Question 45·Hard·Advanced Math

0 / 700

\frac{4x^3 - 5x^2 + 2x - 7}{2x - 3}

The rational expression above can be rewritten in the form

ax^2 + bx + c + \frac{d}{2x - 3},

where $a$ , $b$ , $c$ , and $d$ are constants. What is the value of $4(a + b + c + d)$ ?

Your answer

A rational expression written as a polynomial plus a remainder fraction gives you a useful equation when you multiply by its denominator. Clear the denominator before using its zero: that input isolates the remainder, even though it cannot go into the original fraction. Then use $x=1$ to find the sum of the polynomial’s coefficients. The trap is forgetting the remainder when finding the requested sum.

Hints

Hint 1
Multiply the proposed rewrite by $2x-3$ . The result is a polynomial identity, an equation between polynomials that holds for every input. Which input makes the entire quadratic part disappear, leaving only the remainder?
Hint 2
At $x=1$ , the quadratic $ax^2+bx+c$ becomes $a+b+c$ because every power of $1$ is $1$ . Substitute $1$ into the cleared identity, and keep the remainder $d$ in that equation.
Hint 3
The question asks for $a+b+c+d$ , not just the sum of the quadratic’s coefficients. Once you know $d$ and have an equation for $a+b+c$ , combine them before multiplying by $4$ .

Step-by-step

Use two inputs in the cleared identity
Step 1
Clear the denominator
Let $p(x)=4x^3-5x^2+2x-7$ be the numerator, the top of the given fraction. Multiply the proposed rewrite by $2x-3$ :
$p(x)=(2x-3)(ax^2+bx+c)+d.$
This is a polynomial identity: the polynomials agree wherever the fraction is defined, so they agree at every input. Use the cleared identity, not the original fraction, when the denominator is zero.
Step 2
Find the remainder
The factor $2x-3$ is zero at $x=\frac32$ , so the identity gives $p\!\left(\frac32\right)=d$ . Type $p(x)=4x^3-5x^2+2x-7$ , then $p\!\left(\frac32\right)$ . Desmos shows $-1.75$ , or $-\frac74$ using its fraction button. So $d=-\frac74$ .
Step 3
Get an equation for the coefficient sum
Write $S=a+b+c$ for the coefficient sum. At $x=1$ , the quadratic $ax^2+bx+c$ becomes $S$ , while $2x-3$ becomes $-1$ . Substitute into the identity:
$p(1)=-S+d.$
Add $p(1)$ in Desmos. It shows $-6$ , so $-6=-S+d$ . Keep the $d$ ; it has not disappeared.
Step 4
Isolate the coefficient sum
From $-6=-S+d$ , subtract $d$ from both sides:
$-6-d=-S.$
Multiply by $-1$ :
$6+d=S.$
So the coefficient sum is $6+d$ , not $6$ .
Step 5
Evaluate the requested expression
The target includes both $S$ and $d$ . Substitute $S=6+d$ :
$4(S+d)=4(2d+6).$
Type $4\left(2p\!\left(\frac32\right)+6\right)$ in Desmos. It shows $10$ , so $4(a+b+c+d)=10$ . Grid in 10.

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Question 45·Hard·Advanced Math

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