Question

expand each binomial.
(2d + 3)^6
\bigcirc 64d^6 - 576d^5 - 2160d^4 - 4320d^3 - 4860d^2 - 2916d - 729
\bigcirc 64d^6 + 729d^5 + 2916d^4 + 4860d^3 + 4320d^2 + 2160d + 576
\bigcirc 64d^6 + 576d^5 + 2160d^4 + 4320d^3 + 4860d^2 + 2916d + 729
\bigcirc 64d^6 + 729

question 5
1 pts
expand each binomial.
(2x - y)^5
\bigcirc 32x^5 + 80x^4y + 80x^3y^2 - 40x^2y^3 + 10xy^4 + y^5
\bigcirc 32x^5 - y^5
\bigcirc 32x^5 - 80x^4y + 80x^3y^2 - 40x^2y^3 + 10xy^4 - y^5
\bigcirc 32x^5 - 80x^4y^5 + 80x^3y^4 - 40x^2y^3 + 10xy^2 - y

Explanation:

First Binomial: \((2d + 3)^6\)

Step 1: Recall the Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\), \(a = 2d\), \(b = 3\), and \(n = 6\).

Step 2: Calculate each term for \(k=0\) to \(k = 6\)

• For \(k = 0\):

\(\binom{6}{0}(2d)^{6}(3)^{0}=1\times64d^{6}\times1 = 64d^{6}\)

• For \(k = 1\):

\(\binom{6}{1}(2d)^{5}(3)^{1}=\frac{6!}{1!5!}\times32d^{5}\times3=6\times32d^{5}\times3 = 576d^{5}\)

• For \(k = 2\):

\(\binom{6}{2}(2d)^{4}(3)^{2}=\frac{6!}{2!4!}\times16d^{4}\times9 = 15\times16d^{4}\times9=2160d^{4}\)

• For \(k = 3\):

\(\binom{6}{3}(2d)^{3}(3)^{3}=\frac{6!}{3!3!}\times8d^{3}\times27 = 20\times8d^{3}\times27 = 4320d^{3}\)

• For \(k = 4\):

\(\binom{6}{4}(2d)^{2}(3)^{4}=\frac{6!}{4!2!}\times4d^{2}\times81=15\times4d^{2}\times81 = 4860d^{2}\)

• For \(k = 5\):

\(\binom{6}{5}(2d)^{1}(3)^{5}=\frac{6!}{5!1!}\times2d\times243 = 6\times2d\times243=2916d\)

• For \(k = 6\):

\(\binom{6}{6}(2d)^{0}(3)^{6}=1\times1\times729 = 729\)

Step 3: Combine the terms

\((2d + 3)^6=64d^{6}+576d^{5}+2160d^{4}+4320d^{3}+4860d^{2}+2916d + 729\)

Second Binomial: \((2x - y)^5\)

Step 1: Recall the Binomial Theorem (with \(a = 2x\), \(b=-y\), \(n = 5\))

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

Step 2: Calculate each term for \(k = 0\) to \(k=5\)

• For \(k = 0\):

\(\binom{5}{0}(2x)^{5}(-y)^{0}=1\times32x^{5}\times1 = 32x^{5}\)

• For \(k = 1\):

\(\binom{5}{1}(2x)^{4}(-y)^{1}=\frac{5!}{1!4!}\times16x^{4}\times(-y)=5\times16x^{4}\times(-y)=- 80x^{4}y\)

• For \(k = 2\):

\(\binom{5}{2}(2x)^{3}(-y)^{2}=\frac{5!}{2!3!}\times8x^{3}\times y^{2}=10\times8x^{3}\times y^{2}=80x^{3}y^{2}\)

• For \(k = 3\):

\(\binom{5}{3}(2x)^{2}(-y)^{3}=\frac{5!}{3!2!}\times4x^{2}\times(-y^{3}) = 10\times4x^{2}\times(-y^{3})=-40x^{2}y^{3}\)

• For \(k = 4\):

\(\binom{5}{4}(2x)^{1}(-y)^{4}=\frac{5!}{4!1!}\times2x\times y^{4}=5\times2x\times y^{4}=10xy^{4}\)

• For \(k = 5\):

\(\binom{5}{5}(2x)^{0}(-y)^{5}=1\times1\times(-y^{5})=-y^{5}\)

Step 3: Combine the terms

\((2x - y)^5=32x^{5}-80x^{4}y + 80x^{3}y^{2}-40x^{2}y^{3}+10xy^{4}-y^{5}\)

Answer:

s
For \((2d + 3)^6\): \(\boldsymbol{64d^{6}+576d^{5}+2160d^{4}+4320d^{3}+4860d^{2}+2916d + 729}\) (the third option)

For \((2x - y)^5\): \(\boldsymbol{32x^{5}-80x^{4}y + 80x^{3}y^{2}-40x^{2}y^{3}+10xy^{4}-y^{5}}\) (the third option)