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Class 9  >  Mathematics (Maths) Class 9  >  Test: Remainder Theorem (Old Syllabus) Download as PDF

Test: Remainder Theorem (Old Syllabus)


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15 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Remainder Theorem (Old Syllabus)

Test: Remainder Theorem (Old Syllabus) for Class 9 2023 is part of Mathematics (Maths) Class 9 preparation. The Test: Remainder Theorem (Old Syllabus) questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Remainder Theorem (Old Syllabus) MCQs are made for Class 9 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Remainder Theorem (Old Syllabus) below.
Solutions of Test: Remainder Theorem (Old Syllabus) questions in English are available as part of our Mathematics (Maths) Class 9 for Class 9 & Test: Remainder Theorem (Old Syllabus) 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: Remainder Theorem (Old Syllabus) | 15 questions in 15 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: Remainder Theorem (Old Syllabus) - Question 1
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If x² - 7x + a has a remainder 1 when divided by x + 1, then

Test: Remainder Theorem (Old Syllabus) - Question 2
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The remainder when the polynomial p(x) = x3 -3x2 +2x-1 is divided by x-2 is

Test: Remainder Theorem (Old Syllabus) - Question 3
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What is remainder when x3 – 2x2 + x + 1 is divided by (x -1)?

Test: Remainder Theorem (Old Syllabus) - Question 4
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The remainder when x4 – 3x2 + 5x – 7 is divided by x + 1 is:
Detailed Solution for Test: Remainder Theorem (Old Syllabus) - Question 4

Correct Answer :- a

Explanation : x + 1 = 0

x = -1

x4 – 3x2 + 5x – 7

(-1)4 - 3(-1)2 + 5(-1) - 7 = 0

1 - 3 - 5 - 7 

= -14

Test: Remainder Theorem (Old Syllabus) - Question 5
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If x² - 7x + a has a remainder 1 when divided by x + 1, then
Test: Remainder Theorem (Old Syllabus) - Question 6
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The remainder when x3 + x2 - 2x +1 is divided by (x+1) is
Detailed Solution for Test: Remainder Theorem (Old Syllabus) - Question 6

for finding the remainder we need to use remainder theorom

let p(x)=x3+x2-2x+1

Here, the divisor is x+1 for dividend p(x), we need to equate it to 0 and put that value of x in p(x).

−x+1=0

∴x=−1

p(−1)=(-1)3+(−1)2-2(−1)+1=3

Hence, remainder=3

Test: Remainder Theorem (Old Syllabus) - Question 7
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For a polynomial p(x) = 2x4 - 3x3 + 2x2 + 2x-1 what is the remainder when it’s divided by x+4?
Test: Remainder Theorem (Old Syllabus) - Question 8
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Find remainder when x3+1 is divided by x+1
Test: Remainder Theorem (Old Syllabus) - Question 9
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Using Remainder theorem, find the remainder when 3x4 - 4x3 - 3x - 1 by x - 1
Detailed Solution for Test: Remainder Theorem (Old Syllabus) - Question 9

By remainder theorem if x-1 = 0 then x=1 using it in equation we get p(x)= 3x⁴-4x³-3x-1 p(1)= 3x1⁴-4x1³-3x1-1 p(1)= 3-4-3-1 p(1)= -5

Test: Remainder Theorem (Old Syllabus) - Question 10
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In the division of a cubic polynomial p(x) by a linear polynomial, the remainder is p(-2).So, the divisor must be
Test: Remainder Theorem (Old Syllabus) - Question 11
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If x³ + 9x +5 is divided by x, then remainder is
Test: Remainder Theorem (Old Syllabus) - Question 12
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Find the remainder when P(x) = x2 - 2x is divided by x - 2
Test: Remainder Theorem (Old Syllabus) - Question 13
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Find p(1/3) for p(t) = t2 – t + 2
Detailed Solution for Test: Remainder Theorem (Old Syllabus) - Question 13

1/9-1/3+2 =1/9-3/9-18/9 =16/9

Test: Remainder Theorem (Old Syllabus) - Question 14
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Using Remainder Theorem find the remainder when x3 - x2 + x - 1 is divided by x - 1
Test: Remainder Theorem (Old Syllabus) - Question 15
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P(x) is a polynomial in x, ‘a’ is a real number. If (x-a) is a factor of p(x), then p (a) must be​
Detailed Solution for Test: Remainder Theorem (Old Syllabus) - Question 15

 If p(x) is a polynomial of degree n which is greater than or equal to one and a is any real number which will be the divisor, then there will be two conditions fulfilled: If p (a) =0, then x-a is a factor of that polynomial p(x). x-a would be the factor of the polynomial if the r(x) i.e. remainder is 0.

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