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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.?
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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.
Community Answer
The polynomials ax³ + 3x² − 13 and 2x³ − 5x + a are divided by x + 2. ...
X+2=0×=-2a×-8 +3×4 -13-8a+12-13-8a-12×-8+10+a-16+10 +a-6+a-8a-1=-6+a-8a-a=-6+1-9a=-5a=5÷9
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