Question 4·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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The expression $12(3x^2-2x-1)^2+38(3x^2-2x-1)+30$ can be rewritten as $(4(3x^2-2x-1)+a)(3(3x^2-2x-1)+b)$ , where $a$ and $b$ are positive integers. It can also be rewritten as $(4(3x^2-2x-1)+c)(3(3x^2-2x-1)+d)$ , where $c$ and $d$ are positive nonintegers. What is the value of $3a+2c$ ?

Your answer

A repeated expression that is squared and then appears again on its own makes a quadratic in a hidden quantity. Treat the whole repeated chunk as one variable. Graph that quadratic in Desmos and use its zeros to find the constants in its factors. The trap is to assign both zeros to the same factorization without checking which constants must be integers.

Hints

Hint 1
Call the repeated chunk $t$ . Then the long expression becomes a quadratic in $t$ . Graph it with Desmos’s $x$ standing for $t$ , not for the original $x$ .
Hint 2
An x-intercept is where the graphed quadratic equals $0$ . At each intercept, one of the factors must equal $0$ . What constant makes the first factor, $4t$ plus a constant, equal $0$ ?
Hint 3
Either intercept could belong to the first factor. Use the integer versus noninteger conditions to decide which intercept gives $a$ and which gives $c$ .

Step-by-step

Graph the quadratic in the repeated chunk
Step 1
Turn the repeated chunk into one variable
Let $t=3x^2-2x-1$ . Replacing each copy of that whole chunk gives $12t^2+38t+30$ . This is a quadratic in $t$ , so you can find its factors without solving for the original $x$ .
Step 2
Read the quadratic’s zeros
Type $f(x)=12x^2+38x+30$ , using Desmos’s $x$ for your temporary $t$ . Click both x-intercepts, where the graph equals $0$ . Desmos shows $-1.5$ and about $-1.667$ . These are zeros of the quadratic in $t$ , not solutions for the original $x$ .
Step 3
Confirm both rounded zeros exactly
The intercept near $-1.667$ suggests $-\frac{5}{3}$ , but a rounded coordinate doesn’t establish an exact fraction. Type $f(-\frac{5}{3})$ ; Desmos prints $0$ , confirming it. To confirm the other zero is exactly $-1.5=-\frac{3}{2}$ , type $f(-\frac{3}{2})$ ; Desmos prints $0$ again.
Step 4
Assign the zeros to the first factors
At a zero of a product, one factor must be $0$ . Set the first factor in the integer form to $0$ :
$4t+a=0$
Subtract $4t$ :
$a=-4t$
Use $t=-\frac{3}{2}$ to get the integer $a=-4(-\frac{3}{2})=6$ . The other zero gives the noninteger $c=-4(-\frac{5}{3})=\frac{20}{3}$ .
Step 5
Check the other constants
Each form has a second factor, so its unused zero must make that factor $0$ . For the integer form, $3(-\frac{5}{3})+b=0$ gives $b=5$ . For the noninteger form, $3(-\frac{3}{2})+d=0$ gives $d=\frac{9}{2}$ . Both pairings satisfy the stated conditions.
Step 6
Calculate the requested combination
The question asks for $3a+2c$ , not a combination using $b$ or $d$ . Type $3(6)+2(\frac{20}{3})$ . Desmos displays $31.333\ldots$ ; use its fraction button to see the exact value $\frac{94}{3}$ . Grid in 94/3.

Question 4·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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