College

What is the product?

[tex]\[
\left(7x^2\right)\left(2x^3+5\right)\left(x^2-4x-9\right)
\][/tex]

A. [tex]\(14x^5 - x^4 - 46x^3 - 58x^2 - 20x - 45\)[/tex]

B. [tex]\(14x^6 - 56x^5 - 91x^4 - 140x^3 - 315x^2\)[/tex]

C. [tex]\(14x^7 - 56x^6 - 126x^5 + 35x^4 - 140x^3 - 315x^2\)[/tex]

D. [tex]\(14x^{12} - 182x^6 + 35x^4 - 455x^2\)[/tex]

Answer :

We want to multiply the three expressions:

[tex]$$
7x^2,\quad 2x^3 + 5,\quad \text{and} \quad x^2 - 4x - 9.
$$[/tex]

First, multiply the first two factors:

[tex]$$
7x^2(2x^3 + 5) = 7x^2 \cdot 2x^3 + 7x^2 \cdot 5 = 14x^5 + 35x^2.
$$[/tex]

Next, multiply this result by the third factor. We have:

[tex]$$
(14x^5 + 35x^2)(x^2 - 4x - 9).
$$[/tex]

Now, distribute each term:

1. Multiply [tex]$14x^5$[/tex] by every term in the trinomial:
[tex]$$
\begin{aligned}
14x^5 \cdot x^2 &= 14x^7, \\
14x^5 \cdot (-4x) &= -56x^6, \\
14x^5 \cdot (-9) &= -126x^5.
\end{aligned}
$$[/tex]

2. Multiply [tex]$35x^2$[/tex] by every term in the trinomial:
[tex]$$
\begin{aligned}
35x^2 \cdot x^2 &= 35x^4, \\
35x^2 \cdot (-4x) &= -140x^3, \\
35x^2 \cdot (-9) &= -315x^2.
\end{aligned}
$$[/tex]

Combine all these products:

[tex]$$
14x^7 - 56x^6 - 126x^5 + 35x^4 - 140x^3 - 315x^2.
$$[/tex]

Thus, the product is:

[tex]$$
\boxed{14x^7 - 56x^6 - 126x^5 + 35x^4 - 140x^3 - 315x^2}.
$$[/tex]