Express in partial fraction
in Other Math Topics by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

(5x²+17x+15)/((x+2)²(x-3))≡

A/(x+2)+B/(x+2)²+C/(x-3).

Therefore:

A(x+2)(x-3)+B(x-3)+C(x+2)²=

Ax²-Ax-6A+Bx-3B+Cx²+4Cx+4C=

(A+C)x²+(-A+B+4C)x+(-6A-3B+4C)≡

       5x²+                17x+                  15.

So, A+C=5, C=5-A; -A+B+4C=17; -6A-3B+4C=15.

4C=17+A-B=15+6A+3B, 2=5A+4B, B=(2-5A)/4.

We now have B and C in terms of A, so substitute:

-A+(2-5A)/4+20-4A=17, 

-4A+2-5A+80-16A=68.

14=25A, A=14/25; C=5-A=111/25; B=(2-5A)/4=-1/5.

We can write these as decimals: A=0.56, B=-0.2, C=4.44.

Partial fractions:

0.56/(x+2)-0.2/(x+2)²+4.44/(x-3).

by Top Rated User (711k points)

Related questions

1 answer
asked Nov 15, 2015 in Other Math Topics by iretomiwa Level 1 User (240 points) | 116 views
1 answer
asked Sep 28, 2016 in Other Math Topics by anonymous | 122 views
1 answer
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
84,410 questions
89,222 answers
1,989 comments
7,503 users