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

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Nimo N. Share
Mar 8, 2018

Answer:

See below.

Explanation:

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

Given: # (x^2)/(x^2 + x + 2) #. Note that the degrees of the numerator and denominator are the same. And, as a bonus, the denominator is not factorable with real numbers.

The discriminant for the denominator:
# color(red)( b^2 - 4ac = (1)^2 - 4(1)(2) = 1 - 8 = -7 # # , \ # tells the story.

As in reference A), above, a case such as ours, the procedure is to do long division to obtain a fraction that might be written in partial fraction form.

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 #
# x^2 + x + 2 bar(")" x^2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2 + x + 2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ "---------------"#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - x - 2 #

So,
# color(red)( (x^2)/(x^2 + x + 2) = 1 - (x + 2)/(x^2 + x + 2) #

Concentrate on the fraction, then we can put it back together. Since the denominator can not be factored with real numbers, we leave it alone and write:
# (x + 2)/(x^2 + x + 2) = (Ax + B)/(x^2 + x + 2) #.

Then, solve for A and B.
The denominators are the same, so the numerators must be equal.
# (x + 2) = (Ax + B) #
The only way that can happen is if A = 1, and B = 2.

Putting it back together:
# color(brown)( (x^2)/(x^2 + x + 2) = 1 - (1x + 2)/(x^2 + x + 2) #.
This expression looks familiar, but it is all we can do!

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