**Question:**

What must be added to $3 x^{3}+x^{2}-22 x+9$ so that the result is exactly divisible by $3 x^{2}+7 x-6$

**Solution:**

Let, $p(x)=3 x^{3}+x^{2}-22 x+9$ and $q(x)=3 x^{2}+7 x-6$

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

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

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

$\Rightarrow f(x)=3 x^{3}+x^{2}-22 x+9(a x+b)$

$\Rightarrow=3 x^{3}+x^{2}+x(a-22)+b+9$

We know that,

$q(x)=3 x^{2}+7 x-6$

$=3 x^{2}+9 x-2 x-6$

= 3x(x + 3) - 2(x + 3)

= (3x - 2)(x + 3)

So, f(x) is divided by q(x) if (3x - 2) and (x + 3) are the factors of f(x)

From, factor theorem

f(2/3) = 0 and f(-3) = 0

let, 3x - 2 = 0

3x = 2

x = 2/3

$\Rightarrow f(2 / 3)=3(2 / 3)^{3}+(2 / 3)^{2}+(2 / 3)(a-22)+b+9$

= 3(8/27) + 4/9 + 2/3a − 44/3 + b + 9

= 12/9 + 2/3a − 44/3 + b + 9

$=\frac{12+6 a-132+9 b+81}{9}$

Equate to zero

$=\frac{12+6 a-132+9 b+81}{9}=0$

⟹ 6a + 9b - 39 = 0

⟹ 3(2a + 3b - 13) = 0

⟹ 2a + 3b - 13 = 0 .... 1

Similarly,

Let, x + 3 = 0

⟹ x = - 3

$\Rightarrow f(-3)=3(-3)^{3}+(-3)^{2}+(-3)(a-22)+b+9$

= – 81 + 9 - 3a + 66 + b + 9

= – 3a + b + 3

Equate to zero

- 3a + b + 3 = 0

Multiply by 3

- 9a + 3b + 9 = 0 ... 2

Substact eq 1 from 2

⟹ -9a + 3b + 9 -2a - 3b + 13 = 0

⟹ -11a + 22 = 0

⟹ -11a = -22

⟹ a = 22/11

⟹ a = 2

Substitute a value in eq 1

⟹ - 3(2) + b = - 3

⟹ - 6 + b = - 3

⟹ b = - 3 + 6

⟹ b = 3

Put the values in r(x)

r(x) = ax + b

= 2x + 3

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

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