Use the Binomial Theorem to expand $$ (x+3)^{5} $$ $$ \begin{array}{l}x^{5}+3 x^{4}+9 x^{3}+27 x^{2}+81 x+243 \ \frac{x^{5}+15 x^{4}+90 x^{3}+270 x^{2}+405 x+243}{x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x+1} \ x^{2}+243\end{array} $$

Question

UpStudy AI · Accepted Answer

To expand $$ (x+3)^{5} $$ using the Binomial Theorem, we can use the formula:

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

In this case, $$ a = x $$, $$ b = 3 $$, and $$ n = 5 $$. Thus, we have:

$$
(x + 3)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \cdot 3^k
$$

Now, we will calculate each term in the expansion:

- For $$ k = 0 $$:
  $$
  \binom{5}{0} x^{5-0} \cdot 3^0 = 1 \cdot x^5 \cdot 1 = x^5
  $$

- For $$ k = 1 $$:
  $$
  \binom{5}{1} x^{5-1} \cdot 3^1 = 5 \cdot x^4 \cdot 3 = 15x^4
  $$

- For $$ k = 2 $$:
  $$
  \binom{5}{2} x^{5-2} \cdot 3^2 = 10 \cdot x^3 \cdot 9 = 90x^3
  $$

- For $$ k = 3 $$:
  $$
  \binom{5}{3} x^{5-3} \cdot 3^3 = 10 \cdot x^2 \cdot 27 = 270x^2
  $$

- For $$ k = 4 $$:
  $$
  \binom{5}{4} x^{5-4} \cdot 3^4 = 5 \cdot x^1 \cdot 81 = 405x
  $$

- For $$ k = 5 $$:
  $$
  \binom{5}{5} x^{5-5} \cdot 3^5 = 1 \cdot 1 \cdot 243 = 243
  $$

Now, we can combine all these terms together:

$$
(x + 3)^5 = x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243
$$

Thus, the expansion of $$ (x + 3)^5 $$ is:

$$
\boxed{x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243}
$$
The expansion of $$ (x + 3)^5 $$ is $$ x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243 $$.

Use the Binomial Theorem to expand

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Algebra Questions