further mathematics:partial fractions?

Question

hey guys i got this assignment on partial fractions. am good in maths and all these calculations but i missed the class and i tried to check the topic on ma text books but i didnt c it..so i dont hav any idea how am going to solve them..plz help mi xplain them and if possible u can help mi solve them. da questions r..
1) 3x+2/(x+1)(x^2+x+2)
2) (x-1)(x^2+x+1)/(x^2+1)(3x^2+4x+5)
3) 3/(x-2)(x^2+5)

thenx guyz.....

2) (x-1)(x^2+x+1)/(x^2+1)(3x^2+4x+5)

ok i dont no y the question in 2) iz not complete...after 4x is+5 (3x^2+4x+5)

Answer 1

I hope this helps.

Partial fraction decomposition means taking a single fraction whose denominator is factorable, and rewriting it as a sum of fractions whose denominators contain the individual factors.

I'll do 1), and hope you can follow suit for 2) and 3).

1) 3x+2 / (x+1)(x^2 + x + 2).

When you decompose a fraction into partial fractions, you want one fraction per factor in the denominator. If one of those factors has a power, like (x+2)^3, then you would want a fraction for each power of the factor: (x+2), (x+2)^2, and (x+2)^3. This is not an issue for your questions, since none of the factors have a power greater than one.

For each linear factor, you set up an unknown constant for the numerator. For each quadratic factor, you set up an unknown linear factor for the numerator.

For example, in

3x+2 / (x+1)(x^2 + x + 2)

you need two fractions (one per factor). They would be written as:

A / (x+1) + (Bx + C) / (x^2 +x + 2),

where A, B, and C are (presently) unknown values. Your job is to find such values so that

3x+2 / (x+1)(x^2 + x + 2) = A / (x+1) + (Bx + C) / (x^2 +x + 2) (*)

for every x for which all the fractions are defined.

To find A, B, and C, most texts advise the following. Clear out the fractions in the equation above (*) by multiplying throughout by the least common denominator (which is the original denominator of the first fraction). Cancel what you can, and you should get

3x + 2 = A(x^2 + x + 2) + (Bx + C)(x + 1).

Next, bust open the parentheses on the right by distributing to get

3x + 2 = Ax^2 + Ax + 2A + Bx^2 + Bx + Cx + C.

Next, regroup the terms on the right in descending order by powers of x:

3x + 2 = Ax^2 + Bx^2 + Ax + Bx + Cx + 2A + C.

You regroup so that you can factor out the powers of x from like terms, i.e.:

3x + 2 = (A + B)x^2 + (A + B + C)x + 2A + C.

The only way this can be true for all values of x is if all the coefficients of x on the left match all the coefficients of x on the right.

Since there is no x^2 on the left, the coefficient of x^2 on the right must equal zero, i.e.
A + B = 0.

Since the coefficient of x on the left is 3, so must the coefficient of x on the right, i.e.
A + B + C = 3

Since the constant term on the left is 2, so must the constant term on the right, i.e.
2A + C = 2.

What you get is a system of three equations in three variables:

A + B = 0
A + B + C = 3
2A + C = 2

Solve this system for A, B, and C. I'll asume you can, and get

A = - 1/2
B = 1/2
C = 3

Your unknown coefficients are now known,so your partial fraction decomposition of

3x + 2 / (x + 1)(x^2 + x + 2)

is

-1/2 / (x + 1) + (1/2 x + 3)/(x^2 + x + 2).

Having fractions inside of fractions is so pointless, so I would multiply the top and bottom of each fraction by 2 to get a cleaner answer of

-1 / 2(x+1) + (x + 6) / 2(x^2 + x + 2).




Crazy, but with practice, it becomes easy.


Geez, I hope I didn't screw up...

Answer 2

I´ll show you the first
(3x+2)/(x+1)(x^2+x+2) (a)

1)The second factor has no real roots so it can´t be factorized
so
(a)= m/(x+1) +(px+q)/(x^2+x+1)
common denominator( x+1)(x^2+x+2)so
3x+2= m(x^2+x+2)+(px+q)*(x+1)

3x+2= (m+p)x^2 +(m+p+q)x+2m+q

m+p=0
m+p+q=3
2m+q=2
q=3 ,m=-1/2 and p=1/2
your partial fractions are (-1/2)/(x+1)+ (1/2x+3)/(x2+x+2)
The degree of the denominator must be less than the degree of the numerator
I think you can take it now.