Super-Hard SAT Advanced Math Question 6 | Free SAT Prep

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Question 6·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For constants $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ , the following equation is true for all values of $x$ for which the expressions are defined:

\begin{aligned} \frac{ax^5+bx^4+cx^3+dx^2+ex+f}{(x-2)^2(x+3)} &=x^2-2x+1+\frac{5}{x-2}\\ &\quad-\frac{4}{(x-2)^2}+\frac{3}{x+3} \end{aligned}

What is the value of $a+c+e$ ?

Your answer

When a question gives a rational expression whose numerator has unknown coefficients, avoid expanding unless the question truly requires every coefficient. Instead, translate the requested coefficient combination into polynomial values. Evaluating a polynomial at $1$ adds all coefficients, while evaluating at $-1$ changes the signs of the odd-power terms. Adding or subtracting these values can isolate a targeted group of coefficients efficiently.

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Name the numerator polynomial

Let $P(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ . The requested expression contains exactly the coefficients of the odd-power terms.

Compare opposite inputs

Write $P(1)$ and $P(-1)$ , then subtract them. Notice which coefficients cancel.

Evaluate without expanding

Multiply each side of the given equation by $(x-2)^2(x+3)$ , then substitute $x=1$ and $x=-1$ . Both values are allowed because neither makes a denominator zero.

Desmos Guide

Use algebra first

Algebra is faster because the key idea is that $P(1)-P(-1)$ isolates the odd-power coefficients. Desmos can quickly verify the two needed values.

Define the equivalent numerator expression

Enter the following definitions:

\begin{aligned} D(x)&=(x-2)^2(x+3)\\ R(x)&=x^2-2x+1+\frac{5}{x-2}-\frac{4}{(x-2)^2}+\frac{3}{x+3}\\ P(x)&=D(x)R(x) \end{aligned}

Check the two inputs

Create a table with $x$ values $1$ and $-1$ , and enter $P(x)$ in the second column. Use the two displayed outputs in $\frac{P(1)-P(-1)}{2}$ to verify the coefficient sum.

Step-by-step Explanation

Use values that separate odd and even powers

For the numerator polynomial $P(x)$ ,

\begin{aligned} P(1)&=a+b+c+d+e+f\\ P(-1)&=-a+b-c+d-e+f \end{aligned}

Therefore,

P(1)-P(-1)=2(a+c+e).

Find $P(1)$ from the equivalent expression

At $x=1$ , the denominator $(x-2)^2(x+3)$ equals $4$ . The right-hand side equals

0-5-4+\frac{3}{4}=-\frac{33}{4}.

Thus,

P(1)=4\left(-\frac{33}{4}\right)=-33.

Find $P(-1)$ from the equivalent expression

At $x=-1$ , the denominator $(x-2)^2(x+3)$ equals $18$ . The right-hand side equals

4-\frac{5}{3}-\frac{4}{9}+\frac{3}{2}=\frac{61}{18}.

Thus,

P(-1)=18\left(\frac{61}{18}\right)=61.

Isolate the requested coefficient sum

Substitute the two polynomial values:

2(a+c+e)=P(1)-P(-1)=-33-61=-94.

Dividing by $2$ gives $a+c+e=-47$ .

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation