# What must be added to x3 − 3x2 − 12x + 19 so

Question:

What must be added to $x^{3}-3 x^{2}-12 x+19$ so that the result is exactly divisible by $x^{2}+x-6$

Solution:

Here, $p(x)=x^{3}-3 x^{2}-12 x+19$

$g(x)=x^{2}+x-6$

by division algorithm, when p(x) is divided by g(x) , the remainder will be a linear expression in x

Let, r(x) = ax + b is added to p(x)

⟹ f(x) = p(x) + r(x)

$=x^{3}-3 x^{2}-12 x+19+a x+b$

$f(x)=x^{3}-3 x^{2}+x(a-12)+19+b$

We know that, $g(x)=x^{2}+x-6$

First, find the factors for g(x)

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

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

= (x + 3) (x - 2) are the factors

From, factor theorem when (x + 3) and (x - 2) are the factors of f(x) the f(-3) = 0 and f(2) = 0

Let, x + 3 = 0

⟹ x = -3

Substitute the value of x in f(x)

$f(-3)=(-3)^{3}-3(-3)^{2}+(-3)(a-12)+19+b$

= -27 - 27 - 3a + 24 + 19 + b

= -3a + b + 1 ...... 1

Let, x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

$f(2)=(2)^{3}-3(2)^{2}+(2)(a-12)+19+b$

= 8 - 12 + 2a - 24 + b

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

Solve equations 1 and 2

- 3a + b = -1

2a + b = 9

(-) (-) (-)

- 5a = - 10

a = 2

substitute the value of a in eq 1

⟹ -3(2) + b = -1

⟹ - 6 + b = - 1

⟹ b = - 1 + 6

⟹ b = 5

∴ r(x) = ax + b

= 2x + 5

$\therefore x^{3}-3 x^{2}-12 x+19$ is divided by $x^{2}+x-6$ when it is added by $2 x+5$