Question

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

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 \((p - 9)(6p + 5)\))