write each polynomial is standard form. pascals triangle can be used as a tool, if desired.
12. ( p(x)=(x + 1)^{5} ) 13. ( p(x)=(x - y)^{3} )
14. ( p(x)=(x - 1)^{4} ) 15. ( p(x)=(x + 4)^{3} )
16. ( p(x)=(x + 2)^{4} ) 17. ( p(x)=(3x + 1)^{3} )

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Accepted Answer

# Explanation:
## Step1: Recall the binomial expansion formula
The binomial expansion of $(x + a)^n=\sum_{k = 0}^{n}\binom{n}{k}x^{n - k}a^{k}$, where $\binom{n}{k}=\frac{n!}{k!(n - k)!}$. For $n = 3$, $(x + a)^3=x^{3}+3ax^{2}+3a^{2}x + a^{3}$. In the case of $p(x)=(x - y)^3$, we have $a=-y$.
## Step2: Substitute $a=-y$ into the binomial expansion formula
$$
\begin{align*}
(x+(-y))^{3}&=x^{3}+3(-y)x^{2}+3(-y)^{2}x+(-y)^{3}\
&=x^{3}- 3yx^{2}+3y^{2}x - y^{3}
\end{align*}
$$

## Step3: Recall the binomial expansion formula for $n = 4$
From the table, $(x + a)^4=x^{4}+4ax^{3}+6a^{2}x^{2}+4a^{3}x + a^{4}$. For $p(x)=(x + 2)^4$, we have $a = 2$.
## Step4: Substitute $a = 2$ into the binomial expansion formula
$$
\begin{align*}
(x + 2)^4&=x^{4}+4\times2x^{3}+6\times2^{2}x^{2}+4\times2^{3}x+2^{4}\
&=x^{4}+8x^{3}+24x^{2}+32x + 16
\end{align*}
$$

## Step5: Recall the binomial expansion formula for $n = 3$ again
For $p(x)=(3x + 1)^3$, where $x$ in the binomial formula is $3x$ and $a = 1$. Using $(u + v)^3=u^{3}+3u^{2}v+3uv^{2}+v^{3}$, with $u = 3x$ and $v = 1$
$$
\begin{align*}
(3x+1)^{3}&=(3x)^{3}+3\times(3x)^{2}\times1+3\times(3x)\times1^{2}+1^{3}\
&=27x^{3}+27x^{2}+9x + 1
\end{align*}
$$

# Answer:
13. $p(x)=x^{3}-3yx^{2}+3y^{2}x - y^{3}$
16. $p(x)=x^{4}+8x^{3}+24x^{2}+32x + 16$
17. $p(x)=27x^{3}+27x^{2}+9x + 1$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the binomial expansion formula

Step2: Substitute \(a=-y\) into the binomial expansion formula

Step3: Recall the binomial expansion formula for \(n = 4\)

Step4: Substitute \(a = 2\) into the binomial expansion formula

Step5: Recall the binomial expansion formula for \(n = 3\) again

Answer: