factor
$6p^2 -17p - 45$
$\bigcirc$ $(6p-9)(p+5)$
$\bigcirc$ $(2p+9)(3p-5)$
$\bigcirc$ $(2p-9)(3p+5)$
$\bigcirc$ $(p-9)(6p+5)$

Question

Accepted Answer

# Explanation:
## Step1: Expand each option
- Option 1: Expand $(6p - 9)(p + 5)$
$$
\begin{align*}
(6p - 9)(p + 5)&=6p\times p+6p\times5 - 9\times p-9\times5\
&=6p^{2}+30p - 9p - 45\
&=6p^{2}+21p - 45
\end{align*}
$$
- Option 2: Expand $(2p + 9)(3p - 5)$
$$
\begin{align*}
(2p + 9)(3p - 5)&=2p\times3p+2p\times(-5)+9\times3p + 9\times(-5)\
&=6p^{2}-10p + 27p - 45\
&=6p^{2}+17p - 45
\end{align*}
$$
- Option 3: Expand $(2p - 9)(3p + 5)$
$$
\begin{align*}
(2p - 9)(3p + 5)&=2p\times3p+2p\times5-9\times3p - 9\times5\
&=6p^{2}+10p - 27p - 45\
&=6p^{2}-17p - 45
\end{align*}
$$
- Option 4: Expand $(p - 9)(6p + 5)$
$$
\begin{align*}
(p - 9)(6p + 5)&=p\times6p+p\times5-9\times6p - 9\times5\
&=6p^{2}+5p - 54p - 45\
&=6p^{2}-49p - 45
\end{align*}
$$

# Answer:
C. $(2p - 9)(3p + 5)$ (assuming the third option is labeled as C, if the original options are labeled as A, B, C, D with A being $(6p - 9)(p + 5)$, B being $(2p + 9)(3p - 5)$, C being $(2p - 9)(3p + 5)$, D being $(p - 9)(6p + 5)$)

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QUESTION IMAGE

Question

Explanation:

Step1: Expand each option

Answer: