Question

Resolve (x³-6x²+10x-2)/x²-5x+6 into partial fraction.​

Answer 1

Answer:

Step-by-step explanation:

$\frac{X^{2}-6x^{2} + 10x -2 }{x^{2}-5x + 6 } = x - 1 + \frac{-x + 4}{x^{2 } - 5x + 6} \\We have \frac{-x+4}{x^{2}- 5x + 6} = \frac{x + 4 }{(x - 2)(x-3)}$

$so \\lets \\\frac{-x + 4}{(x-2)(x-3)} = \frac{A}{x-2} = \frac{B}{x-3} Then \\- x + 4 = A (x-3) + B (x-2)$

$Putting \\x-3 = 0 \\or x=3 we will get = \frac{-x+4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3} then\\\\-x+4 = A (x-3) + B(x-2)$

$Putting\\x-2 \\or x=2\\2=A(2-3)=A=-2$

$Therefore \frac{-x +4 }{(x-2) (x-3)} = -\frac{2}{x-2} + \frac{1}{x-3} \\Hence \frac{x^{3}-6x^{3} + 10x - 2 }{x^{2} -5x + 6}= x-1 - \frac{2}{x-2} + \frac{1}{x-3}$