MHB Partial Fractions: Struggling to Remember? Help Here!

[Simon green]

struggling to remember anything about partial fractions, can anybody help me with this?

6x-5
(x-4) (x²+3)

 

[Prove It]

struggling to remember anything about partial fractions, can anybody help me with this?

6x-5
(x-4) (x²+3)

Each of the smaller fractions needs to have a numerator of degree one less than the denominator. So you will want to try $\displaystyle \begin{align*} \frac{A}{x - 4} + \frac{B\,x + C}{x^2 + 3} \equiv \frac{6\,x - 5}{ \left( x - 4 \right) \left( x^2 + 3 \right) } \end{align*}$ as your partial fraction decomposition.

 

[Simon green]

I see, and where do you go after that? It's been an awful long time since I've done any of this and i just want to refresh my memory

 

[MarkFL]

I see, and where do you go after that? It's been an awful long time since I've done any of this and i just want to refresh my memory

After we state:

$$\frac{6x-5}{(x-4)(x^2+3)}=\frac{A}{x-4}+\frac{Bx+C}{x^2+3}$$

We then multiply through by $(x-4)(x^2+3)$ to get:

$$6x-5=A(x^2+3)+(Bx+C)(x-4)$$

Which we then arrange as:

$$0x^2+6x+(-5)=(A+B)x^2+(C-4B)x+(3A-4C)$$

Equating coefficients, we obtain the system:

$$A+B=0$$

$$C-5B=6$$

$$3A-4C=-5$$

We have 3 equations in 3 unknowns, so we solve the system. :D

 

[BestMethod]

(6x-5)/(x-4)(x²+3)=

a/(x-4) + (bx+c)/(x²+3)

Now,to find a,multiplying both sides of the equality by (x-4);

(6x-5)/(x²+3) = a + (x-4)(bx+c)/(x²+3). (1)

Now,setting x=4 or x-4=0,makes the expression on the right vanish,so

(24-5)/(16+4)=a+0

a=19/20

To find c,set x=0 in (1),so we get rid of b.Then,

-5/-12=(19/20)/-4 + c/3

So,c can be easily calculated.

To find b,plugging any number but 4&0 into x,say 1,

-1/12=a/-3 +(b+c)/4

We have found a&b,then c can be easily calculated.