Answer:
[tex] \frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)} [/tex]
Step-by-step explanation:
Given the rational expression: [tex] \frac{2x + 3}{x^2 - 5x + 4} [/tex], to express this in simplified form, we would need to apply the concept of partial fraction.
Step 1: factorise the denominator
[tex] x^2 - 5x + 4 [/tex]
[tex] x^2 - 4x - x + 4 [/tex]
[tex] (x^2 - 4x) - (x + 4) [/tex]
[tex] x(x - 4) - 1(x - 4) [/tex]
[tex] (x- 1)(x - 4) [/tex]
Thus, we now have: [tex] \frac{2x + 3}{(x- 1)(x - 4)} [/tex]
Step 2: Apply the concept of Partial Fraction
Let,
[tex] \frac{2x + 3}{(x- 1)(x - 4)} [/tex] = [tex] \frac{A}{x- 1} + \frac{B}{x - 4} [/tex]
Multiply both sides by (x - 1)(x - 4)
[tex] \frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4) [/tex] = [tex] (\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4) [/tex]
[tex] 2x + 3 = A(x - 4) + B(x - 1) [/tex]
Step 3:
Substituting x = 4 in [tex] 2x + 3 = A(x - 4) + B(x - 1) [/tex]
[tex] 2(4) + 3 = A(4 - 4) + B(4 - 1) [/tex]
[tex] 8 + 3 = A(0) + B(3) [/tex]
[tex] 11 = 3B [/tex]
[tex] \frac{11}{3} = B [/tex]
[tex] B = \frac{11}{3} [/tex]
Substituting x = 1 in [tex] 2x + 3 = A(x - 4) + B(x - 1) [/tex]
[tex] 2(1) + 3 = A(1 - 4) + B(1 - 1) [/tex]
[tex] 2 + 3 = A(-3) + B(0) [/tex]
[tex] 5 = -3A [/tex]
[tex] \frac{5}{-3} = \frac{-3A}{-3} [/tex]
[tex] A = -\frac{5}{3} [/tex]
Step 4: Plug in the values of A and B into the original equation in step 2
[tex] \frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4} [/tex]
[tex] \frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)} [/tex]
