Expand the following using the Binomial Theorem and Pascal’s triangle. Show your work. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning. a2b3; a5b3; ab8; b8; a4b4; a8; ab7; a6b5

As per the question expand the bimonoidal theorem and the pascal triangle. Showing the (x+2)6 (x-4)4 (2x+3)5 (2x-3y)4.

Learn more about the use the binomial theorem.

link.

The answer is below

Step-by-step explanation:

Expansion using pascal triangle:

a)  (x + 2)⁶ = x⁶2⁰ + 6(x⁵)(2)¹ + 15(x⁴)(2²) + 20(x³)(2³) + 15(x²)(2⁴) + 6(x)(2⁵) + 1(2⁶)

                 = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64

b) (x-4)⁴ = x⁴ + 4(x³)(-4) + 6(x²)(-4)² + 4(x)(-4)³ + 1(x⁰)(-4)⁴

              =x⁴-16x³+96x²-256x+256

c) (2x + 3)⁵ = (2x)⁵ + 5(2x)⁴(3) + 10(2x)³(3)² + 10(2x)²(3)³ + 5(2x)(3)⁴ + 1(2x)⁰(3)⁵ =

                  = 32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243

d) (2x-3y)⁴ = 1(2x)⁴(-3y)⁰ + 4(2x)³(-3y) + 6(2x)²(-3y)² + 4(2x)(-3y)³ + 1(2x)⁰(-3y)⁴

                  = 16x⁴- 96x³ + 216x² - 216x + 81

Expansion using binomial where

a)  (x + 2)⁶ = C(6,0)[x⁶2⁰] + C(6,1)[(x⁵)(2)¹] + C(6,2)[(x⁴)(2²)] + C(6,3)[(x³)(2³)] + C(6,4)[(x²)(2⁴)] + C(6,5)[(x)(2⁵)] + C(6,6)[(2⁶)]

                = x⁶2⁰ + 6(x⁵)(2)¹ + 15(x⁴)(2²) + 20(x³)(2³) + 15(x²)(2⁴) + 6(x)(2⁵) + 1(2⁶)

                 = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64

b) (x-4)⁴ = C(4,0)[x⁴] + C(4,1)[(x³)(-4)] + C(4,2)[(x²)(-4)²] + C(4,3)[(x)(-4)³] + C(4,4)[(x⁰)(-4)⁴]

              = x⁴ + 4(x³)(-4) + 6(x²)(-4)² + 4(x)(-4)³ + 1(x⁰)(-4)⁴

              =x⁴-16x³+96x²-256x+256

c) (2x + 3)⁵   = C(5,0)[(2x)⁵] + C(5,1)[(2x)⁴(3)] + C(5,2)[(2x)³(3)²] + C(5,3)[(2x)²(3)³] + C(5,4)[(2x)(3)⁴] + C(5,5)[(2x)⁰(3)⁵]

                   = (2x)⁵ + 5(2x)⁴(3) + 10(2x)³(3)² + 10(2x)²(3)³ + 5(2x)(3)⁴ + 1(2x)⁰(3)⁵

                  = 32x⁵ + 240x⁴ + 720x³ + 1080x² + 810x + 243

d) (2x-3y)⁴ = C(4,0){(2x)⁴(-3y)⁰} + C(4,1)[(2x)³(-3y)] + C(4,2)[(2x)²(-3y)²] + C(4,3)[(2x)(-3y)³] + C(4,4)[(2x)⁰(-3y)⁴]

                  = 1(2x)⁴(-3y)⁰ + 4(2x)³(-3y) + 6(2x)²(-3y)² + 4(2x)(-3y)³ + 1(2x)⁰(-3y)⁴

                  = 16x⁴- 96x³ + 216x² - 216x + 81

In the expansion of (3a + 4b)⁸, the only possible variable terms are a⁵b³, b⁸, a⁴b⁴, a⁸, ab⁷ because for each of them, the sum of there powers is eight. If the sum of the powers is not 8 then it is not correct.

For a²b³, the sum of the power is 5, for ab⁸ the sum of power is 9 and for a⁶b⁵ the sum of the power is 11 therefore thy are not correct.

-4

step-by-step explanation:

do 6-10=

if u need it to be negative.

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.