10x÷x^2+x-6 = Please find partial fraction

Question

10x ÷ x^2 + x-6 =

Please find partial fraction decomposition. 

Thank you kindly! It is greatly appreciated!

Answer 1

You need to use parentheses.

 

10x / (x²+x-6) = 10x/[(x+3)(x-2)] = A/(x+3) + B/(x-2)

 

Multiply by (x+3)(x-2) to obtain 10x = A(x-2) + B(x+3)

 

                                             10x + 0  = (A+B)x + (3B-2A)

 

                                               A + B = 10

                                             -2A + 3B = 0

 

2A + 2B = 20

-2A + 3B =0

 

Adding the equations, we have 5B = 20.  So, B = 4

                                                                     

Since A+B = 10 and B = 4, we get A = 6.

 

So, 10x/(x²+x-6) = 6/(x+3) + 4/(x-2)

 

Answer 2

FIRST, the quadratic must be enclosed properly in parentheses (multi-term denominator).

 

your expression equals 10x ÷ [(x+3)(x-2)] = A/(x+3) + B/(x-2)

 

Clearing fractions, we get 10x = A(x-2) + B(x+3)

                                              = x(A+B) + (3B-2A)

 

Therefore 10 = A+B    and   B-2A = 0  by equating coefficients of 'like' terms, to make this an identity.

 

Can you now solve for A & B?