Question

Find value of a if x^3 - 3x^2 + ax - 4 and 3x^3 + ax^2 - 8x - 26 are divided by x - 2 the remainder is same

Answer 1

5

Answer 2

To find the value of 'a', we use the Remainder Theorem.

Let the first polynomial be $P(x) = x^3 - 3x^2 + ax - 4$. Let the second polynomial be $Q(x) = 3x^3 + ax^2 - 8x - 26$.

According to the Remainder Theorem, if a polynomial $F(x)$ is divided by $(x-c)$, the remainder is $F(c)$. In this case, the divisor is $(x-2)$, so $c=2$.

1. Calculate the remainder for $P(x)$:

   The remainder when $P(x)$ is divided by $(x-2)$ is $P(2)$.

   $P(2) = (2)^3 - 3(2)^2 + a(2) - 4$

   $P(2) = 8 - 3(4) + 2a - 4$

   $P(2) = 8 - 12 + 2a - 4$

   $P(2) = -4 + 2a - 4$

   $P(2) = 2a - 8$

2. Calculate the remainder for $Q(x)$:

   The remainder when $Q(x)$ is divided by $(x-2)$ is $Q(2)$.

   $Q(2) = 3(2)^3 + a(2)^2 - 8(2) - 26$

   $Q(2) = 3(8) + a(4) - 16 - 26$

   $Q(2) = 24 + 4a - 16 - 26$

   $Q(2) = 8 + 4a - 26$

   $Q(2) = 4a - 18$

3. Equate the remainders:

   The problem states that the remainders are the same.

   So, $P(2) = Q(2)$

   $2a - 8 = 4a - 18$

4. Solve for 'a':

   Subtract $2a$ from both sides:

   $-8 = 4a - 2a - 18$

   $-8 = 2a - 18$

   Add $18$ to both sides:

   $-8 + 18 = 2a$

   $10 = 2a$

   Divide by $2$:

   $a = \frac{10}{2}$

   $a = 5$