Question

If (x^3+ax^2+bx+6)has (x-2)as a factor and leaves a remainder 3 when divided by (x-3),find the value of a and b

Answer 1

★ QUADRATIC EQUATIONS ★

Let the given polynomial be p(x) = x³ + ax² + bx + 6.

Given: p(x) is divisible by (x – 2),
           ∴ p(2) = 0

Now, Put x = 2 in p(x)

So, p(2) = (2)³ + a(2)² + b(2) + 6
  ⇒ 0 = 8 + 4a + 2b + 6
  ⇒ 4a + 2b = - 14
  ⇒ 2a + b = - 7  → (1)

Now, It is also given that when p(x) is divided by (x – 3) the remainder is 3.

Again, Put x = 3 in p(x)

So, p(3) = (3)³ + a(3)² + b(3) + 6
   ⇒ 3 = 27 + 9a + 3b + 6
   ⇒ 9a + 3b = - 30
   ⇒ 3a + b = - 10  → (2)

Now, Subtract (2) from (1)
 
              2a + b = - 7
          −  3a + b = - 10 
          ---------------------
We get, - a = 3

∴ a = – 3

Now, Substitute a = - 3 in (1),

We get, 2(– 3) + b = –7
∴ b = –1

Answer 2

Answer:

answer

Step-by-step explanation:

Let the given polynomial be p(x) = x³ + ax² + bx + 6.

Given: p(x) is divisible by (x – 2),

          ∴ p(2) = 0

Now, Put x = 2 in p(x)

So, p(2) = (2)³ + a(2)² + b(2) + 6

 ⇒ 0 = 8 + 4a + 2b + 6

 ⇒ 4a + 2b = - 14

 ⇒ 2a + b = - 7  → (1)

Now, It is also given that when p(x) is divided by (x – 3) the remainder is 3.

Again, Put x = 3 in p(x)

So, p(3) = (3)³ + a(3)² + b(3) + 6

  ⇒ 3 = 27 + 9a + 3b + 6

  ⇒ 9a + 3b = - 30

  ⇒ 3a + b = - 10  → (2)

Now, Subtract (2) from (1)

 

             2a + b = - 7

         −  3a + b = - 10  

         ---------------------

We get, - a = 3

∴ a = – 3

Now, Substitute a = - 3 in (1),

We get, 2(– 3) + b = –7

∴ b = –1

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