Question 8·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For the expression $P(x)=-11(5x-8)^2+7(3x-4)^2+9(5x-8)(3x-4)$ , suppose $P(x)$ can be rewritten as $\frac{a}{18}(2x-3)^2+\frac{b}{18}(2x-3)+\frac{c}{18}$ , where $a$ , $b$ , and $c$ are constants. Which choice gives the value of $a+b+c$ ?

A coefficient sum is easiest to find by making the expression being squared equal $1$ : its square then equals $1$ too. Here that expression is $2x-3$ , not $x$ . Use a Desmos regression to find the needed $x$ , then evaluate the given expression there. Keep track of the denominator $18$ ; the value you evaluate is only one eighteenth of the requested sum.

Hints

Hint 1
A coefficient is a number attached to a power of a variable. If the inside expression $2x-3$ equals $1$ , its square also equals $1$ . What does the rewritten form become?
Hint 2
Setting $x=1$ won't make the inside expression equal $1$ : it makes $2x-3=-1$ . Set the inside expression equal to $1$ instead, then find the value of $x$ that works.
Hint 3
At that input, the value of $P(x)$ is one eighteenth of the coefficient sum. Evaluate all three terms, including the mixed-product term containing both parentheses, before multiplying by $18$ .

Step-by-step

Make the inside expression equal 1
Step 1
Make the coefficient sum appear
The rewritten form uses $2x-3$ as its inside expression. To get a coefficient sum, make that inside expression equal $1$ . Then its square is $1$ too. Substitute into the rewritten form:
$P(x)=\frac{a}{18}(1)^2+\frac{b}{18}(1)+\frac{c}{18}.$
Combine the numerators:
$P(x)=\frac{a+b+c}{18}\quad\text{when }2x-3=1.$
Step 2
Find the input that makes it 1
Type $2x_1-3\sim1$ in Desmos. The $\sim$ asks Desmos to find the unknown $x_1$ ; under PARAMETERS, it shows $x_1=2$ . So use $x=2$ when evaluating $P(x)$ . Using $x=1$ would make $2x-3=-1$ , not $1$ .
Step 3
Evaluate the given expression
Name the given expression $p(x)$ in Desmos, then type $p(2)$ . Desmos shows $p(2)=20$ , so $P(2)=20$ . At $x=2$ , both parentheses equal $2$ , so the mixed-product term is positive and must be included.
Step 4
Restore the factor of 18
The relation from Step 1 says $P(2)$ is one eighteenth of the requested sum. Multiply both sides by $18$ :
$a+b+c=18P(2).$
Stopping at $20$ would leave the denominator in place.
Step 5
Find the requested sum
Type $18p(2)$ on a new Desmos line. It shows $360$ , so $a+b+c$ is $360$ . Choice D.

Lessons that teach this

SAT Advanced AlgebraAdvancedCoreUse polynomial identities, factors, and unknown coefficients

Question 8·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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