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

Solution available with step-by-step explanation

Super-Hard SAT Math Question 74 | Free SAT Prep | Aniko

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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}

When you see a difference of powers (especially cubes), avoid expanding immediately. First, rewrite the expression using a substitution like $A$ and $B$ , check whether the denominator matches a factor such as $A-B$ , and then apply the correct factoring identity (for cubes: $A^3-B^3=(A-B)(A^2+AB+B^2)$ ). Only after canceling should you do the algebra to combine like terms.

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Look for a factoring pattern in the numerator

The numerator has the form something cubed minus something cubed. There is a standard factoring identity for this.

Name the two repeated expressions

Set $A=2x^2+x-1$ and $B=x^2+6x-7$ . Then compare $A-B$ to the denominator.

After canceling, simplify efficiently

Once you cancel the common factor, you will need $A^2$ , $AB$ , and $B^2$ . Compute each one carefully and then add like terms.

Desmos Guide

Enter the original expression as a function

Define

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

(Desmos will show it as undefined at $x=2$ and $x=3$ .)

Enter the candidate simplified expression

Define

g(x)=7x^4+29x^3+10x^2-99x+57.

Compare by graphing the difference

Define $h(x)=f(x)-g(x)$ . If $g(x)$ is equivalent to the original, then $h(x)$ should lie on $y=0$ everywhere the original expression is defined (it may be undefined at $x=2$ and $x=3$ ).

Spot-check with a table

Make a table for several $x$ -values like $x=-2, -1, 0, 1, 4$ (avoid $2$ and $3$ ). The $f(x)$ and $g(x)$ values should match at those inputs if the expressions are equivalent.

Step-by-step Explanation

Match the expression to a difference of cubes

Let

A=2x^2+x-1 \quad \text{and} \quad B=x^2+6x-7.

Then the numerator is $A^3-B^3$ .

Show the denominator matches $A-B$

Compute $A-B$ :

\begin{aligned} A-B&=(2x^2+x-1)-(x^2+6x-7)\\ &=x^2-5x+6. \end{aligned}

This is exactly the denominator.

Factor and cancel

Use the identity

A^3-B^3=(A-B)(A^2+AB+B^2).

\frac{A^3-B^3}{A-B}=A^2+AB+B^2

(for all $x$ where the original expression is defined).

Now compute each piece:

\begin{aligned} A^2&=(2x^2+x-1)^2=4x^4+4x^3-3x^2-2x+1 \\ AB&=(2x^2+x-1)(x^2+6x-7)=2x^4+13x^3-9x^2-13x+7 \\ B^2&=(x^2+6x-7)^2=x^4+12x^3+22x^2-84x+49. \end{aligned}

Add the results and select the matching choice

Add $A^2+AB+B^2$ :

\begin{aligned} A^2+AB+B^2&=(4x^4+4x^3-3x^2-2x+1)\\ &\quad +(2x^4+13x^3-9x^2-13x+7)\\ &\quad +(x^4+12x^3+22x^2-84x+49)\\ &=7x^4+29x^3+10x^2-99x+57. \end{aligned}

So the equivalent expression is $7x^4+29x^3+10x^2-99x+57$ .

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Desmos Guide

Step-by-step Explanation

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Step-by-step Explanation