What must be added to x- 3x- 12+ 19 so that the result is exactly divisibly by x- 6?

 


Let p (x) = x3-3x2-12x+19 and q (x) = x2+x-6


By division algorithm, when p (x) is divided by q (x), the remainder is a linear expression in x.


So, let r (x) = ax + b is added to p (x) so that p (x) + r (x) is divisible by q (x).


Let,


f (x) = p (x) + r (x)


        = x3 - 3x2 - 12x + 19 + ax + b


        = x3 – 3x2 + x (a – 12) + b + 19


We have,


q (x) = x2+x-6


        = (x + 3) (x – 2)


Clearly, q (x) is divisible by (x – 2) and (x + 3) i.e. (x – 2) and (x + 3) are factors of q (x)


We have,


f (x) is divisible by q (x)


(x – 2) and (x + 3) are factors of f (x)


From factor theorem,


If (x – 2) and (x + 3) are factors of f (x) then f (2) = 0 and f (-3) = 0 respectively.


f (2) = 0


(2)3 – 3 (2)2 + 2 (a – 12) + b + 19 = 0


⇒ 8 – 12 + 2a – 24 + b + 19 = 0


⇒ 2a + b – 9 = 0    (i)


Similarly,


f (-3) = 0


(-3)3 – 3 (-3)2 + (-3) (a – 12) + b + 19 = 0


⇒ -27 – 27 – 3a + 36 + b + 19 = 0


⇒ b – 3a + 1 = 0    (ii)


Subtract (i) from (ii), we get


b – 3a + 1 – (2a + b – 9) = 0 – 0


⇒ b – 3a + 1 – 2a – b + 9 = 0


⇒ - 5a + 10 = 0


⇒ 5a = 10


⇒ a = 2


Put a = 2 in (ii), we get


b – 3 × 2 + 1 = 0


⇒ b – 6 + 1 = 0


⇒ b – 5 = 0


⇒ b = 5


Therefore, r (x) = ax + b


                          = 2x + 5


Hence, x3 – 3x – 12x + 19 is divisible by x2 + x – 6 when 2x + 5 is added to it.

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