Find values of the constants A, B, and C such that
(16x^2+48x−216)/(x^3−36x)=A/(x−6)+B/(x)+C/(x+6).
just place the right side over a common denominator and equate the coefficients of the polynomials. You will get three equations in A,B,C. Solve them to get (9,6,1)
To find the values of the constants A, B, and C, we need to perform partial fraction decomposition on the right side of the equation.
First, let's factor the denominator, x^3 - 36x, as x(x-6)(x+6).
The partial fraction decomposition will have the form:
(16x^2 + 48x - 216)/(x^3 - 36x) = A/(x-6) + B/(x) + C/(x+6)
Now, we need to find the values of A, B, and C.
To do this, we multiply both sides of the equation by the denominator (x)(x - 6)(x + 6) to get rid of the fractions:
(16x^2 + 48x - 216) = A(x)(x + 6) + B(x-6)(x+6) + C(x)(x-6)
Expand the right side:
16x^2 + 48x - 216 = Ax(x + 6) + B(x^2 - 36) + C(x^2 - 36x)
Distribute and simplify:
16x^2 + 48x - 216 = Ax^2 + 6Ax + Bx^2 - 36B + Cx^2 - 36Cx
Combine like terms:
16x^2 + 48x - 216 = (A + B + C)x^2 + (6A - 36B - 36C)x - 36B
Now, we can equate the coefficients of like terms on both sides of the equation.
For the x^2 terms:
A + B + C = 16
For the x terms:
6A - 36B - 36C = 48
For the constant terms:
-36B = -216
Solving these three equations will give us the values of A, B, and C.
From the constant term equation, we find that B = 6.
Substituting this value of B into the x term equation, we have:
6A - 36(6) - 36C = 48
6A - 216 - 36C = 48
6A - 36C = 264
Now, let's substitute the value of B = 6 and solve the x^2 term equation:
A + 6 + C = 16
A + C = 10
Solving these two equations simultaneously will give us the values of A and C.
From A + C = 10, we can substitute A = 10 - C into 6A - 36C = 264:
6(10 - C) - 36C = 264
60 - 6C - 36C = 264
-42C = 204
C = -204/ -42
C = 17/3
Substituting this value of C into A + C = 10, we find:
A + 17/3 = 10
A = 10 - 17/3
A = 23/3
So, the values of the constants A, B, and C are:
A = 23/3
B = 6
C = 17/3