Question

6. If the polynomials x + ar? + 5 and x-2r? +a are divided by (x + 2) leave the same
remainder, find the value of a.​

Answer 1

Answer :

a = -13/3

Solution :

• Note : Remainder theorem : If a polynomial p(x) is divided by (x-c) , then the remainder obtained is given as R = p(c) .

Here ,

The given equations are ;

x + ax² + 5 and x - 2x² + a

Let f(x) = x + ax² + 5

g(x) = x - 2x² + a

Also ,

If x + 2 = 0 , then x = -2 .

Now ,

According to the remainder theorem , if f(x) is divided by (x + 2) then the remainder R will be given as f(-2) .

Thus ,

=> R = f(-2)

=> R = -2 + a•(-2)² + 5

=> R = -2 + 4a + 5

=> R = 4a + 3

Also ,

If g(x) is divided by (x + 2) then the remainder R' will be given as g(-2) .

Thus ,

=> R' = g(-2)

=> R' = -2 - 2(-2)² + a

=> R' = -2 - 8 + a

=> R' = a - 10

Now ,

According to the question , if f(x) and g(x) is divided by (x + 2) leave the same remainder .

Thus ,

=> R = R'

=> 4a + 3 = a - 10

=> 4a - a = -10 - 3

=> 3a = -13

=> a = -13/3

Hence a = -13/3 .

Answer 2

Answer:

<b><i> Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 4x + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)

⇒ p(2) = 2(2)3 +a(2)2 + 3(2) – 5

⇒p(2) = 16 + 4a +6 – 5

⇒p(2) = 17 + 4a

Similarly, q(2) = (2)3 + (2)2 + –4(2) + a

⇒ q(2) = 8 + 4 –8 + a

⇒ q(2) = 4 + a

Since they both leave the same remainder, so p(2) = q(2)

⇒ 17 + 4a = 4 + a

⇒ 13 = -3a

\to\sf\:a=\dfrac{-13}{3}→a=3−13

∴ The value of a is –13/3