The constant term of the expansion is $kz$. Set it equal to $45$ and see what that implies about $45/k$.

Rewrite $b^2-180a$ as $b^2-4a\cdot 45$, and substitute the expressions you found for $a$, $b$, and $45$. А-n-i-⁠к-ο.a​i

Pick integers so that $kz=45$ (for example, $k=3$, $z=15$), then choose simple nonzero integers for $h$ and $q$ and compute $\frac{b^2-180a}{a}$.

Define variables (as numbers you can change):

• h=2
• q=3
• k=3

Then define z=45/k.

(Choose a value of k that makes z an integer, such as 1, 3, 5, 9, 15, or 45.)

Enter:

• a=h*q
• b=h*z+k*q
• D=b^2-180*a
• E=(h*z-k*q)^2

Compute D-E. If it equals 0 for your values, then $b^2-180a$ matches $(hz-kq)^2$ for that case, suggesting it is always an integer (a square). Try a few different integer choices for h, q, and k (keeping z=45/k an integer). © anіkο.ai

Compute D/a. With some valid integer choices (for example h=2, q=3, k=3), you will see D/a is not an integer, showing statement III does not have to be true.

Expand:

Match coefficients with $ax^2+bx+45$:

• $a=hq$
• $b=hz+kq$
• $45=kz$

From $45=kz$, divide both sides by $k$:

Since $z$ is an integer, statement I must be true.

Notice $180a=4\cdot 45\cdot a$, so

Substitute $a=hq$, $b=hz+kq$, and $45=kz$:

Because $hz-kq$ is an integer, $(hz-kq)^2$ is an integer. So statement II must be true. Ѕοurce:‌ а‍nikо⁠.а‌і

Statement III is

This quotient does not have to be an integer.

For example, choose integers with $kz=45$: let $h=2$, $q=3$, $k=3$, so $z=15$.

Then

• $a=hq=6$
• $b=hz+kq=2(15)+3(3)=39$
• $b^2-180a=39^2-180(6)=1521-1080=441$

So

which is not an integer. Therefore, statement III does not have to be true.

Statements I and II must be integers, but statement III does not have to be.

Correct choice: I and II only