To express and simplify [tex]\((y - 3)^4\)[/tex], we will expand the expression step-by-step using the binomial theorem.
The binomial theorem states that:
[tex]$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$[/tex]
For our specific problem, we have [tex]\(a = y\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(n = 4\)[/tex]. Therefore:
[tex]$(y - 3)^4 = \sum_{k=0}^{4} \binom{4}{k} y^{4-k} (-3)^k$[/tex]
Now, let's expand this step-by-step:
1. For [tex]\(k = 0\)[/tex]:
[tex]$\binom{4}{0} y^{4-0} (-3)^0 = 1 \cdot y^4 \cdot 1 = y^4$[/tex]
2. For [tex]\(k = 1\)[/tex]:
[tex]$\binom{4}{1} y^{4-1} (-3)^1 = 4 \cdot y^3 \cdot (-3) = -12y^3$[/tex]
3. For [tex]\(k = 2\)[/tex]:
[tex]$\binom{4}{2} y^{4-2} (-3)^2 = 6 \cdot y^2 \cdot 9 = 54y^2$[/tex]
4. For [tex]\(k = 3\)[/tex]:
[tex]$\binom{4}{3} y^{4-3} (-3)^3 = 4 \cdot y \cdot (-27) = -108y$[/tex]
5. For [tex]\(k = 4\)[/tex]:
[tex]$\binom{4}{4} y^{4-4} (-3)^4 = 1 \cdot 1 \cdot 81 = 81$[/tex]
Combining all these terms, we get:
[tex]$(y - 3)^4 = y^4 - 12y^3 + 54y^2 - 108y + 81$[/tex]
Thus, the expanded and simplified form of [tex]\((y - 3)^4\)[/tex] is:
[tex]$y^4 - 12y^3 + 54y^2 - 108y + 81$[/tex]
