What must be added to $x^{3}-3 x^{2}-12 x+19$ so that the result is exactly divisible by $x^{2}+x-6$
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 $2 x+5$
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