Find constants A and B such that 1 / (x + 3)(x - 2) = A / x + 3 + B / x - 2.

Question

Find constants A and B such that [latex]\frac{1}{(x + 3)(x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2} [/latex].

Accepted Answer

If we multiply both sides of the equation by [latex] (x + 3)(x - 2)[/latex], we get

[latex]1 = A(x - 2) + B(x + 3) [/latex]

We rewrite this as

[latex]0x + 1 = (A + B)x + (-2A + 3B)[/latex]

Upon comparing coefficients of like powers of x, we get a system of linear equations in the unknowns A and B

[latex]A + B = 0 \ -2A + 3B = 1[/latex]

Row reducing the corresponding augmented matrix gives

[latex]\left [ \begin{matrix} 1 & 1 \ -2 & 3 \end{matrix} \left | \begin{matrix} 0 \ 1 \end{matrix} ight. ight ] hicksim \left [ \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \left | \begin{matrix} -1/5 \ 1/5 \end{matrix} ight. ight ][/latex]

Hence,

[latex] \frac{1}{(x + 3)(x - 2)} = \frac{{-1}/{5}}{x + 3} + \frac{{1}/{5}}{x - 2}[/latex]

It is easy to perform the addition on the right to verify this is correct.

Question 2.4.4: Find constants A and B such that 1 / (x + 3)(x - 2) = A / x ...

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