how to work out partial fraction decomposition when given
1/x(x^2+1) and 2x-1/(x+4)(x-3)
if you could give me an idea of the formula and what changes i need to make to the formula when the denominator changes that would help and please use idiot speak because the text books are confusing me
i think i worked out the second one but i still cant do the first
is it = 9/7(x+4)+5/7(x-3)???
let 1/x(x^2+1)
= A/x + (Bx+C)/(x^2+1)
= [A(x^2+1) + x(Bx+C)]/[x(x^2+1)]
= [Ax^2 + A + Bx^2 + Cx}/[x(x^2+1)]
= [x^2(A+B) + Cx + A]/[x(x^2+1)]
there is no x term in the original numerator, so C = 0
there is no x^2 in the original numerator, so A+B = 0,
The constant was 1, so A = 1
then B = -1, since A+B=0
so 1/x(x^2+1)
= 1/x -x/(x^2 + 1)
careful how you write the second,
it should be
9/(7(x+4))+5/(7(x-3))
What happened to the grocer who stacked all of the liquid detergent no the high shelf ?
To perform partial fraction decomposition, we want to express a rational function with a complicated denominator as the sum of simpler fractions. The general form of a partial fraction decomposition of a rational function is:
P(x)
-------------- = A/x + (Bx + C)/(x^2 + D^2) + ...
Q(x)(x - r1)(x - r2)(x - r3)...
Where P(x) and Q(x) are polynomials, r1, r2, r3, etc. are the distinct roots of the denominator (Q(x)) and A, B, C, etc. are constants.
Now, let's apply this general formula to the given examples:
1. Partial fraction decomposition for 1 / (x(x^2 + 1)):
The denominator is x(x^2 + 1), which doesn't have any linear factors. So, we need to factorize the denominator by completing the square:
x^2 + 1 = (x^2 + 1^2)
Notice that x = 0 is a simple root since it only appears once in the factorization.
Therefore, we can rewrite the original expression as:
1 / (x(x^2 + 1)) = A/x + (Bx + C)/(x^2 + 1)
To find the constant values A, B, and C, we can multiply each side of the equation by the denominator (x(x^2 + 1)) and simplify:
1 = A(x^2 + 1) + x(Bx + C)
Next, substitute different values or perform algebraic manipulation to find the coefficients A, B, and C. For instance, substituting x = 0:
1 = A(0^2 + 1) + 0(B(0) + C)
1 = A + 0
A = 1
Now, choose other convenient values for x to determine B and C. For example, substituting x = 1:
1 = (1)(1^2 + 1) + 1(B(1) + C)
1 = 2 + B + C
B + C = -1 ----(1)
Substituting x = -1:
1 = (-1)(-1^2 + 1) - 1(B(-1) + C)
1 = 2 - B + C
-B + C = -1 ----(2)
Now, we have a system of linear equations (equations (1) and (2)). Solving this system, we find the values of B and C. Subtracting equation (1) from equation (2):
(B + C) - (-B + C) = -1 - (-1)
2B = -2
B = -1
Substituting B = -1 back into equation (1):
(-1) + C = -1
C = 0
Therefore, the partial fraction decomposition of 1 / (x(x^2 + 1)) is:
1 / (x(x^2 + 1)) = 1/x - (x - 0) / (x^2 + 1)
= 1/x - x / (x^2 + 1)
2. Partial fraction decomposition for (2x - 1) / ((x + 4)(x - 3)):
In this case, the denominator is already factored as (x + 4)(x - 3), which means we can apply the formula directly:
(2x - 1) / ((x + 4)(x - 3)) = A/(x + 4) + B/(x - 3)
To find A and B, we can multiply each side of the equation by the denominator and simplify:
2x - 1 = A(x - 3) + B(x + 4)
Now, substitute different values for x to determine A and B. For example, substituting x = 3:
2(3) - 1 = A(3 - 3) + B(3 + 4)
5 = 7B
B = 5/7
Substituting x = -4:
2(-4) - 1 = A(-4 - 3) + B(-4 + 4)
-9 = -7A
A = 9/7
Therefore, the partial fraction decomposition of (2x - 1) / ((x + 4)(x - 3)) is:
(2x - 1) / ((x + 4)(x - 3)) = 9/7(x + 4) + 5/7(x - 3)
So, your partial fraction decomposition for the second expression is correct! However, the partial fraction decomposition for the first expression is 1/x - x / (x^2 + 1), not 9/7(x + 4) + 5/7(x - 3).