Arithmetic with Polynomials and Rational Expressions: Applying the Binomial Theorem (CCSS.A-APR.5)

Use Pascal's Triangle row 3: \$1,3,3,1$. The expansion is $x^3 + 3x^2y + 3xy^2 + y^3$. Choice B uses \$1,2,2,1$ (row 2). Choice C omits the $xy^2$ term. Choice D has incorrect exponents (terms like $xy$ and $y^2$ do not sum to degree 3).

In $(x+y)^5$, the term $x^{5-k}y^k$ has coefficient $\binom{5}{k}$. For $x^2y^3$, $5-k=2\Rightarrow k=3$, so the coefficient is $\binom{5}{3}=10$. The distractors come from misreads: 5 is an edge coefficient, 20 resembles $\binom{6}{3}$ or a doubled value, and 30 misapplies multiplication to the binomial coefficient.

Pascal's Triangle row 5 is \$1,5,10,10,5,1$. Thus $(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5$. Choice A uses row 4 (missing terms). Choice B uses row 6 (too large coefficients). Choice D drops the $xy^4$ term and misstates the middle coefficient.

In $(x+y)^3$, the term $x^{3-k}y^k$ has coefficient $\binom{3}{k}$. For $xy^2$, $3-k=1\Rightarrow k=2$, so the coefficient is $\binom{3}{2}=3$. The distractors reflect common errors: 1 is an edge coefficient, 2 uses $\binom{3}{1}$ instead of $\binom{3}{2}$, and 6 overcounts.

In $(x+y)^4$, the term $x^{4-k}y^k$ has coefficient $\binom{4}{k}$. For $x^2y^2$, $4-k=2\Rightarrow k=2$, so the coefficient is $\binom{4}{2}=6$. The other options come from using the wrong row or doubling the correct middle coefficient.

By the Binomial Theorem (or Pascal's Triangle), the coefficients for $(x+y)^5$ are \$1,5,10,10,5,1$, giving $x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5$. Choice A mistakenly uses $20$ for the middle instead of two $10$s. Choice C uses the $n=4$ row (\$1,4,6,4,1$). Choice D incorrectly changes the $xy^4$ coefficient from $5$ to $6$.

The general term is $\binom{5}{k}x^{5-k}y^k$. For $x^2y^3$, take $5-k=2\Rightarrow k=3$, so the coefficient is $\binom{5}{3}=10$. The distractors reflect miscounts: $5=\binom{5}{1}$, $20$ is a common wrong 'middle' guess, and $15$ would come from an incorrect $\binom{6}{2}$ type mistake.

Write the general term as $\binom{4}{k}x^{4-k}y^k$. For $xy^3$, we need $4-k=1\Rightarrow k=3$, giving coefficient $\binom{4}{3}=4$. The others come from wrong choices of $k$: $3=\binom{3}{1}$ (using the wrong row), $6=\binom{4}{2}$, and $8$ is not a binomial coefficient for $n=4$.

Pascal's Triangle row for $n=3$ is \$1,3,3,1$, so $(x+y)^3=x^3+3x^2y+3xy^2+y^3$. Choice A uses the $n=2$ coefficients. Choice B doubles the middle coefficient incorrectly. Choice D omits the $y^3$ term.

Using $\binom{5}{k}x^{5-k}y^k$, for $x^4y$ we have $5-k=4\Rightarrow k=1$, so the coefficient is $\binom{5}{1}=5$. The other options reflect common mistakes: $1$ would be for an end term, $10=\binom{5}{2}$ (wrong $k$), and $4=\binom{4}{1}$ (wrong row).