Solve $2x + 3 = 0$ to find the $x$-value to test in the polynomial. When you plug that value in, carefully simplify step by step.

For $x - 2$, what value of $x$ makes it zero? Plug that into the polynomial and see if the result is 0.

After testing each candidate binomial, decide for each one separately whether it is a factor, then match that to the answer choices (I only, II only, both, or neither).

Type f(x) = 4x^3 - 12x^2 - 7x + 30 into Desmos to define the polynomial.

Solve $2x + 3 = 0$ to get the test value $x = -1.5$. In Desmos, type f(-1.5) and check the output; if it is 0, then $2x + 3$ is a factor.

For $x - 2$, the test value is $x = 2$. In Desmos, type f(2) and check the output; if it is 0, then $x - 2$ is a factor.

Based on what Desmos shows for f(-1.5) and f(2), decide for each binomial whether it is a factor (output 0) or not (nonzero), then pick the answer choice that matches which of I and II are factors.

A quick way to test if a binomial is a factor of a polynomial is:

• If the binomial is $x - a$, plug in $x = a$.
• If the result is 0, then $x - a$ is a factor of the polynomial.

For a binomial like $2x + 3$, first solve $2x + 3 = 0$ to find the $x$-value to plug in.

First, find the $x$-value that makes $2x + 3 = 0$:

• $2x + 3 = 0 \Rightarrow 2x = -3 \Rightarrow x = -\dfrac{3}{2}$.

Now plug $x = -\dfrac{3}{2}$ into the polynomial $4x^{3} - 12x^{2} - 7x + 30$:

• $x^{3} = \left(-\dfrac{3}{2}\right)^{3} = -\dfrac{27}{8}$, so $4x^{3} = 4\cdot\left(-\dfrac{27}{8}\right) = -\dfrac{27}{2}$.
• $x^{2} = \left(-\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$, so $-12x^{2} = -12\cdot\dfrac{9}{4} = -27$.
• $-7x = -7\cdot\left(-\dfrac{3}{2}\right) = \dfrac{21}{2}$.

Combine the terms:

• $4x^{3} - 12x^{2} - 7x + 30 = -\dfrac{27}{2} - 27 + \dfrac{21}{2} + 30$.
• First combine the fraction terms: $-\dfrac{27}{2} + \dfrac{21}{2} = -\dfrac{6}{2} = -3$.
• Now combine everything: $-3 - 27 + 30 = 0$.

Since plugging in $x = -\dfrac{3}{2}$ gives 0, $2x + 3$ is a factor of the polynomial.

For $x - 2$, the zero is $x = 2$.

Plug $x = 2$ into the polynomial $4x^{3} - 12x^{2} - 7x + 30$:

• $4x^{3} = 4\cdot 2^{3} = 4\cdot 8 = 32$.
• $-12x^{2} = -12\cdot 2^{2} = -12\cdot 4 = -48$.
• $-7x = -7\cdot 2 = -14$.

Now add all terms:

• $4x^{3} - 12x^{2} - 7x + 30 = 32 - 48 - 14 + 30$.
• Combine: $(32 + 30) - (48 + 14) = 62 - 62 = 0$.

Since plugging in $x = 2$ gives 0, $x - 2$ is also a factor of the polynomial.

We tested both binomials using their zeros:

• For I ($2x + 3$), plugging in $x = -\dfrac{3}{2}$ made the polynomial equal 0, so I is a factor.
• For II ($x - 2$), plugging in $x = 2$ also made the polynomial equal 0, so II is a factor.

Therefore, both I and II are factors of $4x^{3} - 12x^{2} - 7x + 30$, so the correct answer is I and II.