What must be added to $x^{3}-6 x^{2}-15 x+80$ so that the result is exactly divisible by $x^{2}+x-12$
Let, $p(x)=x^{3}-6 x^{2}-15 x+80$
$q(x)=x^{2}+x-12$
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 subtracted from p(x), so that p(x) - q(x) is divisible by q(x) let f(x) = p(x) - q(x)
$q(x)=x^{2}+x-12$
$=x^{2}+4 x-3 x-12$
$=x(x+4)(-3)(x+4)$
$=(x+4),(x-3)$
clearly, (x - 3) and (x + 4) are factors of q(x)
so, f(x) will be divisible by q(x) if (x - 3) and (x + 4) are factors of q(x)
from, factor theorem
f(-4) = 0 and f(3) = 0
$\Rightarrow \mathrm{f}(3)=33-6(3)^{2}-3(\mathrm{a}+15)+80-\mathrm{b}=0$
= 27 - 54 - 3a - 45 + 80 - b
= -3a - b + 8 .... 1
Similarly,
f(- 4) = 0
⟹ f(- 4)
$\Rightarrow(-4)^{3}-6(-4)^{2}-(-4)(a+15)+80-b=0$
⟹ - 64 - 96 - 4a + 60 + 80 - b = 0
⟹ 4a - b - 20 = 0 .... 2
Substract eq 1 and 2
⟹ 4a - b - 20 - 8 + 3a + b = 0
⟹ 7a - 28 = 0
⟹ a = 28/7
⟹ a = 4
Put a = 4 in eq 1
⟹ - 3(4) - b = - 8
⟹ - b - 12 = - 8
⟹ - b = - 8 + 12
⟹ b = - 4
Substitute a and b values in r(x)
⟹ r(x) = ax + b
= 4x - 4
Hence, p(x) is divisible by q(x), if r(x) = 4x - 4 is subtracted from it