+0

# Algebra problem

0
19
1
+743

Find constants A and B such that

(x + 17)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1)

for all x such that $x \neq -1$ and $x \neq 2$. Give your answer as the ordered pair (A,B).

Feb 15, 2024

#1
+396
+2

This is pretty simple partial fraction decomposition.

Our first indication that partial fraction decomposition works is $$(x-2)(x+1)={x}^{2}-x-2$$

$$\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}$$, and we make the denominators equal, $$\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}$$, and $$x+17=A(x+1)+B(x-2)$$. To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.

Feb 15, 2024

#1
+396
+2
Our first indication that partial fraction decomposition works is $$(x-2)(x+1)={x}^{2}-x-2$$
$$\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}$$, and we make the denominators equal, $$\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}$$, and $$x+17=A(x+1)+B(x-2)$$. To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.