One of the factors of 20pq²+ 4pq+ 45pr²−60pqr−6rp is :

The correct option is 2q−3r.

Let us factorize the given algebraic expression step-by-step.

The given expression is:

$20p{q}^2+4pq+45p{r}^2-60pqr-6rp$

First, notice that $p$ is a common factor in all the terms of the expression. Factor out $p$:

$=p(20q^2+4q+45r^2-60qr-6r)$

Now, let us group and rearrange the terms inside the parentheses to highlight algebraic identities:

$=p\left[\right(20q^2-60qr+45r^2)+(4q-6r\left)\right]$

Notice that in the first group, $20q^2-60qr+45r^2$, we can factor out $5$:

$20q^2-60qr+45r^2=5(4q^2-12qr+9r^2)$

The expression inside the parenthesis, $4q^2-12qr+9r^2$, is a perfect square of the form $a^2-2ab+b^2={(a-b)}^2$, where $a=2q$ and $b=3r$:

$4q^2-12qr+9r^2={(2q-3r)}^2$

Now, factor out $2$ from the second group $(4q-6r)$:

$4q-6r=2(2q-3r)$

Substitute these factored terms back into the expression:

$=p[5{(2q-3r)}^2+2(2q-3r\left)\right]$

Now, factor out the common term $(2q-3r)$ from inside the square brackets:

$=p(2q-3r)\left[5\right(2q-3r)+2]$

Simplify inside the brackets:

$=p(2q-3r)(10q-15r+2)$

Thus, the factors of the given expression are $p$, $(2q-3r)$, and $(10q-15r+2)$.

Therefore, one of the factors listed in the options is 2q−3r.