How do you write the partial fraction decomposition of the rational expression $\frac{x^{2}}{(x - 1) (x + 2)}$ $x^2/((x-1)(x+2))$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2}}{(x - 1) (x + 2)}$ $x^2/((x-1)(x+2))$ ?

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Answer 1 · 2018-03-08T12:42:15

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

Explanation:

We need to write these in terms of each factors.

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

$x^{2} = A (x + 2) + B (x - 1)$ $x^2=A(x+2)+B(x-1)$

Putting in $x = - 2$ $x=-2$ :
${(- 2)}^{2} = A (- 2 + 2) + B (- 2 - 1)$ $(-2)^2=A(-2+2)+B(-2-1)$
$4 = - 3 B$ $4=-3B$
$B = - \frac{4}{3}$ $B=-4/3$

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

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

Answer 2 · 2018-03-08T12:53:37

$1 + \frac{1}{3} \cdot \frac{1}{x - 1} - \frac{4}{3} \cdot \frac{1}{x + 2}$ $1+1/3*1/(x-1)-4/3*1/(x+2)$

Explanation:

$\frac{x^{2}}{(x - 1) (x + 2)}$ $x^2/[(x-1)(x+2)]$

= $\frac{(x - 1) (x + 2) + x^{2} - (x - 1) (x + 2)}{(x - 1) (x + 2)}$ $[(x-1)(x+2)+x^2-(x-1)(x+2)]/[(x-1)(x+2)]$

= $1 - \frac{(x - 1) (x + 2) - x^{2}}{(x - 1) (x + 2)}$ $1-[(x-1)(x+2)-x^2]/[(x-1)(x+2)]$

= $1 - \frac{x - 2}{(x - 1) (x + 2)}$ $1-(x-2)/[(x-1)(x+2)]$

Now, I decomposed fraction into basic ones,

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

After expanding denominator,

$A \cdot (x + 2) + B \cdot (x - 1) = x - 2$ $A*(x+2)+B*(x-1)=x-2$

Set $x = - 2$ $x=-2$ , $- 3 B = - 4$ $-3B=-4$ , so $B = \frac{4}{3}$ $B=4/3$

Set $x = 1$ $x=1$ , $3 A = - 1$ $3A=-1$ , so $A = - \frac{1}{3}$ $A=-1/3$

Hence,

$\frac{x - 2}{(x - 1) (x + 2)} = - \frac{1}{3} \cdot \frac{1}{x - 1} + \frac{4}{3} \cdot \frac{1}{x + 2}$ $(x-2)/[(x-1)(x+2)]=-1/3*1/(x-1)+4/3*1/(x+2)$

Thus,

$\frac{x^{2}}{(x - 1) (x + 2)}$ $x^2/[(x-1)(x+2)]$

= $1 - \frac{x - 2}{(x - 1) (x + 2)}$ $1-(x-2)/[(x-1)(x+2)]$

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

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2}}{(x - 1) (x + 2)}$ $x^2/((x-1)(x+2))$ ?

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