Question 126·Hard·Advanced Math

The denominator is zero at $x=2$ and $x=-1$, so don’t use those inputs. Type $x_1=[0,1,3,4]$; Desmos displays a list of four allowed inputs, one for each unknown constant.

Name the given expression $f(x)$ in Desmos, then type $f(x_1)$. Desmos shows $[-1,0,17,17.4]$. These are the values the stated form must match at the four selected inputs.

Type $f(x_1)\sim ax_1+b+\frac{c}{x_1-2}+\frac{d}{x_1+1}$. The $\sim$ asks Desmos to fit the constants across the list. Under PARAMETERS, it shows $a=3$, $b=3$, $c\approx5.333$, and $d\approx-1.333$. Here $a,b,c,d$ correspond to $C,D,A,B$.

Type $a+b+c+d$ on the next line. Desmos shows $10$, so the requested sum $A+B+C+D$ is $10$. Choice C.

The two forms are equal wherever the original denominator is nonzero. Multiply both sides by $(x-2)(x+1)$ to remove the fractions:

Match only the coefficients needed for the sum. You don’t need $A$ and $B$ separately.

Multiply the denominator factors:

Now you can compare coefficients, the numbers multiplying matching powers of $x$, on the two sides.

Only $Cx$ times $x^2$ makes an $x^3$ term. Match its coefficient to the given $3x^3$:

The $x^2$ terms come from $Cx(-x)$ and $D(x^2)$. The given numerator has no $x^2$ term, so its coefficient is $0$:

Add $C$ to both sides:

Use $C=3$:

The $x$ terms from $(Cx+D)(x^2-x-2)$ have coefficient $-2C-D$. The fractions’ numerators add $A+B$, so match the given coefficient $-5$:

Substitute $C=D=3$:

Add $9$ to both sides:

Add the two sums you found:

The value of $A+B+C+D$ is $10$. Choice C.