Hi Davidche and Secretkiller,
I have used the long-division and cover up method to solve the
questions.
However, these two questions can also be solved by using the
comparison method or substitution method ie with the use of long
division for example 2 too.
Question
1
Express x^2/(9-x^2)
in the form A + B/(3-x) + C/(3+x)
and find the values
of A, B and C.
[The trick on this
question is on the constant A]
Answer
Use Long
Division,
x^2/(9-x^2) = -1 +
9/(x^2-1)
= -1 +
9/[(x-3)(x+3)] = A + B/(3-x) + C/(3+x)
Use Cover up
method,
= -1 + 1.5/(3-x) +
1.5/(3+x)
Hence, A = -1, B =
1.5 and C = 1.5
Question
2
Express in partial
fractions
(2x^3 - 3x^2 - 7x +
4)/[(x-1)(x+1)]
Answer
(2x^3 - 3x^2 - 7x +
4)/[(x-1)(x+1)]
= (2x^3 - 3x^2 - 7x
+ 4)/(x^2-1)
Use long
Division,
= 2x - 3 +
(1-5x)/(x^2 -1)
= 2x - 3 +
(1-5x)/[(x -1)(x+1)] = 2x-3 + A/(x -1) + B/(x+1)
= 2x -3 -2/(x-1) –
3/(x+1)
Thank you for your
kind attention.
Regards
ahm97sic
PS :
The objective of
Example 1 is to let students to know/realise that
it is not correct to
express in
partial fractions
as
x^2/9-x^2 = x^2/[(3-x)(3+x)] = A/(3-x) +
B/(x+3)
it should be
x^2/(9-x^2) = -1 +
9/(x^2-1) = -1 + 9/[(x-3)(x+3)] = A + B/(3-x)
and C/(3+x)
The
objective of example 2 is let the students to know / realize the
trick involved when they use the long division, there is a trick
involved. You must do the long division for this question to
realise the trick.
Example 2 also
requires the concept learnt in example 1 too.
Dear
Davidche, you have solved for the values of A, B and C without
realizing the trick for example 1 and hence your answer to
second example is incorrect as you have not done the long division
for example 2 before you proceed to express it in partial
fractions.
Dear Secretkiller,
well done ! Your answer is perfectly correct.