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 :

To find the product of the expression [tex]\((7x^2)(2x^3 + 5)(x^2 - 4x - 9)\)[/tex], we can break it down into smaller steps and then combine the results. Let's solve it step by step:

### Step 1: Multiply the first two polynomials
First, we will multiply [tex]\((7x^2)\)[/tex] and [tex]\((2x^3 + 5)\)[/tex].

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

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

So, the product of the first two polynomials is:

[tex]\[
14x^5 + 35x^2
\][/tex]

### Step 2: Multiply the result of Step 1 with the third polynomial
Now, we need to multiply [tex]\((14x^5 + 35x^2)\)[/tex] by [tex]\((x^2 - 4x - 9)\)[/tex].

[tex]\[
(14x^5 + 35x^2)(x^2 - 4x - 9) = 14x^5 \cdot (x^2 - 4x - 9) + 35x^2 \cdot (x^2 - 4x - 9)
\][/tex]

First, expand [tex]\(14x^5 \cdot (x^2 - 4x - 9)\)[/tex]:

- [tex]\(14x^5 \cdot x^2 = 14x^7\)[/tex]
- [tex]\(14x^5 \cdot (-4x) = -56x^6\)[/tex]
- [tex]\(14x^5 \cdot (-9) = -126x^5\)[/tex]

So, the first part becomes:

[tex]\[
14x^7 - 56x^6 - 126x^5
\][/tex]

Next, expand [tex]\(35x^2 \cdot (x^2 - 4x - 9)\)[/tex]:

- [tex]\(35x^2 \cdot x^2 = 35x^4\)[/tex]
- [tex]\(35x^2 \cdot (-4x) = -140x^3\)[/tex]
- [tex]\(35x^2 \cdot (-9) = -315x^2\)[/tex]

So, the second part becomes:

[tex]\[
35x^4 - 140x^3 - 315x^2
\][/tex]

### Step 3: Combine the results
Combine all the terms from Step 2:

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

Now we have the final product. Let's check if it matches any of the listed options. Comparing with the given choices:

The correct result is:
[tex]\[
14x^7 - 56x^6 - 126x^5 + 35x^4 - 140x^3 - 315x^2
\][/tex]

This corresponds with the third choice in the list. So, that is the correct answer.