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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 :

Let's find the product of the polynomial expression [tex]\((7x^2)(2x^3 + 5)(x^2 - 4x - 9)\)[/tex].

### Step 1: Multiply [tex]\(7x^2\)[/tex] with [tex]\((2x^3 + 5)\)[/tex]

First, distribute [tex]\(7x^2\)[/tex] across the terms in the binomial:

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

- Multiply [tex]\(7x^2\)[/tex] and [tex]\(5\)[/tex]:
[tex]\[
7x^2 \cdot 5 = 35x^2
\][/tex]

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

### Step 2: Multiply [tex]\((14x^5 + 35x^2)\)[/tex] with [tex]\((x^2 - 4x - 9)\)[/tex]

Now distribute each term in [tex]\((14x^5 + 35x^2)\)[/tex] across each term in [tex]\((x^2 - 4x - 9)\)[/tex].

#### Multiply [tex]\(14x^5\)[/tex] with [tex]\((x^2 - 4x - 9)\)[/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]

#### Multiply [tex]\(35x^2\)[/tex] with [tex]\((x^2 - 4x - 9)\)[/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]

### Step 3: Combine all the terms

Now, let's write out all the terms we have from our multiplication:

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

Make sure to combine terms with the same powers of [tex]\(x\)[/tex], if any. In this problem, each term has a unique power of [tex]\(x\)[/tex], so there's no simplification needed.

So, the expanded product of the expression is:

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

This is the final answer.

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