How do you express $\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)}$ $(3x^2 - 4x - 2) / [(x-1)(x-2)]$ in partial fractions?

Question

How do you express $\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)}$ $(3x^2 - 4x - 2) / [(x-1)(x-2)]$ in partial fractions?

1 Answer

Impact of this question

2963 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-07T21:16:51

$\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)} = 3 + \frac{3}{x - 1} + \frac{2}{x - 2}$ $(3x^2-4x-2)/((x-1)(x-2)) = 3 + 3/(x-1) + 2/(x-2)$

Explanation:

$\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)}$ $(3x^2-4x-2)/((x-1)(x-2))$

$= \frac{3 x^{2} - 4 x - 2}{x^{2} - 3 x + 2}$ $=(3x^2-4x-2)/(x^2-3x+2)$

$= \frac{3 x^{2} - 9 x + 6 + 5 x - 8}{x^{2} - 3 x + 2}$ $=(3x^2-9x+6+5x-8)/(x^2-3x+2)$

$= \frac{3 (x^{2} - 3 x + 2) + 5 x - 8}{x^{2} - 3 x + 2}$ $=(3(x^2-3x+2)+5x-8)/(x^2-3x+2)$

$= 3 + \frac{5 x - 8}{x^{2} - 3 x + 2}$ $=3 + (5x-8)/(x^2-3x+2)$

$= 3 + \frac{5 x - 8}{(x - 1) (x - 2)}$ $=3 + (5x-8)/((x-1)(x-2))$

$= 3 + \frac{A}{x - 1} + \frac{B}{x - 2}$ $=3 + A/(x-1) + B/(x-2)$

Use Heaviside's cover-up method to find:

$A = \frac{5 (1) - 8}{(1) - 2} = \frac{- 3}{- 1} = 3$ $A=(5(1)-8)/((1)-2) = (-3)/(-1) = 3$

$B = \frac{5 (2) - 8}{(2) - 1} = \frac{2}{1} = 2$ $B=(5(2)-8)/((2)-1) = 2/1 = 2$

So:

$\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)} = 3 + \frac{3}{x - 1} + \frac{2}{x - 2}$ $(3x^2-4x-2)/((x-1)(x-2)) = 3 + 3/(x-1) + 2/(x-2)$

Science

Math

Humanities

... and beyond

How do you express $\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)}$ $(3x^2 - 4x - 2) / [(x-1)(x-2)]$ in partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you express 3x2−4x−2(x−1)(x−2) (3x^2 - 4x - 2) / [(x-1)(x-2)] in partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you express $\frac{3 x^{2} - 4 x - 2}{(x - 1) (x - 2)}$ $(3x^2 - 4x - 2) / [(x-1)(x-2)]$ in partial fractions?