# If (x^2-5x-6)/(x^3+ax^2+bx+c simplifies to 1/(x+2), then a + b + c = ?

Nov 9, 2017

$- 21.$

#### Explanation:

Given that, $\frac{{x}^{2} - 5 x - 6}{{x}^{3} + a {x}^{2} + b x + c} = \frac{1}{x + 2} .$

By Cross Multiplication, we get,

${x}^{3} + a {x}^{2} + b x + c = \left({x}^{2} - 5 x - 6\right) \left(x + 2\right) , i . e . ,$

$= \left(x - 6\right) \left(x + 1\right) \left(x + 2\right) ,$

$= x \left({x}^{2} - 5 x - 6\right) + 2 \left({x}^{2} - 5 x - 6\right) ,$

$= \left({x}^{3} - 5 {x}^{2} - 6 x\right) + \left(2 {x}^{2} - 10 x - 12\right) .$

$\Rightarrow {x}^{3} + a {x}^{2} + b x + c = {x}^{3} - 3 {x}^{2} - 16 x - 12.$

Comparing the co-efficients of polynomials on both sides, we have

$a = - 3 , b = - 16 , \mathmr{and} , c = - 12.$

$\therefore a + b + c = - 21 ,$ as desired.

Enjoy Maths.!