Question

The polynomials ar? + 3x - 13 and
2x2 - 5x + a are divided by x-2. The remainder
in each case is the same. Find the value of 'a'.​

Answer 1

Answer:

Correct question;

If the polynomials (ax^2 + 3x - 13) and

(2x^2 - 5x + a) are divided by (x-2) then the remainders in each case are same.

Find the value of 'a'.

Solution;

Let ,

The first polynomial be;

P1(x) = ax^2 + 3x - 13

And the second polynomial be;

P2(x) = 2x^2 - 5x + a

Note:

Remainder theorem:- If a polynomial P(x) is divided by (x-a) then the reminder is given by;

R = P(a).

Case (1):

When the first polynomial P1(x) is divided by (x-2) , then the reminder is given by; R1 = P1(2)

=> R1 = a(2)^2 + 3(2) - 13

=> R1 = 4a + 6 - 13

=> R1 = 4a - 7

Case (2):

When the second polynomial P2(x) is divided by (x-2) , then the reminder is given by; R2 = P2(2)

=> R2 = 2(2^2) - 5(2) + a

=> R2 = 8 - 10 + a

=> R2 = a - 2

Also, it is given that the reminders are same in each cases.

Thus, R1 = R2

=> 4a - 7 = a - 2

=> 4a - a = 7 - 2

=> 3a = 5

=> a = 5/3

Hence, the required value of a is 5/3.