Respuesta :

Answer:

Part A:

[tex]\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807[/tex]

Part B:

The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.

It means:

[tex]\text{Sum of polynomials } \Rightarrow \text{ It will always be a polynomial}[/tex]  

[tex]\text{Subtraction of polynomials } \Rightarrow \text{ It will always be a polynomial}[/tex]  

[tex]\text{Multiplication of polynomials } \Rightarrow \text{ It will always be a polynomial}[/tex]  

But when it is about division:

[tex]\text{Division of polynomials } \Rightarrow \text{ It will not always/sometimes be a polynomial}[/tex]  

Example of subtraction of polynomials:

[tex](2x^2+2x+3) - (x^2+5x+2)[/tex]

[tex]x^2-3x+1[/tex]

Step-by-step explanation:

First, it is very important to define what is a polynomial in standard form:

It is when the terms are ordered from the highest degree to the lowest degree.

Therefore I can give:

[tex]x^5-5x^4+3x^3-3x^2+7x+20[/tex]

but,

[tex]x^5+3x^3-3x^2+7x+20-5x^4[/tex] is not in standard form.

For this question, I can simply give the answer: [tex]x^5-5x^4+3x^3-3x^2+7x+20[/tex] and it is correct.

But I will create a fifth-degree polynomial using this formula

[tex]$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$[/tex]

Also, note that

[tex]$\binom{n}{k}=\frac{n!}{(n-k)!k!}$[/tex]

For [tex]a=x \text{ and } b=7[/tex]

[tex]$\left(x + 7\right)^{5}=\sum_{k=0}^{5} \binom{5}{k} \left(x\right)^{5-k} \left(7\right)^k$[/tex]

[tex]\text{Solving for } k \text{ values: } 0, 1, 2, 3, 4 \text{ and } 5[/tex]

Sorry but I will not type every step for each value of [tex]k[/tex]

The first one is enough.

For [tex]k=0[/tex]

[tex]$\binom{5}{0} \left(x\right)^{5-0} \left(7\right)^{0}=\frac{5!}{(5-0)! 0!}\left(x\right)^{5} \left(7\right)^{0}=\frac{5!}{5!} \cdot x^5= x^{5}$[/tex]

Doing that for [tex]k[/tex] values:

[tex]\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807[/tex]

Answer:

Ty for the free pointsd!

Step-by-step explanation: