How do you express $\frac{17 x - 50}{x^{2} - 6 x + 8}$ $(17x-50)/(x^(2)-6x+8)$ in partial fractions?

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Answer 1 · 2018-06-13T21:58:39

$\frac{17 x - 50}{x^{2} - 6 x + 8} = \frac{8}{x - 2} + \frac{9}{x - 4}$ $(17x-50)/(x^2-6x+8) = 8/(x-2)+9/(x-4)$

Note that:

$x^{2} - 6 x + 8 = (x - 2) (x - 4)$ $x^2-6x+8 = (x-2)(x-4)$

So:

$\frac{17 x - 50}{x^{2} - 6 x + 8} = \frac{A}{x - 2} + \frac{B}{x - 4}$ $(17x-50)/(x^2-6x+8) = A/(x-2)+B/(x-4)$

Multiplying both sides by $x^{2} - 6 x + 8$ $x^2-6x+8$ this becomes:

$17 x - 50 = A (x - 4) + B (x - 2)$ $17x-50 = A(x-4)+B(x-2)$

Putting $x = 2$ $x=2$ , we find:

$- 16 = 34 - 50 = 17 (2) - 50 = A ((2) - 4) = - 2 A$ $-16 = 34-50 = 17(color(blue)(2))-50 = A((color(blue)(2))-4) = -2A$

Hence $A = 8$ $A=8$

Putting $x = 4$ $x=4$ , we find:

$18 = 17 (4) - 50 = B ((4) - 2) = 2 B$ $18 = 17(color(blue)(4))-50 = B((color(blue)(4))-2) = 2B$

Hence $B = 9$ $B=9$

How do you express (17x-50)/(x^(2)-6x+8)17x−50x2−6x+8(17x-50)/(x^(2)-6x+8) in partial fractions?