The polynomials ax³ + 3x² − 13 and 2x³ − 5x + a are divided by x + 2. ...
Problem:
The polynomials ax³ 3x² − 13 and 2x³ − 5x a are divided by x 2. If the remainder in each case is the same, find the value of a.
Solution:
Given polynomials are ax³ + 3x² − 13 and 2x³ − 5x a.
Step 1: Find the Remainder
When a polynomial P(x) is divided by (x-a), the remainder is P(a).
Therefore, when ax³ + 3x² − 13 is divided by (x-2), the remainder is:
R1 = a(2)³ + 3(2)² - 13 = 8a + 12 - 13 = 8a - 1
Similarly, when 2x³ − 5x a is divided by (x-2), the remainder is:
R2 = 2(2)³ - 5(2) a = 16 - 10a
Since the remainders are the same:
R1 = R2
8a - 1 = 16 - 10a
18a = 17
a = 17/18
Step 2: Verify the Solution
To verify the solution, we need to divide both polynomials by (x-2) and check if the remainder is the same for both.
When ax³ + 3x² − 13 is divided by (x-2), we get:
ax² + 5ax + 17a - 39
When 2x³ − 5x a is divided by (x-2), we get:
2x² + ax + 4
Now, we need to substitute a = 17/18 and check if the remainders are the same:
For the first polynomial:
R1 = a(2)³ + 3(2)² - 13 = (17/18)(8) + 3(4) - 13 = 1/18
For the second polynomial:
R2 = 2(2)² + (17/18)(2) + 4 = 1/18
Since both remainders are the same, the solution is verified.
Step 3: Conclusion
The value of a is 17/18.