For I, $x-2$, what value of $x$ makes it zero? For II, $4x-5$, what value of $x$ makes it zero? These values are what you should plug into the polynomial.

After you find the relevant $x$-values, substitute each into $8x^{3}-14x^{2}-19x+30$ and simplify carefully. If the result is $0$, that expression is a factor.

Pay special attention when working with fractions like $\dfrac{5}{4}$ and with negative signs; small sign errors can change whether you get $0$ or not.

In Desmos, type f(x) = 8x^3 - 14x^2 - 19x + 30 to define the polynomial.

In the input line, type f(2) and check the output. If the result is 0, then $x-2$ is a factor; if it is not 0, then $x-2$ is not a factor.

Find the root of $4x-5$ (solve $4x-5=0$) and then type f(root_value) in Desmos (for example, f(5/4)). If that value is 0, then $4x-5$ is a factor.

You can also type (x - 2)(4x - 5)(2x + 3) as another function and compare its graph to f(x), or expand it using Desmos’s tools to confirm it matches $8x^3 - 14x^2 - 19x + 30$ exactly.

To check if a linear expression is a factor of a polynomial, use the Factor Theorem:

• $x-a$ is a factor of a polynomial $P(x)$ if and only if $P(a)=0$.
• For an expression like $4x-5$, first solve $4x-5=0$ to get its root $x=\dfrac{5}{4}$, then check $P\!\left(\dfrac{5}{4}\right)$.

We will apply this to $P(x)=8x^{3}-14x^{2}-19x+30$.

Expression I is $x-2$. Its root is $x=2$.

Compute $P(2)$:

Now simplify:

• $64-56=8$
• $-38+30=-8$
• $8+(-8)=0$

So $P(2)=0$, which means $x-2$ is a factor of $P(x)$.

Expression II is $4x-5$. Find its root by solving $4x-5=0$:

Now compute $P\!\left(\dfrac{5}{4}\right)$:

First find the powers:

• $x^{3}=\left(\dfrac{5}{4}\right)^{3}=\dfrac{125}{64}$, so $8x^{3}=8\cdot\dfrac{125}{64}=\dfrac{125}{8}$
• $x^{2}=\left(\dfrac{5}{4}\right)^{2}=\dfrac{25}{16}$, so $-14x^{2}=-14\cdot\dfrac{25}{16}= -\dfrac{350}{16}= -\dfrac{175}{8}$
• $-19x=-19\cdot\dfrac{5}{4}= -\dfrac{95}{4}$

Now add all terms:

Combine step by step:

• $\dfrac{125}{8}-\dfrac{175}{8}= -\dfrac{50}{8}=-\dfrac{25}{4}$
• $-\dfrac{25}{4}-\dfrac{95}{4}=-\dfrac{120}{4}=-30$
• $-30+30=0$

So $P\!\left(\dfrac{5}{4}\right)=0$, which means $4x-5$ is also a factor of $P(x)$.

Since both tests gave $0$:

• $x-2$ is a factor.
• $4x-5$ is a factor.

We can even check the full factorization:

This matches the original polynomial, confirming that both expressions I and II are factors, so the correct choice is I and II.