Question 74·200 Super-Hard SAT Math Questions·Advanced Math

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Which choice is equivalent to the expression below for all real numbers $x$ such that the expression is defined?

\frac{(2x^2+x-1)^3-(x^2+6x-7)^3}{x^2-5x+6}

A difference of cubes over a quadratic may hide a shared factor: subtract the two expressions being cubed before expanding. On a timed test, you can use Desmos to eliminate choices at convenient allowed inputs, then check the survivor by graph overlap. Matching at a few inputs is not the same as matching at every allowed input. Keep track of values the original denominator excludes.

Hints

Hint 1
A denominator cannot be zero. Factor the original denominator before you test any inputs. Are $0$ and $1$ allowed values to use for quick comparisons?
Hint 2
At $x=0$ , a polynomial equals its constant term, because every term containing $x$ becomes zero. Compare the original expression’s value there with the constants in the choices.
Hint 3
If two choices survive one input, try another allowed input. Then use graph overlap to check that the remaining choice matches along the whole curve, not merely at the inputs you tested.

Step-by-step

Approach 1: Eliminate choices in Desmos, then check the graph
Step 1
Find the excluded inputs
Factor the original denominator:
$x^2-5x+6=(x-2)(x-3).$
It is zero at $x=2$ and $x=3$ , so the fraction is undefined there. Both $0$ and $1$ are allowed inputs for testing choices.
Step 2
Use the constant terms to eliminate two choices
Type the original fraction as $f(x)=\frac{(2x^2+x-1)^3-(x^2+6x-7)^3}{x^2-5x+6}$ , then type $f(0)$ . Desmos prints $57$ . At $x=0$ , a polynomial equals its constant term, so only the two choices ending in $+57$ can match.
Step 3
Separate the choices that survived
Keep those lines and type $f(1)$ ; Desmos prints $4$ . For the surviving choice with $+99x$ , type $7(1)^4+29(1)^3+10(1)^2+99(1)+57$ . Desmos prints $202$ , so that choice fails at $x=1$ . One matching input would not prove equivalence, but one mismatch rules a choice out.
Step 4
Check the remaining choice across the graph
Add $y=7x^4+29x^3+10x^2-99x+57$ . Its curve sits on the graph of $f$ across the window: full-curve overlap, not a meeting at one point. The matching expression is $7x^4+29x^3+10x^2-99x+57$ for every input the original allows; $x=2$ and $x=3$ remain excluded. Choice C.

Approach 2: See why the fraction simplifies
Step 1
Connect the denominator to the two cubes
Let $A=2x^2+x-1$ and $B=x^2+6x-7$ , the two bases being cubed. Subtract them:
$A-B=(2x^2+x-1)-(x^2+6x-7).$
Flip the signs in the second parentheses:
$A-B=2x^2+x-1-x^2-6x+7.$
Combine like terms:
$A-B=x^2-5x+6.$
So the denominator is exactly $A-B$ .
Step 2
Factor the difference of cubes
The difference-of-cubes identity applies to the numerator:
$A^3-B^3=(A-B)(A^2+AB+B^2).$
The middle products cancel when you multiply the right side out. Watch the plus sign before $AB$ : using a minus there produces a tempting wrong choice.
Step 3
Cancel the whole shared factor
Because the denominator is $A-B$ , replace the numerator with its factored form:
$\frac{A^3-B^3}{A-B}=\frac{(A-B)(A^2+AB+B^2)}{A-B}.$
Cancel the whole factor $A-B$ on allowed inputs:
$\frac{A^3-B^3}{A-B}=A^2+AB+B^2.$
Canceling does not make the original fraction defined at $x=2$ or $x=3$ .
Step 4
Expand the first square
Multiply $A$ by itself to find the first part of the quotient:
$A^2=(2x^2+x-1)^2=4x^4+4x^3-3x^2-2x+1.$
A square multiplies every term by every term, so keep the $-2x$ from the two products involving $x$ and $-1$ .
Step 5
Expand the middle product
Multiply the two bases for the $+AB$ term:
$AB=(2x^2+x-1)(x^2+6x-7)=2x^4+13x^3-9x^2-13x+7.$
The minus signs matter: the terms with only $x$ combine to $-13x$ .
Step 6
Expand the second square
Multiply $B$ by itself for the last part:
$B^2=(x^2+6x-7)^2=x^4+12x^3+22x^2-84x+49.$
The two products involving $6x$ and $-7$ contribute $-84x$ .
Step 7
Combine matching powers
Add the coefficients of each like term, from $x^4$ down to the constant. Type the five sums shown in Desmos; it prints $7$ , $29$ , $10$ , $-99$ , and $57$ , in that order. So
$A^2+AB+B^2=7x^4+29x^3+10x^2-99x+57.$
In particular, the $x$ coefficient is $-2-13-84=-99$ . This expression matches the original wherever the original is defined. Choice C.

Question 74·200 Super-Hard SAT Math Questions·Advanced Math

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