Question

what should be added to 1/x^2-7x+12to get 2/x^2-6x+8

Answer 1

$\frac{1}{ {x}^{2} - 7x + 12 } \\ \\ \frac{1}{(x - 4)(x - 3)}$
$\frac{1}{ {x}^{2} - 6x + 8 } \\ \\ \frac{1}{(x - 4)(x - 2)}$


let S should be added to
$\frac{1}{(x - 4)(x - 3)}$
so
S =
$\frac{1}{( x - 4)(x - 2)} \: - \frac{1}{(x - 4)(x - 3) } \\ \\ \frac{(x - 3 ) - (x - 2)}{(x - 4)(x - 3)(x - 2)} \\ \\ \frac{ - 1}{(x - 4)(x - 3)(x - 2)}$

-1 / (x^2-6x+8)(x-3)
-1/x^3-6x^2+8x-3x^2+18x-24

Answer is
$\frac{ - 1}{ {x}^{3} \: - 9x^{2} + 26x - 24 }$

Answer 2

Step-by-step explanation:

Let the number to be added be 'a'

so,

$\frac{1}{x {}^{2} - 7x + 12 } + a = \frac{2}{x {}^{2} - 6x + 8} \\ a = \frac{2}{x {}^{2} - 6x + 8} - \frac{1}{x {}^{2} - 7x + 12 } \\ a = \frac{2}{(x - 2)(x - 4)} - \frac{1}{(x - 3)(x - 4)} \\ a = \frac{2(x - 3) - 1(x - 2)}{(x - 2)(x - 3)(x - 4)} \\ a = \frac{2x - 6 - x + 2}{(x - 2)(x - 3)(x - 4)} \\ a = \frac{(x - 4)}{(x - 2)(x - 3)(x - 4) } \\ a = \frac{1}{(x - 2)(x - 3)} \\ a = \frac{1}{x {}^{2} - 5x + 6}$