Question

find the partial fraction decomposition. (\frac{6x + 43}{(x - 2)^2(x + 3)}) choose the correct pattern for the decomposition below, and fill in any answer boxes to complete your choice. (\bigcirc) a. (\frac{square}{(x - 2)^2} + \frac{square}{(x + 3)}) (\bigcirc) b. (\frac{square}{(x - 2)} + \frac{square}{(x - 2)^2} + \frac{square x}{(x + 3)}) (\bigcirc) c. (\frac{square}{(x - 2)} + \frac{square}{(x - 2)^2} + \frac{square}{(x + 3)}) (\bigcirc) d. (\frac{square}{(x - 2)} + \frac{square x + square}{(x + 3)})

Explanation:

Step1: Recall Partial Fraction Rules

For a repeated linear factor \((ax + b)^n\), the partial fractions have terms \(\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \dots + \frac{A_n}{(ax + b)^n}\). For a non - repeated linear factor \((cx + d)\), the partial fraction is \(\frac{B}{cx + d}\).

The denominator is \((x - 2)^2(x + 3)\). Here, \((x - 2)\) is a repeated linear factor (with multiplicity 2) and \((x + 3)\) is a non - repeated linear factor.

So the partial fraction decomposition pattern should have \(\frac{A}{x - 2}+\frac{B}{(x - 2)^2}+\frac{C}{x + 3}\), which corresponds to option C.

Answer:

C. \(\frac{\square}{(x - 2)}+\frac{\square}{(x - 2)^2}+\frac{\square}{(x + 3)}\)

Now, to find the values of the constants:

We start with \(\frac{6x + 43}{(x - 2)^2(x + 3)}=\frac{A}{x - 2}+\frac{B}{(x - 2)^2}+\frac{C}{x + 3}\)

Multiply both sides by \((x - 2)^2(x + 3)\) to get:

\(6x+43=A(x - 2)(x + 3)+B(x + 3)+C(x - 2)^2\)

Step1: Find B (let \(x = 2\))

Substitute \(x = 2\) into the equation:

\(6(2)+43=A(2 - 2)(2 + 3)+B(2 + 3)+C(2 - 2)^2\)

\(12 + 43=0+5B + 0\)

\(55 = 5B\)

Divide both sides by 5: \(B = 11\)

Step2: Find C (let \(x=-3\))

Substitute \(x=-3\) into the equation:

\(6(-3)+43=A(-3 - 2)(-3 + 3)+B(-3 + 3)+C(-3 - 2)^2\)

\(-18 + 43=0+0 + C(25)\)

\(25 = 25C\)

Divide both sides by 25: \(C = 1\)

Step3: Find A (compare coefficients of \(x^2\))

Expand the right - hand side:

\(A(x - 2)(x + 3)=A(x^{2}+3x-2x - 6)=A(x^{2}+x - 6)\)

\(B(x + 3)=11x+33\)

\(C(x - 2)^2=1\times(x^{2}-4x + 4)=x^{2}-4x + 4\)

The right - hand side is \(A(x^{2}+x - 6)+11x + 33+x^{2}-4x + 4=(A + 1)x^{2}+(A + 11-4)x+(-6A + 33 + 4)\)

The left - hand side is \(6x+43\), which has no \(x^{2}\) term. So the coefficient of \(x^{2}\) on the right - hand side must be 0.

\(A + 1=0\)

So \(A=-1\)

So the partial fraction decomposition is \(\frac{-1}{x - 2}+\frac{11}{(x - 2)^2}+\frac{1}{x + 3}\)

Final Answer (for the filled - in option C):

C. \(\frac{-1}{(x - 2)}+\frac{11}{(x - 2)^2}+\frac{1}{(x + 3)}\)