Crosswords0 min ago
Solving Partial Fractions
Hello
Can someone help me please I am really struggling with partial fractions I have 2 questions to complete and would appreciate some help, my questions are:
x2-3x+6/x(x-2)(x-1) and x2+9x+8/x2+x-6 I know there are some clever people on here if anyone can help with the steps aswell please.
Thanks in advance David
Can someone help me please I am really struggling with partial fractions I have 2 questions to complete and would appreciate some help, my questions are:
x2-3x+6/x(x-2)(x-1) and x2+9x+8/x2+x-6 I know there are some clever people on here if anyone can help with the steps aswell please.
Thanks in advance David
Answers
Best Answer
No best answer has yet been selected by david7. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Start by writing the expression in the form you require (introducing unknowns A, B & C).
Then multiply the numerator and denominator of each fraction accordingly, in order to create a common denominator.
Then group like terms (from the numerators) together. i.e. collect up x-squared, x and numeric terms.
Next, compare the coefficients of each term, to create three simultaneous linear equations.
Solve those equations.
Job done (but you should 'multiply out' your answer, in order to check it).
Here's the first one done for you:
http://i40.tinypic.com/2utlv10.jpg
Chris
Then multiply the numerator and denominator of each fraction accordingly, in order to create a common denominator.
Then group like terms (from the numerators) together. i.e. collect up x-squared, x and numeric terms.
Next, compare the coefficients of each term, to create three simultaneous linear equations.
Solve those equations.
Job done (but you should 'multiply out' your answer, in order to check it).
Here's the first one done for you:
http://i40.tinypic.com/2utlv10.jpg
Chris
-- answer removed --
Sorry- you don't need to bother factorising the numerator.
I will start you off with what i think is the right method.
First, factorise the denominator into (x+3)(x-2).
x² +9x +8
_________
(x+3)(x-2)
Set this as Ξ to A/(x+3) + B/(x-2) + C
Proceeding in the way Chris did, by putting every term over (x+3)(x-2):
x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)
Expanding:
x² +9x +8 Ξ Ax-2A +Bx +3B +Cx² + Cx - 6C
Comparing coefficients:
(of x²).............C=1
(of x)..... ... A+B+C=9, so A+B=8
(of constants)...8=-2A +3B-6C = -2A+3B-6, so 14= 3B-2A
Solving these simultaneous equations gives, I think, A=2, B=6 and C=1
Put these values now into
A/(x+3) + B/(x-2) + C
It's a bit late at night for this but I hope you can follow the method and check my working. Also, I suggest you try substituting in some values of x and make sure it works
I will start you off with what i think is the right method.
First, factorise the denominator into (x+3)(x-2).
x² +9x +8
_________
(x+3)(x-2)
Set this as Ξ to A/(x+3) + B/(x-2) + C
Proceeding in the way Chris did, by putting every term over (x+3)(x-2):
x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)
Expanding:
x² +9x +8 Ξ Ax-2A +Bx +3B +Cx² + Cx - 6C
Comparing coefficients:
(of x²).............C=1
(of x)..... ... A+B+C=9, so A+B=8
(of constants)...8=-2A +3B-6C = -2A+3B-6, so 14= 3B-2A
Solving these simultaneous equations gives, I think, A=2, B=6 and C=1
Put these values now into
A/(x+3) + B/(x-2) + C
It's a bit late at night for this but I hope you can follow the method and check my working. Also, I suggest you try substituting in some values of x and make sure it works
@factor
I like your solution. Another way which saves a bit of work is to start from your equation:
x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)
and instead of multiplying this out, notice that if x=2 or 3 that the second and third terms are zero. So put x=2 in the above gives:
4+18+8=5B, so B=6
Putting x=-3 gives:
9-27+8=-5A, so A=2
and then putting x=0 and substituting for A and B gives:
8=-4+18-6C, so C=1
I like your solution. Another way which saves a bit of work is to start from your equation:
x² +9x +8 Ξ A(x-2) +B(x+3) + C(x+3)(x-2)
and instead of multiplying this out, notice that if x=2 or 3 that the second and third terms are zero. So put x=2 in the above gives:
4+18+8=5B, so B=6
Putting x=-3 gives:
9-27+8=-5A, so A=2
and then putting x=0 and substituting for A and B gives:
8=-4+18-6C, so C=1
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