Question 16·200 Super-Hard SAT Math Questions·Advanced Math

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Consider the expression

11(4x-3)^2-7(4x+1)^2+6(4x-3)(4x+1)

The expression can be rewritten as $\frac{a}{4}x^2+\frac{b}{4}x+\frac{c}{4}$ , where $a$ , $b$ , and $c$ are constants. Which choice is equal to $a+b+c$ ?

A coefficient sum is the result of adding a polynomial’s coefficients. Setting $x=1$ makes every power of $x$ equal $1$ , so it reveals that sum without expanding the expression. Here, each coefficient is divided by $4$ : evaluate the original expression at $1$ in Desmos, then multiply by $4$ . The tempting mistake is to stop at the value Desmos gives.

Hints

Hint 1
In a polynomial, setting $x=1$ turns both $x^2$ and $x$ into $1$ . Apply that to the rewritten form. What fraction of $a+b+c$ do you get?
Hint 2
Name the original expression $f(x)$ in Desmos, then evaluate $f(1)$ . That output is not yet the coefficient sum. What operation undoes the division by $4$ ?

Step-by-step

Evaluate at one, then scale
Step 1
Find what substituting one reveals
The rewritten form matches the original expression at every $x$ . Set $x=1$ . Since $1^2=1$ , the coefficient sum, meaning $a+b+c$ , appears as $\frac{a}{4}(1)^2+\frac{b}{4}(1)+\frac{c}{4}=\frac{a+b+c}{4}$ .At $x=1$ , the expression gives one-fourth of the sum you need.
Step 2
Evaluate the original expression
Type the original as $f(x)=11(4x-3)^2-7(4x+1)^2+6(4x-3)(4x+1)$ in Desmos, then type $f(1)$ . Desmos shows $-134$ . At $x=1$ , the square term $-7(4x+1)^2$ becomes $-7(5)^2=-175$ , so keep its negative sign. The value $-134$ equals $\frac{a+b+c}{4}$ .
Step 3
Recover the coefficient sum
To undo the division, multiply $\frac{a+b+c}{4}=-134$ by $4$ :
$a+b+c=4(-134).$
Add $4f(1)$ in Desmos. It prints $-536$ , the requested coefficient sum. Choice A.

Question 16·200 Super-Hard SAT Math Questions·Advanced Math

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