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.

maybe showing the graph would

step-by-step explanation:

-387129 iq

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.