How do you express #x^2/(x^2 + x +2)# in partial fractions?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

6
Sep 19, 2017

Answer:

Partial fractions of #x^2/(x^2+x+2)# are #1-(x+2)/(x^2+x+2#

Explanation:

As the denominator #x^2+x+2# is quadratic and its determinant #(-b+-sqrt(b^2-4ac))/(2a)# is not rational (as #sqrt(b^2-4ac)=sqrt(-7)# is not rational), its partial fractions will be of type #(Ax+B)/(x^2+x+2)#.

But, degree of numerator is #2# hence let us write #x^2/(x^2+x+2)# as

#x^2/(x^2+x+2)=1-(x+2)/(x^2+x+2#

Hence partial fractions of #x^2/(x^2+x+2)# are #1-(x+2)/(x^2+x+2#

Was this helpful? Let the contributor know!
1500