After you factor out the greatest common factor, you will have something like $kx(9x^2 - 4)$. How can you factor $9x^2 - 4$ further using the pattern $a^2 - b^2 = (a-b)(a+b)$?

Once $18x^3 - 8x$ is fully factored, line it up with $sx(3x-2)(3x+2)$. Which part of your factored expression corresponds to $s$?

In Desmos, type f(x) = 18x^3 - 8x to define the original function.

Type g(x) = s*x*(3x - 2)*(3x + 2). Desmos will prompt you to create a slider for s; add the slider so you can change the value of $s$.

Move the slider for $s$ until the graph of $g(x)$ lies exactly on top of the graph of $f(x)$ for all visible $x$-values. The value of $s$ at that point is the constant that makes the expressions equivalent.

Look at the terms $18x^3$ and $-8x$.

• Both terms have a factor of $x$.
• Both coefficients, $18$ and $8$, are divisible by $2$.

So the greatest common factor is $2x$.

Factor $2x$ out of the expression:

Now you have $18x^3 - 8x$ written as $2x(9x^2 - 4)$. Next, factor the quadratic inside the parentheses.

Inside the parentheses you have $9x^2 - 4$.

Notice that $9x^2$ and $4$ are perfect squares:

• $9x^2 = (3x)^2$
• $4 = 2^2$

This matches the pattern $a^2 - b^2 = (a-b)(a+b)$ with $a = 3x$ and $b = 2$.

So:

Substitute this back into your factored expression:

The problem tells you that

From the previous step, you found that

Compare these two factored forms term by term:

• Both have $x(3x - 2)(3x + 2)$.
• The only difference is the constant in front: $s$ versus $2$.

Therefore, $s = 2$.