How do you express #(9x)/(9x^2+3x-2)# in partial fractions?

Question

Serenity Johnson · Accepted Answer

# 2/(3x + 2 ) + 1/(3x - 1 ) #

first step here is to factor the denominator # 9x^2 + 3x - 2 = (3x + 2 )(3x - 1 ) # since these factors are linear , the numerators will be constants #( 9x)/((3x + 2 )(3x - 1 )) = A/(3x +2 ) + B/(3x - 1 ) # now multiply through by (3x + 2 )(3x - 1 ) hence : 9x = A(3x - 1 ) + B(3x + 2 ).......................(1) the next step is to find values for A and B. Note that if x#= 1/3#then the term with A will be zero and if x # = -2/3 # the term with B will be zero. let # x = 1/3 color(black)(" in") (1) : 3 = 3B rArr B = 1 # let # x = -2/3color(black)(" in") (1) : -6 = -3A rArr A = 2# # rArr (9x)/(9x^2 + 3x - 2 ) = 2/(3x + 2 ) + 1/(3x - 1 )#

Caleb Alexander · Answer

undefined To express $\frac{9x}{9x^2+3x-2}$ in partial fractions, we first factor the denominator as $(3x-1)(3x+2)$. Then, we write the expression as:

$$
\frac{9x}{9x^2+3x-2} = \frac{A}{3x-1} + \frac{B}{3x+2}
$$

Next, we multiply both sides by the common denominator $(3x-1)(3x+2)$ to clear the fractions:

$$
9x = A(3x+2) + B(3x-1)
$$

Expanding and grouping like terms, we get:

$$
9x = (3A + 3B)x + (2A - B)
$$

By equating coefficients, we find:

$$
\begin{cases}
3A + 3B = 9\
2A - B = 0
\end{cases}
$$

Solving this system of equations gives $A = 3$ and $B = 6$.

Thus, the expression $\frac{9x}{9x^2+3x-2}$ can be expressed as:

$$
\frac{3}{3x-1} + \frac{6}{3x+2}
$$