What must be added to 3x ^ 3 + x ^ 2 - 22x + 9 so that the result is exactly divisible by

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Let p (x) = 3x3 + x2 - 22x + 9 and q (x) = 3x2 + 7x - 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)

              = 3x3 + x2 – 22x + 9 + (ax + b)

              = 3x3 + x2 + x (a – 22) + b + 9

We have,

q (x) = 3x2 + 7x – 6

q (x) = 3x (x + 3) – 2 (x + 3)

q (x) = (3x – 2) (x + 3)

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

Therefore, f(x) will be divisible by q(x), if (3x – 2) and (x + 3) are factors of f(x).

i.e. f (2/3) = 0 and f (-3) = 0    [∵ 3x – 2 = 0, x = 2/3 and x + 3 = 0, x = -3]

f (2/3) = 0

⇒ 6a + 9b – 39 = 0

⇒ 3 (2a + 3b – 13) = 0

⇒ 2a + 3b – 13 = 0     (i)

Similarly,

f (-3) = 0

⇒ 3 (-3)3 + (-3)2 + (-3) (a – 2x) + b + 9 = 0

⇒ -81 + 9 – 3a + 66 + b + 9 = 0

⇒ b – 3a + 3 = 0

⇒ 3 (b – 3a + 3) = 0

⇒ 3b – 9a + 9 = 0     (ii)

Subtract (i) from (ii), we get

3b – 9a + 9 – (2a + 3b – 13) = 0

3b – 9a + 9 – 2a – 3b + 13 = 0

⇒ -11a + 22 = 0

⇒ a = 2

Putting value of a in (i), we get

⇒ b = 3

Putting the values of a and b in r (x) = ax + b, we get

r (x) = 2x + 3

Hence, p (x) is divisible by q (x) if r (x) = 2x + 3 is divisible by it.

By division algorithm, when p(x) = 3x3 + x2 − 22x + 9  is divided by `3x^2 + 7x - 6,`the reminder is a linear polynomial. So, let r(x) = ax + b be added to p(x) so that the result is divisible by q(x)

Let

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

` = 3x^2 + x^2 - 22x + 9 ax +b`

` = 3x^2 + x^2 +(a- 22) x + 9 + b`

We have

\[q\left( x \right) = 3 x^2 + 7x - 6\]

\[ = 3 x^2 + 9x - 2x - 6\]

\[ = 3x\left( x + 3 \right) - 2\left( x + 3 \right)\]

\[ = \left( 3x - 2 \right) \left( x + 3 \right)\]

Clearly, 

\[\left( 3x - 2 \right)\] and  \[\left( x + 3 \right)\]

are factors of q(x).

Therefore, f(x) will be divisible by q(x) if (3x - 2)and (x + 3)are factors of f(x), i.e.,

`f (2/3)`and f(−3) are equal to zero.

Now,

\[f\left( \frac{2}{3} \right) = 0\]

\[ \Rightarrow 3 \left( \frac{2}{3} \right)^3 + \left( \frac{2}{3} \right)^2 + \left( a - 22 \right)\left( \frac{2}{3} \right) + 9 + b = 0\]

\[ \Rightarrow 3 \times \frac{8}{27} + \frac{4}{9} + \frac{2a}{3} - \frac{44}{3} + 9 + b = 0\]

\[ \Rightarrow \frac{8}{9} + \frac{4}{9} - \frac{44}{3} + 9 + \frac{2a}{3} + b = 0\]

\[ \Rightarrow \frac{8 + 4 - 132 + 81}{9} + \frac{2a}{3} + b = 0\]

\[ \Rightarrow - \frac{39}{9} + \frac{2a}{3} + b = 0\]

\[ \Rightarrow \frac{2a}{3} + b = \frac{13}{3}\]

\[ \Rightarrow 2a + 3b = 13 . . . . . . . . \left( i \right)\]

And

\[f\left( - 3 \right) = 0\]

\[ \Rightarrow 3 \left( - 3 \right)^3 + \left( - 3 \right)^2 + \left( a - 22 \right)\left( - 3 \right) + 9 + b = 0\]

\[ \Rightarrow - 81 + 9 - 3a + 66 + 9 + b = 0\]

\[ \Rightarrow - 3a + b = - 3 \]

\[ \Rightarrow b = - 3 + 3a . . . . . . . . . \left( ii \right)\]

Substituting the value of b from (ii) in (i), we get,

\[2a + 3\left( 3a - 3 \right) = 13\]

\[ \Rightarrow 2a + 9a - 9 = 13\]

\[ \Rightarrow 11a = 13 + 9\]

\[ \Rightarrow 11a = 22\]

\[ \Rightarrow a = 2\]

Now, from (ii), we get 

\[b = - 3 + 3\left( 2 \right) = - 3 + 6 = 3\]

So, we have a = 2  and  b = 3

Hence, p(x) is divisible by q(x), if  2x + 3is added to it.