96.9k views
16 votes
Write out the form of the partial fraction decomposition of the function (as in this example). Do not determine the numerical values of the coefficients. (a) x − 20 x2 + x − 20 Incorrect: Your answer is incorrect. (b) x2 x2 + x + 20 Incorrect: Your answer is incorrect.

User Ntalbs
by
5.2k points

1 Answer

6 votes

Answer:


(x - 20)/( x^2 + x - 20) = (5)/(9(x-4))+(4)/((x+5))

Explanation:

Given


(x - 20)/( x^2 + x - 20)

Required

Decompose into partial fraction

Start by factorizing the denominator:


x^2 + x - 20 =x^2 + 5x -4x- 20


x^2 + x - 20 =x(x + 5)-4(x+5)


x^2 + x - 20 =(x-4)(x+5)

So, we have:


(x - 20)/( x^2 + x - 20) = (A)/(x-4)+(B)/(x+5)

Take LCM


(x - 20)/( x^2 + x - 20) = (A(x+5)+B(x-4))/((x-4)(x+5))

Cancel out the denominator


x - 20 = A(x+5)+B(x-4)}

Open brackets


x - 20 = Ax+5A+Bx-4B

Collect Like Terms


x - 20 = Ax+Bx+5A-4B


x - 20 = (A+B)x+5A-4B

By comparison;


(A +B)x = x ==>
A + B = 1


5A - 4B = 1

Make A the subject in
A + B = 1


A = 1 - B

Substitute 1 - B for A in
5A - 4B = 1


5(1-B) -4B = 1


5-5B -4B = 1


5-9B = 1

Collect Like Terms


-9B = 1 - 5


-9B = - 4


B = (4)/(9)


A = 1 - B


A = 1 - (4)/(9)


A = (9-4)/(9)


A = (5)/(9)

So:


(x - 20)/( x^2 + x - 20) = (A)/(x-4)+(B)/(x+5)


(x - 20)/( x^2 + x - 20) = (5)/(9(x-4))+(4)/((x+5))

User Timothy Lee
by
5.0k points