To expand the expression [tex]\((-x - 3)^5\)[/tex], we need to apply the binomial theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}
\][/tex]
In this case, [tex]\( a = -x \)[/tex] and [tex]\( b = -3 \)[/tex], and [tex]\( n = 5 \)[/tex]. Using the binomial expansion, we have:
[tex]\[
(-x - 3)^5 = \sum_{k=0}^{5} \binom{5}{k} (-x)^{5-k} (-3)^{k}
\][/tex]
Let's break it down step by step:
1. For [tex]\( k = 0 \)[/tex]:
[tex]\[
\binom{5}{0} (-x)^{5-0} (-3)^{0} = 1 \cdot (-x)^5 \cdot 1 = -x^5
\][/tex]
2. For [tex]\( k = 1 \)[/tex]:
[tex]\[
\binom{5}{1} (-x)^{5-1} (-3)^1 = 5 \cdot (-x)^4 \cdot (-3) = 5 \cdot x^4 \cdot (-3) = -15x^4
\][/tex]
3. For [tex]\( k = 2 \)[/tex]:
[tex]\[
\binom{5}{2} (-x)^{5-2} (-3)^2 = 10 \cdot (-x)^3 \cdot 9 = 10 \cdot (-x)^3 \cdot 9 = 10 \cdot -x^3 \cdot 9 = -90x^3
\][/tex]
4. For [tex]\( k = 3 \)[/tex]:
[tex]\[
\binom{5}{3} (-x)^{5-3} (-3)^3 = 10 \cdot (-x)^2 \cdot (-27) = 10 \cdot x^2 \cdot (-27) = -270x^2
\][/tex]
5. For [tex]\( k = 4 \)[/tex]:
[tex]\[
\binom{5}{4} (-x)^{5-4} (-3)^4 = 5 \cdot (-x)^1 \cdot 81 = 5 \cdot -x \cdot 81 = -405x
\][/tex]
6. For [tex]\( k = 5 \)[/tex]:
[tex]\[
\binom{5}{5} (-x)^{5-5} (-3)^5 = 1 \cdot x^0 \cdot (-243) = -243
\][/tex]
Now, summing all these terms together:
[tex]\[
-x^5 - 15x^4 - 90x^3 - 270x^2 - 405x - 243
\][/tex]
Thus, the expanded form of [tex]\((-x - 3)^5\)[/tex] is:
[tex]\[
-x^5 - 15x^4 - 90x^3 - 270x^2 - 405x - 243
\][/tex]
So, the correct choice is:
A. [tex]\(-x^5 - 15 x^4 - 90 x^3 - 270 x^2 - 405 x - 243\)[/tex]
