Question

Find $A$ and $B$ such that

Answer 1

Assume

$\dfrac{3x+5}{x^2-x-42}=\dfrac{3x+5}{(x-7)(x+6)}=\dfrac A{x-7}+\dfrac B{x+6}$

Combining the fractions on the right hand side gives

$\dfrac{3x+5}{(x-7)(x-6)}=\dfrac{A(x-6)+B(x-7)}{(x-7)(x-6)}$

The fractions will be equal as long as the their numerator are equal:

$3x+5=A(x-6)+B(x-7)$

$3x+5=(A+B)x+(-6A-7B)$

Polynomials are equal to one another if the coefficients of like terms are equal:

$\begin{cases}A+B=3\\-6A-7B=5\end{cases}$

Solving this, you'd get $A=26$ and $B=-23$, so that your answer would be (26, -23).

- - -

Instead of solving the system of equations above, there is a trick that involves picking $x$ so that some terms disappear and solving for either $A,B$ is much faster.

At the point where we have

$3x+5=A(x-6)+B(x-7)$

notice that setting $x=7$ will eliminate $B$. Doing so, we get

$3(7)+5=A(7-6)\implies26=A$

while setting $x=6$ would give

$3(6)+5=B(6-7)\implies23=-B\implies B=-23$

Answer 2

Answer:

We factor the denominator in the left-hand side to get

$\[\frac{3x+5}{(x-7)(x+6)}= \frac{A}{x - 7} + \frac{B}{x + 6}.\]$

We then multiply both sides by

$(x - 7)(x + 6)$, to get $\[3x + 5 = A(x + 6) + B(x - 7).\]$

We can solve for $A$ and $B$ by substituting suitable values of $x$.

For example, setting $x = 7$, the equation becomes $26 = 13A$,

so $A = 2$.

Setting $x = -6$, the equation becomes $-13 = -13B$,

so $B = 1$

Therefore, $(A,B) = \boxed{(2,1)}$.