In the expression is a constant. The expression is equivalent to . What is the value of ?

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SAT Equivalent Expressions Question 168 | Free SAT Prep | Aniko

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Question 168·Hard·Equivalent Expressions

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In the expression

4\bigl(5x^2 + kx - 6\bigr) + 12x\bigl(7 - k\bigr),

$k$ is a constant. The expression is equivalent to $20x^2 - 116x - 24$ . What is the value of $k$ ?

For problems where an expression with a parameter is said to be equivalent to a given polynomial, focus on matching coefficients instead of plugging in $x$ -values. First, fully expand and combine like terms to write the expression in standard form $ax^2 + bx + c$ in terms of the parameter. Then equate corresponding coefficients (especially the ones involving the parameter) to form a simple linear equation, and solve it carefully, watching signs and basic arithmetic. This is usually faster and more reliable than testing answer choices with random $x$ -values.

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Start by expanding

Use the distributive property to expand both $4(5x^2 + kx - 6)$ and $12x(7 - k)$ . Write out all the terms before combining anything.

Combine like terms carefully

After expanding, group together the $x^2$ terms, the $x$ terms, and the constant terms. Pay close attention to the signs on the $kx$ terms from each product.

Match coefficients of the same power of x

Once simplified, compare your expression to $20x^2 - 116x - 24$ . The $x^2$ and constant terms already match without $k$ . Focus on the coefficient of $x$ to write an equation involving $k$ and then solve it.

Desmos Guide

Enter both expressions

In Desmos, type the original expression as a function with a parameter, for example f(x) = 4(5x^2 + kx - 6) + 12x(7 - k), and then type the target expression g(x) = 20x^2 - 116x - 24.

Create and use a slider for k

Desmos will prompt you to add a slider for k. Turn on the slider, then move it left and right. Watch how the graph of $f(x)$ changes compared to $g(x)$ .

Find the k-value where the graphs coincide

Adjust the slider until the graph of $f(x)$ lies exactly on top of the graph of $g(x)$ for all visible $x$ (there should appear to be only one curve). The value shown for k on the slider at that moment is the solution.

Step-by-step Explanation

Distribute to expand each product

Start by expanding the given expression:

4(5x^2 + kx - 6) + 12x(7 - k)

Distribute inside each set of parentheses:

For $4(5x^2 + kx - 6)$ :
- $4 \cdot 5x^2 = 20x^2$
- $4 \cdot kx = 4kx$
- $4 \cdot (-6) = -24$ So this part becomes $20x^2 + 4kx - 24$ .
For $12x(7 - k)$ :
- $12x \cdot 7 = 84x$
- $12x \cdot (-k) = -12kx$ So this part becomes $84x - 12kx$ .

Now the whole expression is:

20x^2 + 4kx - 24 + 84x - 12kx.

Combine like terms and factor the x-term

Combine the $kx$ terms and the plain $x$ terms:

Combine $4kx$ and $-12kx$ :

4kx - 12kx = -8kx

The $x$ term $84x$ stays as is.
The $x^2$ term $20x^2$ and constant $-24$ also stay the same.

So the expression simplifies to:

20x^2 + (-8kx + 84x) - 24.

Factor $x$ out of the middle terms:

20x^2 + (84 - 8k)x - 24.

Match coefficients with the given quadratic

We are told this expression is equivalent to:

20x^2 - 116x - 24.

For two polynomials to be equivalent for all $x$ , the coefficients of the corresponding powers of $x$ must match:

The $x^2$ -coefficients already match: $20$ and $20$ .
The constant terms already match: $-24$ and $-24$ .
The only part involving $k$ is the coefficient of $x$ .

So set the $x$ -coefficients equal:

84 - 8k = -116.

Solve the linear equation for k

Now solve the equation from the previous step:

84 - 8k = -116.

Subtract $84$ from both sides:

\begin{aligned} -8k &= -116 - 84 \\ -8k &= -200 \end{aligned}

Divide both sides by $-8$ :

k = \frac{-200}{-8} = 25.

So the value of $k$ is $\boxed{25}$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation