After you rewrite the product as $(\text{quadratic})(x - k)$, focus on which multiplication creates the constant term (the term without $x$). Which numbers are being multiplied?

The constant term of $(x - 1)(x - 2)(x - k)$ must equal the constant term of $x^3 - 7x^2 + 14x - 8$. Set those two expressions equal and solve for $k$.

In a new expression line, type factor(x^3 - 7x^2 + 14x - 8). Desmos will display the cubic as a product of three linear factors.

Look at the factored form Desmos shows; it will have the shape $(x - 1)(x - 2)(x - a)$ for some number a. That number a is the same as the $k$ in $(x - 1)(x - 2)(x - k)$.

Graph y = x^3 - 7x^2 + 14x - 8 and read off its three $x$-intercepts. They correspond to the solutions of $(x - 1)(x - 2)(x - k) = 0$; the unknown $k$ is the third intercept (the one that is not 1 or 2).

Start by multiplying the two known factors:

So the given factorization can be rewritten as

When you multiply $(x^2 - 3x + 2)(x - k)$, the constant term (the term without $x$) comes only from multiplying the constants $2$ and $-k$:

So the constant term of $(x - 1)(x - 2)(x - k)$ is $-2k$.

The original polynomial $x^3 - 7x^2 + 14x - 8$ has constant term $-8$.

Since the two expressions are equivalent, their constant terms must be equal:

Divide both sides by $-2$:

So the value of $k$ is $4$.