Expand (x+y)4 and (s+y)3

Question

ScanSolve AI Tutor · Accepted Answer

Here are the expansions of the given expressions:

1) Expand $(x+y)^4$

Use the binomial theorem, which states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ .
For $(x+y)^4$ , we have $a=x$ , $b=y$ , and $n=4$ .
The expansion will have $n+1=5$ terms.
Calculate the binomial coefficients $\binom{4}{k}$ for $k=0, 1, 2, 3, 4$ .
$\binom{4}{0} = 1$ $\binom{4}{1} = 4$ $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ $\binom{4}{3} = 4$ $\binom{4}{4} = 1$
Substitute the coefficients and terms into the binomial expansion formula.
$(x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3 + \binom{4}{4}x^0y^4$ $(x+y)^4 = 1x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 1y^4$
The expanded form is $\boxed{x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4}$ .

2) Expand $(3+y)^3$
Use the binomial theorem.
For $(3+y)^3$ , we have $a=3$ , $b=y$ , and $n=3$ .
The expansion will have $n+1=4$ terms.
Calculate the binomial coefficients $\binom{3}{k}$ for $k=0, 1, 2, 3$ .
$\binom{3}{0} = 1$ $\binom{3}{1} = 3$ $\binom{3}{2} = 3$ $\binom{3}{3} = 1$
Substitute the coefficients and terms into the binomial expansion formula.
$(3+y)^3 = \binom{3}{0}3^3y^0 + \binom{3}{1}3^2y^1 + \binom{3}{2}3^1y^2 + \binom{3}{3}3^0y^3$
Simplify each term.
$(3+y)^3 = 1 \cdot (27) \cdot (1) + 3 \cdot (9) \cdot y + 3 \cdot (3) \cdot y^2 + 1 \cdot (1) \cdot y^3$ $(3+y)^3 = 27 + 27y + 9y^2 + y^3$
The expanded form is $\boxed{27 + 27y + 9y^2 + y^3}$ .

Answer