What must be subtracted from x3-6x2-15x+80 so that the result is exactly divisible by x2+x-12

By divisible algorithm, when  `p(x) = x^3 - 6x^2 - 15x + 80` is divided by  `x^2 + x -12` the reminder is a linear polynomial

Let r(x) = a(x) + b be subtracted from p(x) so that the result is divisible by q(x).

Let

`f(x) = p(x) -q(x)`

` = x^3 - 6x^2 - 15x + 80 - (ax + b)`

` = x^3 -6x^2 - (a+15)x + 80 -b`

We have,

`q(x) = x^2 + x - 12`

` = x^2 + 4x -3x -12`

` = (x+4)(x-3)`

Clearly, (x+4)and (x-3)are factors of q(x), therefore, f(x) will be divisible by q(x) if (x+4) and (x-3) are factors of f(x), i.e. f (−4) and f (3) are equal to zero.

Therefore,

`f(-4) = (-4)^3 -6(-4)^2 - (a+15)(-4) + 80 -b = 0`

                            `-64 -96 + 4a + 60 + 80 -b = 0`

                                                         ` -20 +4a -b = 0`

                                                                  `+4a -b = 20`

and

`f(3) = (3)^2 - (a+15)(3) + 80 -b = 0`

`27 - 25 - 3a - 45 + 80 -b =0`

                                    ` -3a - b = 8`

                                        `3a +b = 8`

Adding (i) and (ii), we get,

 a=4

Putting this value in equation (i), we get,

b=-4

Hence, `x^3 - 6x^2 - 15x + 80` will be divisible by `x^2 + x - 12,` if 4 x − 4 is subtracted from it.

Page 2

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.