Answer:
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 [tex]C(n,r)=\frac{n!}{(n-r)!r!}[/tex]
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.
