Q. 204.6( 11 Votes )

# What must be adde

Answer :

Let p (x) = *x*^{3}-3*x*^{2}-12*x*+19 and q (x) = *x*^{2}+*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)

= *x*^{3 }- 3*x*^{2 }- 12*x *+ 19 + ax + b

= x^{3} – 3x^{2} + x (a – 12) + b + 19

We have,

q (x) = *x*^{2}+*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, x^{3} – 3x – 12x + 19 is divisible by x^{2} + x – 6 when 2x + 5 is added to it.

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