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what must be added to x raise to 3-3x square-12+19 so that the result is exactly divisible by x square +x-6

${Let p(x) =x}^{3} - 3 x^{2} - 12 x + 19 q(x)= x^{2} + x - 6 when {p(x) =x}^{3} - 3 x^{2} - 12 x + 19 {is divided by q(x)=x}^{2} + x - 6, the remainder will be a linear expression Let r(x) = a x + b be added to p(x) so that the result is divisible by q(x) Let f(x)=p(x)+r(x) \Rightarrow f (x) = x^{3} - 3 x^{2} - 12 x + 19 + (a x + b) f (x) = x^{3} - 3 x^{2} - 12 x + 19 + a x + b {q(x)=x}^{2} + x - 6 using factorization we get {q(x)=x}^{2} + 3 x - 2 x - 6 q(x)=x(x+3) - 2 (x + 3) q(x)=(x+3)(x - 2) As (x+3) and (x - 2) factors of q (x) and as f(x) is divisible by q(x) ∴ (x + 3) and (x - 2) are also factors of f(x) \Rightarrow f (- 3) = (- 3)^{3} - 3 (- 3)^{2} - 12 (- 3) + 19 + a (- 3) + b 0 = - 27 - 27 + 36 + 19 - 3 a + b 0 = 1 - 3 a + b - - - - - - (1) f (2) = 0 f (2) = (2)^{3} - 3 (2)^{2} - 12 (2) + 19 + a (2) + b 0 = 8 - 12 - 24 + 19 + 2 a + b 0 = 2 a + b - 9 - - - - - - (2) Solving (1) and (2) we get a=2 b=5 ax+b=2x+5 {thus the expression x}^{3} - 3 x^{2} - 12 x + 19 {is exactly divisible by x}^{2} + x - 6 when 2x+5 is added to it$

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