MHB Partial fractions (5x^2+1)/[(3x+2)(x^2+3)]

[mathlearn]

Trouble here in the below partial fraction (Bug)

$\frac{5x^2+1}{(3x+2)(x^2+3)}$

One factor in the denominator is a quadratic expression

Split this into two parts A&B

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$

${5x^2+1}={A(x^2+3)}+{Bx+c (3x+2)}{}$

and cannot take it from here

 

[Dethrone]

Hi mathlearn,

I believe you have a typo in your last line, it should be:
${5x^2+1}={A(x^2+3)}+(Bx+C) (3x+2)$

One way to move forward is to substitute $x=-2/3$, effectively zeroing the last term (which allows you to solve for $A$). Once we have $A$, unfortunately, we'll have to expand out all the terms and compare the coefficients of both sides of the equation.

 

[BestMethod]

Let's use a better method;

(5x^2+1)/(3x+2)(x^2+3) = A/(3x+2) + (Bx+C)/(x^2+3)

To find A,multiplying both sides of the equality by (3x+2) ;

(5x^2+1)/(x^2+3) =A + (3x+2)(Bx+C)/(x^2+3) (1)

Now,setting x=-2/3 or 3x+2=0,makes the expression on the right containing (3x+2) vanish,so(5(-2/3)^2+1)/((-2/3)^2+3) =A + 0 .Then,

A=29/31To find C,set x=0 in (1),so we get rid of b.Then,

1/3=29/31 +2C/3

So,c can be easily calculated.

To find b,plugging any number but -2/3 & 0 into x,say 1,

6/5=29/31 + 5(B+C)/4

We have found C,then B can be easily calculated.

 

[HallsofIvy]

Actually there are a number of mistakes- though probably mostly typos.

Trouble here in the below partial fraction (Bug)

$\frac{5x^2+1}{(3x+2)(x^2+3)}$

One factor in the denominator is a quadratic expression

Split this into two parts A&B

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$

Okay.

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$

No, the fractions on the right must both have "$(x^2+ 3)(3x+ 2)$" as denominator and the final fraction should have numerator (Bx+c)(3x+ 2) not "Bx+ c(3x+ 2)".

\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$

​

${5x^2+1}={A(x^2+3)}+{Bx+c (3x+2)}{}$

Again that should be (Bx+ c)(3x+ 2).

and cannot take it from here

$5x^2+ 1= Ax^2+ 3A+ 3Bx^2+ 3cx+ 2Bx+ 2c$
$5x^2+ 0x+ 1= (A+ 3B)x^2+ (3c+ 2B)x+ (3A+ 2c)$
Since this is t0 be true for all x, "corresponding coefficients must be equal:
5= A+ 3B, 0= 3c+ 2B, and 1= 3A+ 2c

Subtract 3 times the second equation from twice the first:
10- 0= 2A+ 6B- 6c- 6B= 2A- 6c
Subtract 3(1= 3A+ 2c), 3= 9A- 6c from that:
7= -7A so A= -1. Then -3+ 2c= 1 so 2c= 4 and c= 2.
5= -1+ 3B so 3B= 6, B= 2.

A= -1, B= 2, and c= 2.