# What must be added to x3

Question:

What must be added to $x^{3}-6 x^{2}-15 x+80$ so that the result is exactly divisible by $x^{2}+x-12$

Solution:

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