Question

If p ( x ) = x^3 - ax^2 + bx + 3 leaves a remainder -19 when divided by ( x + 2 ) and
remainder 17 when divided by ( x - 2) ,then prove that a + b = 6 .
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Answer 1

Given that :

when p(x) = x^3 - ax^2 + bx + 3 is divided by (x+2) it leaves a remainder of -19 and

when it is divided by (x-2) then it leaves a remainder of 17.

Let, the quotient be A and B when the given polynomial will be divided by (x+2) and (x-2) respectively.

Now, according to the question :

On putting x = -2 in the given polynomial we will get the remainder -19

p(-2) = -19

=> (-2)^3 -a(-2)^2 + b(-2) + 3 = -19

=> -8 - 4a - 2b + 3 = -19

=> 4a + 2b = 14

=> 2a + b = 7 ... (1)

On putting x = 2 in the given polynomial we will get the remainder 17

p(2) = 17

=> (2)^3 -a(2)^2 + b(2) + 3 = 17

=> 8 - 4a + 2b + 3 = 17

=> -4a + 2b = 6

=> -2a + b = 3 ... (2)

On solving equations (1) and (2) :

2a + b = 7 ... (1)

-2a + b = 3 ... (2)

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On adding :

2b = 10

=> b = 5

On putting the value of b in equation (1) :

2a = 7 - b

=> 2a = 7 - 5

=> 2a = 2

=> a = 1

(Required values are : a = 1 and b = 5)

To prove :

a + b = 6

=> 1 + 5 = 6

=> 6 = 6

L.H.S. = R.H.S.

HENCE PROVED ✔✔

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Hope it helps ☺