High School

Multiply: [tex](x^4 + 1)(3x^2 + 9x + 2)[/tex]

Options:

A. [tex]x^4 + 3x^2 + 9x + 3[/tex]
B. [tex]3x^6 + 9x^5 + 2x^4 + 3x^2 + 9x + 2[/tex]
C. [tex]3x^7 + 9x^6 + 2x^5[/tex]
D. [tex]3x^8 + 9x^4 + 2x^4 + 3x^2 + 9x + 2[/tex]

Answer :

Sure! Let's multiply the polynomials [tex]\((x^4 + 1)\)[/tex] and [tex]\((3x^2 + 9x + 2)\)[/tex] together using the distributive property.

1. Distribute [tex]\(x^4\)[/tex]:
- Multiply [tex]\(x^4\)[/tex] by each term in the second polynomial [tex]\((3x^2 + 9x + 2)\)[/tex]:
- [tex]\(x^4 \cdot 3x^2 = 3x^6\)[/tex]
- [tex]\(x^4 \cdot 9x = 9x^5\)[/tex]
- [tex]\(x^4 \cdot 2 = 2x^4\)[/tex]

So, distributing [tex]\(x^4\)[/tex] gives us: [tex]\(3x^6 + 9x^5 + 2x^4\)[/tex].

2. Distribute 1:
- Multiply 1 by each term in the second polynomial [tex]\((3x^2 + 9x + 2)\)[/tex]:
- [tex]\(1 \cdot 3x^2 = 3x^2\)[/tex]
- [tex]\(1 \cdot 9x = 9x\)[/tex]
- [tex]\(1 \cdot 2 = 2\)[/tex]

So, distributing 1 gives us: [tex]\(3x^2 + 9x + 2\)[/tex].

3. Combine all the terms:
- Combine the results from both distributions:
- [tex]\(3x^6 + 9x^5 + 2x^4 + 3x^2 + 9x + 2\)[/tex]

This is the final expanded form of the polynomial after multiplying [tex]\((x^4 + 1)(3x^2 + 9x + 2)\)[/tex]:

[tex]\[3x^6 + 9x^5 + 2x^4 + 3x^2 + 9x + 2\][/tex]