test: Polynomials - 2 - Question 10

ANSWER :- b

Solution :- (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
Test: Polynomials - 2 - Question 12
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If 10x−4x2−3, then the value of p(0)+p(1) is
Test: Polynomials - 2 - Question 13
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The coefficient of x3 in 2x+x2−5x3+x4 is
Test: Polynomials - 2 - Question 14
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If both x - 2 and are the factors of px2 + 5x + r, then
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
Test: Polynomials - 2 - Question 16
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The value of the polynomial 5x−4x2+35, when x = −1 is
Test: Polynomials - 2 - Question 17
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If x2+kx+6 = (x+2)(x+3), then the value of ‘k’ is
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
Test: Polynomials - 2 - Question 19
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The remainder obtained when the polynomial p(x) is divided by (b – ax) is
Detailed Solution for Test: Polynomials - 2 - Question 19

ANSWER :- b

Solution :- 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
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
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
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
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
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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