suppose f(x)=7x-19/x2-6x+9 then F(x)= A/x-3 + B/(x-3)2 where is A = where is B =

Question

suppose

f(x)=7x-19/x²-6x+9

then

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

where is A =

where is B =

Answer 1

Hi axl,

here is another repeated root (x-3) where since x is not to any power indeed the (x-3) and (x-3)² denominators will only have a constant above them.

to wit, [7x-19]/(x-3)²=A/(x-3)+B/(x-3)²

same as [7x-19]/(x-3)²=[Ax-3A+B]/(x-3)²

need to equate 7x-19 to Ax-3A+B (the simulateneous equation piece)

A=7, B=2

Answer 2

(7x - 19) / (x² - 6x + 9) = [A / (x - 3)] + [B / (x - 3)²]

7x - 19 = A(x - 3)² + B(x - 3)

7x - 19 = A(x² - 6x + 9) + Bx - 3B

7x - 19 = Ax² - 6Ax + 9A + Bx - 3B

7x - 19 = Ax² - 6Ax + Bx + 9A - 3B

-6A + B = 7

9A - 3B = -19

Can you solve for A and B?