Question

10-5 the binomial theorem (pp. 628-635)
use pascals triangle to expand each binomial.

1. \\((4x + 6)^5\\)
2. \\((m - 5n)^6\\)

find the coefficient of the indicated term in each expansion.

1. \\((6x - 3y)^{10}\\), \\(x^4y^6\\) term
2. \\((2y + 3)^{13}\\), \\(8^{\text{th}}\\) term

use the binomial theorem to expand each binomial.

1. \\((2p^2 - 7)^4\\)
2. \\((4m + 3n)^7\\)

Explanation:

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<reasoning>

Expand using Pascal's triangle for questions 36 and 37

\[

$$\begin{aligned} \text{Row 5 of Pascal's Triangle: } & 1, 5, 10, 10, 5, 1 \\ (4x + 6)^5 = & 1(4x)^5(6)^0 + 5(4x)^4(6)^1 + 10(4x)^3(6)^2 + 10(4x)^2(6)^3 + 5(4x)^1(6)^4 + 1(4x)^0(6)^5 \\ = & 1024x^5 + 7680x^4 + 23040x^3 + 34560x^2 + 25920x + 7776 \\ \\ \text{Row 6 of Pascal's Triangle: } & 1, 6, 15, 20, 15, 6, 1 \\ (m - 5n)^6 = & 1(m)^6(-5n)^0 + 6(m)^5(-5n)^1 + 15(m)^4(-5n)^2 + 20(m)^3(-5n)^3 + 15(m)^2(-5n)^4 \\ & + 6(m)^1(-5n)^5 + 1(m)^0(-5n)^6 \\ = & m^6 - 30m^5n + 375m^4n^2 - 2500m^3n^3 + 9375m^2n^4 - 18750mn^5 + 15625n^6 \end{aligned}$$

\]

Find the coefficients for questions 38 and 39

\[

$$\begin{aligned} \text{For 38: } & \text{Term in } (6x - 3y)^{10} \text{ containing } x^4y^6 \\ T = & \binom{10}{6} (6x)^4 (-3y)^6 = 210 \cdot 1296x^4 \cdot 729y^6 = 198,409,440x^4y^6 \\ \text{Coefficient} = & 198,409,440 \\ \\ \text{For 39: } & 8^{\text{th}} \text{ term of } (2y + 3)^{13} \\ T_8 = & \binom{13}{7} (2y)^{6} (3)^7 = 1716 \cdot 64y^6 \cdot 2187 = 240,182,784y^6 \\ \text{Coefficient} = & 240,182,784 \end{aligned}$$

\]

Expand using the Binomial Theorem for questions 40 and 41

\[

$$\begin{aligned} (2p^2 - 7)^4 = & \binom{4}{0}(2p^2)^4(-7)^0 + \binom{4}{1}(2p^2)^3(-7)^1 + \binom{4}{2}(2p^2)^2(-7)^2 + \binom{4}{3}(2p^2)^1(-7)^3 + \binom{4}{4}(2p^2)^0(-7)^4 \\ = & 16p^8 - 224p^6 + 1176p^4 - 2744p^2 + 2401 \\ \\ (4m + 3n)^7 = & \binom{7}{0}(4m)^7(3n)^0 + \binom{7}{1}(4m)^6(3n)^1 + \binom{7}{2}(4m)^5(3n)^2 + \binom{7}{3}(4m)^4(3n)^3 \\ & + \binom{7}{4}(4m)^3(3n)^4 + \binom{7}{5}(4m)^2(3n)^5 + \binom{7}{6}(4m)^1(3n)^6 + \binom{7}{7}(4m)^0(3n)^7 \\ = & 16384m^7 + 86016m^6n + 193536m^5n^2 + 241920m^4n^3 + 181440m^3n^4 + 81648m^2n^5 \\ & + 20412mn^6 + 2187n^7 \end{aligned}$$

\]
</reasoning>

<answer>

No.	Answer
37	\(m^6 - 30m^5n + 375m^4n^2 - 2500m^3n^3 + 9375m^2n^4 - 18750mn^5 + 15625n^6\)
38	\(198,409,440\)
39	\(240,182,784\)
40	\(16p^8 - 224p^6 + 1176p^4 - 2744p^2 + 2401\)
41	\(16384m^7 + 86016m^6n + 193536m^5n^2 + 241920m^4n^3 + 181440m^3n^4 + 81648m^2n^5 + 20412mn^6 + 2187n^7\)

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Answer:

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<reasoning>

Expand using Pascal's triangle for questions 36 and 37

\[

$$\begin{aligned} \text{Row 5 of Pascal's Triangle: } & 1, 5, 10, 10, 5, 1 \\ (4x + 6)^5 = & 1(4x)^5(6)^0 + 5(4x)^4(6)^1 + 10(4x)^3(6)^2 + 10(4x)^2(6)^3 + 5(4x)^1(6)^4 + 1(4x)^0(6)^5 \\ = & 1024x^5 + 7680x^4 + 23040x^3 + 34560x^2 + 25920x + 7776 \\ \\ \text{Row 6 of Pascal's Triangle: } & 1, 6, 15, 20, 15, 6, 1 \\ (m - 5n)^6 = & 1(m)^6(-5n)^0 + 6(m)^5(-5n)^1 + 15(m)^4(-5n)^2 + 20(m)^3(-5n)^3 + 15(m)^2(-5n)^4 \\ & + 6(m)^1(-5n)^5 + 1(m)^0(-5n)^6 \\ = & m^6 - 30m^5n + 375m^4n^2 - 2500m^3n^3 + 9375m^2n^4 - 18750mn^5 + 15625n^6 \end{aligned}$$

\]

Find the coefficients for questions 38 and 39

\[

$$\begin{aligned} \text{For 38: } & \text{Term in } (6x - 3y)^{10} \text{ containing } x^4y^6 \\ T = & \binom{10}{6} (6x)^4 (-3y)^6 = 210 \cdot 1296x^4 \cdot 729y^6 = 198,409,440x^4y^6 \\ \text{Coefficient} = & 198,409,440 \\ \\ \text{For 39: } & 8^{\text{th}} \text{ term of } (2y + 3)^{13} \\ T_8 = & \binom{13}{7} (2y)^{6} (3)^7 = 1716 \cdot 64y^6 \cdot 2187 = 240,182,784y^6 \\ \text{Coefficient} = & 240,182,784 \end{aligned}$$

\]

Expand using the Binomial Theorem for questions 40 and 41

\[

$$\begin{aligned} (2p^2 - 7)^4 = & \binom{4}{0}(2p^2)^4(-7)^0 + \binom{4}{1}(2p^2)^3(-7)^1 + \binom{4}{2}(2p^2)^2(-7)^2 + \binom{4}{3}(2p^2)^1(-7)^3 + \binom{4}{4}(2p^2)^0(-7)^4 \\ = & 16p^8 - 224p^6 + 1176p^4 - 2744p^2 + 2401 \\ \\ (4m + 3n)^7 = & \binom{7}{0}(4m)^7(3n)^0 + \binom{7}{1}(4m)^6(3n)^1 + \binom{7}{2}(4m)^5(3n)^2 + \binom{7}{3}(4m)^4(3n)^3 \\ & + \binom{7}{4}(4m)^3(3n)^4 + \binom{7}{5}(4m)^2(3n)^5 + \binom{7}{6}(4m)^1(3n)^6 + \binom{7}{7}(4m)^0(3n)^7 \\ = & 16384m^7 + 86016m^6n + 193536m^5n^2 + 241920m^4n^3 + 181440m^3n^4 + 81648m^2n^5 \\ & + 20412mn^6 + 2187n^7 \end{aligned}$$

\]
</reasoning>

<answer>

No.	Answer
37	\(m^6 - 30m^5n + 375m^4n^2 - 2500m^3n^3 + 9375m^2n^4 - 18750mn^5 + 15625n^6\)
38	\(198,409,440\)
39	\(240,182,784\)
40	\(16p^8 - 224p^6 + 1176p^4 - 2744p^2 + 2401\)
41	\(16384m^7 + 86016m^6n + 193536m^5n^2 + 241920m^4n^3 + 181440m^3n^4 + 81648m^2n^5 + 20412mn^6 + 2187n^7\)

</answer>

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