After you clear denominators, try plugging in $x = 2$ and $x = -1$. What terms disappear, and how does that help you solve for $A$ and $B$ quickly?

Once $A$ and $B$ are known, plug in two other easy values of $x$ (like $x = 0$ and $x = 1$) into the same identity to get equations that involve only $C$ and $D$.

After you find all four constants, double-check any fraction arithmetic and then carefully add $A + B + C + D$ to get the final value.

Type f(x) = (3x^3 - 5x + 2)/((x - 2)(x + 1)) into Desmos to represent the left-hand side.

After you solve for $A$, $B$, $C$, and $D$ on paper, type g(x) = C*x + D + A/(x - 2) + B/(x + 1) in Desmos, replacing $A$, $B$, $C$, and $D$ with the values you found.

Type h(x) = f(x) - g(x). If your decomposition is correct, the graph of h(x) should be the horizontal line $y = 0$ everywhere it is defined (except at the vertical asymptotes $x = 2$ and $x = -1$). If it is not identically zero, recheck your constants.

Once you are confident your $A$, $B$, $C$, and $D$ are correct, either use Desmos to compute A + B + C + D directly in an expression line or add the four numbers on a calculator to obtain the required value.

We are told

Multiply both sides by $(x - 2)(x + 1)$ to clear the denominator:

This equation must hold for all values of $x$ where both sides are defined.

Choose $x$-values that make factors $(x - 2)$ or $(x + 1)$ equal to 0 to eliminate big chunks of the right side.

• Let $x = 2$:

  • Left side: $3(2)^3 - 5(2) + 2 = 24 - 10 + 2 = 16$.
  • Right side: $(2C + D)\cdot 0 + A(3) + B(0) = 3A$.
  • So $3A = 16$, giving $A = \dfrac{16}{3}$.

• Let $x = -1$:

  • Left side: $3(-1)^3 - 5(-1) + 2 = -3 + 5 + 2 = 4$.
  • Right side: $( -C + D)\cdot 0 + A(0) + B(-3) = -3B$.
  • So $-3B = 4$, giving $B = -\dfrac{4}{3}$.

Now substitute $A = \dfrac{16}{3}$ and $B = -\dfrac{4}{3}$ back into

Choose convenient $x$-values that do not zero out the whole polynomial part.

• Let $x = 0$:

  • Left side: $2$.
  • Right side: $(D)(-2)(1) + A(1) + B(-2) = -2D + A - 2B$.

  So

Plug in $A$ and $B$:

Thus $-2D = 2 - 8 = -6$, so $D = 3$.

• Let $x = 1$:

  • Left side: $3(1)^3 - 5(1) + 2 = 3 - 5 + 2 = 0$.
  • Right side: $(C + D)(-1)(2) + A(2) - B = -2(C + D) + 2A - B$.

  So

Substitute $D = 3$, $A = \dfrac{16}{3}$, $B = -\dfrac{4}{3}$:

Combine the fractions $\frac{32}{3} + \frac{4}{3} = \frac{36}{3} = 12$:

So $-2C + 6 = 0$, giving $C = 3$.

Now we have

• $A = \dfrac{16}{3}$
• $B = -\dfrac{4}{3}$
• $C = 3$
• $D = 3$.

First add $A$ and $B$:

Then add $C$ and $D$:

Finally,

So the value of $A + B + C + D$ is $10$.