Question

write the partial fraction decomposition of the following rational expression.
\\(\frac{8x^2 - 3x + 2}{x(x - 2)(x - 1)}\\)
\\(\frac{8x^2 - 3x + 2}{x(x - 2)(x - 1)} = \square\\)

Explanation:

Step1: Set up partial fraction form

We assume \(\frac{8x^{2}-3x + 2}{x(x - 2)(x - 1)}=\frac{A}{x}+\frac{B}{x - 2}+\frac{C}{x - 1}\)

Multiply both sides by \(x(x - 2)(x - 1)\) to get:
\(8x^{2}-3x + 2=A(x - 2)(x - 1)+Bx(x - 1)+Cx(x - 2)\)

Step2: Find A (let x = 0)

Substitute \(x = 0\) into the equation:
\(8(0)^{2}-3(0)+2=A(0 - 2)(0 - 1)+B(0)(0 - 1)+C(0)(0 - 2)\)
\(2 = A(2)\)
\(A = 1\)

Step3: Find B (let x = 2)

Substitute \(x = 2\) into the equation:
\(8(2)^{2}-3(2)+2=A(2 - 2)(2 - 1)+B(2)(2 - 1)+C(2)(2 - 2)\)
\(32-6 + 2=B(2)(1)\)
\(28 = 2B\)
\(B = 14\)

Step4: Find C (let x = 1)

Substitute \(x = 1\) into the equation:
\(8(1)^{2}-3(1)+2=A(1 - 2)(1 - 1)+B(1)(1 - 1)+C(1)(1 - 2)\)
\(8-3 + 2=C(1)(- 1)\)
\(7=-C\)
\(C=-7\)

Step5: Write the partial fractions

Substitute \(A = 1\), \(B = 14\), \(C=-7\) back into the partial fraction form:
\(\frac{8x^{2}-3x + 2}{x(x - 2)(x - 1)}=\frac{1}{x}+\frac{14}{x - 2}-\frac{7}{x - 1}\)

Answer:

\(\frac{1}{x}+\frac{14}{x - 2}-\frac{7}{x - 1}\)