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:
