The expression is equivalent to , where is a constant and . Which choice is the value of ?

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Super-Hard SAT Math Question 77 | Free SAT Prep | Aniko

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Question 77·200 Super-Hard SAT Math Questions·Advanced Math

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The expression

\frac{x^{19}(x^2-4x+4)}{8x^7}-\frac{2x^{20}(x-2)}{8x^7}+\frac{x^{21}}{8x^7}

is equivalent to $\frac{1}{2}x^k$ , where $k$ is a constant and $x>0$ . Which choice is the value of $k$ ?

When an expression is said to be equivalent to a single-term power like $\frac{1}{2}x^k$ , your goal is to factor out the greatest common factor (especially powers of $x$ ) so that everything else becomes a constant. After factoring, simplify the exponent using subtraction (numerator power minus denominator power), then combine like terms inside the brackets to see if variable terms cancel. Only at the end should you match the remaining power of $x$ to determine $k$ .

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Look for a common factor

All three terms have the same denominator and share a large power of $x$ in the numerator. Factor out the greatest common factor first.

Simplify the power of $x$ outside the brackets

After factoring, you should have something like $\frac{x^{19}}{x^7}$ . Use exponent rules to rewrite it as a single power of $x$ .

Check whether the remaining brackets become a constant

Carefully combine the polynomial pieces inside the brackets; many terms cancel, leaving a constant multiplier.

Desmos Guide

Define the original expression

In Desmos, enter

f(x)=(x^19*(x^2-4x+4))/(8*x^7)-(2*x^20*(x-2))/(8*x^7)+(x^21)/(8*x^7)

Rewrite it as $x^k$

Since the expression is equivalent to $\frac{1}{2}x^k$ , define

g(x)=2*f(x)

Then $g(x)=x^k$ .

Evaluate at a convenient value

Because $x>0$ , choose $x=2$ and compute

g(2)

Record the value Desmos returns.

Match to the answer choices

Compute 2^11, 2^12, 2^13, and 2^14 in Desmos and see which one equals your value for g(2). The matching exponent is $k$ (and the correct choice).

Step-by-step Explanation

Factor out the common factor

Each term has a denominator of $8x^7$ and at least a factor of $x^{19}$ in the numerator, so factor out $\frac{x^{19}}{8x^7}$ :

\frac{x^{19}}{8x^7}\Big[(x^2-4x+4)-2x(x-2)+x^2\Big]

Also simplify the power of $x$ outside:

\frac{x^{19}}{x^7}=x^{12}

So the expression becomes

\frac{x^{12}}{8}\Big[(x^2-4x+4)-2x(x-2)+x^2\Big]

Simplify the bracketed expression

Expand only what is needed:

-2x(x-2)=-2x^2+4x

Now combine like terms:

\begin{aligned} (x^2-4x+4)+(-2x^2+4x)+x^2 &=(x^2-2x^2+x^2)+(-4x+4x)+4 \\ &=0+0+4 \\ &=4 \end{aligned}

Match the form $\frac{1}{2}x^k$

Substitute $4$ back in:

\frac{x^{12}}{8}\cdot 4=\frac{1}{2}x^{12}

So $k=12$ , which corresponds to $12$ .

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Step-by-step Explanation

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