High School

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------------------------------------------------ What is the product?

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

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

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

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

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

Answer :

To find the product of [tex]\(\left(7 x^2\right)\left(2 x^3+5\right)\left(x^2-4 x-9\right)\)[/tex], you need to multiply the expressions together step by step.

### Step 1: Multiply the First Two Expressions
First, start by multiplying [tex]\((7x^2)\)[/tex] with [tex]\((2x^3 + 5)\)[/tex].

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

Distribute [tex]\(7x^2\)[/tex] into the binomial:

[tex]\[ = 7x^2 \times 2x^3 + 7x^2 \times 5 \][/tex]

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

### Step 2: Multiply the Result with the Third Expression
Now, take the result [tex]\((14x^5 + 35x^2)\)[/tex] and multiply it by the third expression [tex]\((x^2 - 4x - 9)\)[/tex].

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

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

1. Multiply [tex]\(14x^5\)[/tex] by each term in the second expression:

[tex]\[
\begin{align*}
& = 14x^5 \times x^2 = 14x^7, \\
& = 14x^5 \times (-4x) = -56x^6, \\
& = 14x^5 \times (-9) = -126x^5.
\end{align*}
\][/tex]

2. Multiply [tex]\(35x^2\)[/tex] by each term in the second expression:

[tex]\[
\begin{align*}
& = 35x^2 \times x^2 = 35x^4, \\
& = 35x^2 \times (-4x) = -140x^3, \\
& = 35x^2 \times (-9) = -315x^2.
\end{align*}
\][/tex]

### Step 3: Combine All Terms
Now, combine all the obtained terms:

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

This gives you the final expanded polynomial.