How do you write the partial fraction decomposition of the rational expression (x+13)/(x^3+2x^2-5x-6)x+13x3+2x25x6?

1 Answer
Mar 26, 2017

The answer is =1/(x-2)-2/(x+1)+1/(x+3)=1x22x+1+1x+3

Explanation:

We need to factorise the denominator

Let f(x)=x^3+2x^2-5x-6f(x)=x3+2x25x6

Then

f(2)=8+8-10-6=0f(2)=8+8106=0

Therefore,

(x-2)(x2) is a factor of f(x)f(x)

We can do a long division

color(white)(aaaa)aaaax^3+2x^2-5x-6x3+2x25x6color(white)(aaaa)aaaa|x-2x2

color(white)(aaaa)aaaax^3-2x^2x32x2color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa|x^2+4x+3x2+4x+3

color(white)(aaaaa)aaaaa0+4x^2-5x0+4x25x

color(white)(aaaaaaa)aaaaaaa+4x^2-8x+4x28x

color(white)(aaaaaaaa)aaaaaaaa+0+3x-6+0+3x6

color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa+3x-6+3x6

color(white)(aaaaaaaaaaaa)aaaaaaaaaaaa+0-0+00

so,

x^3+2x^2-5x-6=(x-2)(x^2+4x+3)x3+2x25x6=(x2)(x2+4x+3)

=(x-2)(x+1)(x+3)=(x2)(x+1)(x+3)

Now, we perform the decomposition into partial fractions

(x+13)/(x^3+2x^2-5x-6)=(x+13)/((x-2)(x+1)(x+3))x+13x3+2x25x6=x+13(x2)(x+1)(x+3)

=A/(x-2)+B/(x+1)+C/(x+3)=Ax2+Bx+1+Cx+3

=(A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1))/((x-2)(x+1)(x+3))=A(x+1)(x+3)+B(x2)(x+3)+C(x2)(x+1)(x2)(x+1)(x+3)

The denominators are the same, we compare the numerators

x+13=A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1)x+13=A(x+1)(x+3)+B(x2)(x+3)+C(x2)(x+1)

Let, x=2x=2, =>,15=15A15=15A, =>,A=1A=1

Let, x=-1x=1, =>, 12=-6B12=6B, =>, B=-2B=2

Let, x=-3x=3, =>, 10=10C10=10C, =>, C=1C=1

So,

(x+13)/(x^3+2x^2-5x-6)=1/(x-2)-2/(x+1)+1/(x+3)x+13x3+2x25x6=1x22x+1+1x+3