`(x + 6)/(x^3 - 3x^2 - 4x + 12)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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`(x+6)/(x^3-3x^2-4x+12)`

To decompose this to partial fractions, factor the denominator.

`x^3 - 3x^2-4x + 12`

`= (x^3-3x^2) + (-4x + 12)`

`= x^2(x-3) - 4(x -3)`

`=(x-3)(x^2-4)`

`=(x-3)(x-2)(x+2)`

Then, write a fraction for each factor. Since the numerators are still unknown, assign a variable for each numerator.

`A/(x-3)`   ,   `B/(x-2)`   and   `C/(x+2)`

Add these three fractions and set it equal to the given fraction.

`(x+6)/((x-3)(x-2)(x+2)) = A/(x-3)+B/(x-2)+C/(x+2)`

To solve for the values of A, B and C, eliminate the fractions in the equation. So, multiply both sides by the LCD.

`(x-3)(x-2)(x+2) *(x+6)/((x-3)(x-2)(x+2)) = (A/(x-3)+B/(x-2)+C/(x+2))*(x-3)(x-2)(x+2)`

`x+6=A(x-2)(x+2) + B(x-3)(x+2) + C(x-3)(x-2)`

Then, plug-in the roots of each factor.

For the factor (x-2), its root is x=2.

`2+6 = A(2-2)(2+2)+B(2-3)(2+2)+C(2-3)(2-2)`

`8=A(0)(4)+B(-1)(4)+C(-1)(0)`

`8=-4B`

`8/(-4)=(-4B)/(-4)`

`-2=B`

For the factor (x + 2), its root is x=-2.

`-2+6= A(-2-2)(-2+2) + B(-2-3)(-2+2)+C(-2-3)(-2-2)`

`4=A(-4)(0)+B(-5)(0)+C(-5)(-4)`

`4=20C`

`4/20=(20C)/20`

`1/5=C`

And for the factor (x-3), its root is x=3.

`3+6=A(3-2)(3+2) + B(3-3)(3+2) + C(3-3)(3-2)`

`9=A(1)(5) + B(0)(5) + C(0)(1)`

`9=5A`

`9/5=(5A)/5`

`9/5=A`

So the partial fraction decomposition of the given rational expression is:

`(9/5)/(x-3) + (-2)/(x-2)+ (1/5)/(x+2)`

And this simplifies to:

`= 9/(5(x-3)) - 2/(x-2) +1/(5(x+2))`

 

To check, express the three fractions with same denominators.

`9/(5(x-3)) - 2/(x-2) +1/(5(x+2)) `

`= 9/(5(x-3))*((x-2)(x+2))/((x-2)(x+2)) - 2/(x-2)*(5(x-3)(x+2))/(5(x-3)(x+2)) + 1/(5(x+2))*((x-3)(x-2))/((x-3)(x-2))`

`= (9(x-2)(x+2))/(5(x-3)(x-2)(x+2))- (10(x-3)(x+2))/(5(x-3)(x-2)(x+2))+((x-3)(x-2))/(5(x-3)(x-2)(x+2))`

`=(9(x^2-4))/(5(x-3)(x-2)(x+2)) - (10(x^2-x-6))/(5(x-3)(x-2)(x+2))+(x^2-5x+6)/(5(x-3)(x-2)(x+2))`

Now that they have same denominators, proceed to add/subtract them.

`= (9(x^2-4) - 10(x^2-x-6) + x^2-5x+6)/(5(x-3)(x-2)(x+2))`

`= (9x^2-36-10x^2+10x+60+x^2-5x+6)/(5(x-3)(x-2)(x+2))`

`=(5x+30)/(5(x-3)(x-2)(x+2))`

`= (5(x+6))/(5(x-3)(x-2)(x+2))`

`=(x+6)/((x-3)(x-2)(x+2))`

 

Therefore,  `(x+6)/((x-3)(x-2)(x+2))=9/(5(x-3)) - 2/(x-2) +1/(5(x+2))` .

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