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

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Answer 1 · 2018-06-13T21:58:39

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

Note that:

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

So:

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

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

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

Putting $x=2$ , we find:

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

Hence $A=8$

Putting $x=4$ , we find:

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

Hence $B=9$

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