Class 9 Exam > Class 9 Questions > . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 an...
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. The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0?
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. The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + ...
Case i: R1 = R2
When the polynomial 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 is divided by x + 1, the remainder R1 is obtained. Similarly, when the polynomial 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 is divided by x – 2, the remainder R2 is obtained.
Given R1 = R2, we have:
(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) = (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2)
Now, we can equate the remainders:
R1 = R2
(-7 - a) = (6 - 2a)
-7 - a = 6 - 2a
a = -13
Case ii: R1 + R2 = 0
Given R1 + R2 = 0, we have:
(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) + (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2) = 0
Now, we can substitute the expressions for R1 and R2:
-x - 13 = 0
x = -13
Case iii: 2R1 + R2 = 0
Given 2R1 + R2 = 0, we have:
2(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) + (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2) = 0
Substitute the expressions for R1 and R2:
-2x - 26 + 6 - 2a = 0
-2x - 20 - 2a = 0
-2(-13) - 20 - 2a = 0
26 - 20 - 2a = 0
6 - 2a = 0
2a = 6
a = 3
Therefore, the value of a is -13 in the case of R1 = R2, -13 in the case of R1 + R2 = 0, and 3 in the case of 2R1 + R2 = 0.
When the polynomial 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 is divided by x + 1, the remainder R1 is obtained. Similarly, when the polynomial 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 is divided by x – 2, the remainder R2 is obtained.
Given R1 = R2, we have:
(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) = (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2)
Now, we can equate the remainders:
R1 = R2
(-7 - a) = (6 - 2a)
-7 - a = 6 - 2a
a = -13
Case ii: R1 + R2 = 0
Given R1 + R2 = 0, we have:
(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) + (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2) = 0
Now, we can substitute the expressions for R1 and R2:
-x - 13 = 0
x = -13
Case iii: 2R1 + R2 = 0
Given 2R1 + R2 = 0, we have:
2(𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7) ÷ (x + 1) + (𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6) ÷ (x – 2) = 0
Substitute the expressions for R1 and R2:
-2x - 26 + 6 - 2a = 0
-2x - 20 - 2a = 0
-2(-13) - 20 - 2a = 0
26 - 20 - 2a = 0
6 - 2a = 0
2a = 6
a = 3
Therefore, the value of a is -13 in the case of R1 = R2, -13 in the case of R1 + R2 = 0, and 3 in the case of 2R1 + R2 = 0.
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. The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0?
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. The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0?.
. The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0?.
Solutions for . The polynomials 𝑥 3 + 2𝑥 2 − 5𝑎𝑥 − 7 and 𝑥 3 + 𝑎𝑥 2 − 12𝑥 + 6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i. R1 = R2 ii. R1 + R2 =0 iii. 2R1 + R2= 0? in English & in Hindi are available as part of our courses for Class 9.
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