Try dividing $4x^3 - 5x^2 + 2x - 7$ by $2x - 3$ using polynomial long division (or synthetic division adapted for $2x - 3$). The result will give you the quadratic part and the remainder.

After you divide, the quotient should be a quadratic $Ax^2 + Bx + C$ and the remainder should be some constant $R$. Compare this with $ax^2 + bx + c + \dfrac{d}{2x - 3}$ to identify $a$, $b$, $c$, and $d$.

Once you know $a$, $b$, $c$, and $d$, you do not need to rewrite the entire expression again—just compute $a + b + c + d$ carefully (watch signs and fractions) and then multiply that sum by $4$.

In one Desmos line, enter the original function: f(x) = (4x^3 - 5x^2 + 2x - 7)/(2x - 3). After you have found your values of $a$, $b$, $c$, and $d$ on paper, enter g(x) = a*x^2 + b*x + c + d/(2x - 3) using your numeric values. If your work is correct, the graphs of $f$ and $g$ should lie exactly on top of each other.

Once you are confident in your $a$, $b$, $c$, and $d$, type a new expression like 4*(a + b + c + d) in Desmos, replacing $a$, $b$, $c$, and $d$ with the numbers you found. The numerical output Desmos shows for this expression is the value of $4(a + b + c + d)$.

The form

means you are writing the original fraction as

(polynomial quotient) $+$ (remainder over the same divisor).

This is exactly what polynomial long division (or synthetic division with care) produces when you divide $4x^3 - 5x^2 + 2x - 7$ by $2x - 3$.

Divide the leading term of the numerator by the leading term of the denominator:

• Leading terms: $4x^3$ and $2x$.
• $4x^3 \div 2x = 2x^2$, so the first term of the quotient is $2x^2$.

Multiply back and subtract:

• Multiply: $2x^2(2x - 3) = 4x^3 - 6x^2$.
• Subtract this from the original numerator (just the first two terms):
  • $(4x^3 - 5x^2) - (4x^3 - 6x^2) = x^2$.

Bring down the next term $+2x$, so you now have a new partial dividend of $x^2 + 2x$.

Now divide $x^2 + 2x$ by $2x - 3$ term by term.

1. Divide leading terms: $x^2 \div 2x = \tfrac{1}{2}x$. This is the next term of the quotient.
2. Multiply: $\tfrac{1}{2}x(2x - 3) = x^2 - \tfrac{3}{2}x$.
3. Subtract from $x^2 + 2x$:
   • $(x^2 + 2x) - (x^2 - \tfrac{3}{2}x) = 2x + \tfrac{3}{2}x = \tfrac{7}{2}x$.
4. Bring down the constant term $-7$ to get $\tfrac{7}{2}x - 7$.

Repeat one more time:

1. Divide leading terms: $\tfrac{7}{2}x \div 2x = \tfrac{7}{4}$. This is the last term of the quotient.
2. Multiply: $\tfrac{7}{4}(2x - 3) = \tfrac{7}{2}x - \tfrac{21}{4}$.
3. Subtract from $\tfrac{7}{2}x - 7$:
   • $\big(\tfrac{7}{2}x - 7\big) - \big(\tfrac{7}{2}x - \tfrac{21}{4}\big) = -7 + \tfrac{21}{4} = -\tfrac{7}{4}$.

So the quotient is $2x^2 + \tfrac{1}{2}x + \tfrac{7}{4}$ and the remainder is $-\tfrac{7}{4}$.

Therefore,

From this, you can read off $a$, $b$, $c$, and $d$.

From the rewritten form

we have:

• $a = 2$
• $b = \tfrac{1}{2}$
• $c = \tfrac{7}{4}$
• $d = -\tfrac{7}{4}$

Now add them:

Finally,

So the value of $4(a + b + c + d)$ is $10$.