Expand (5m+3n)^4?

Question

This is on a bonus quiz so there could be a trick to it.

Answer 1

(25m^2 + 30mn + 9n^2)(5m + 3n)^2
(125m^3 + 150 m^2n + 45mn^2 + 75m^2n + 90 mn^2 + 27 n^3)(5m + 3n) which simplifies to
(125m^3 +225m^2n + 135mn^2 + 27n^3)(5m + 3n)
625m^4 + 1125m^3n + 675m^2n^2 + 135mn^3 + 375m^3n + 675m^2n^2 + 405mn^3 + 81n^4

625m^4 + 1500m^3n + 1350m^2n^2 + 540mn^3 + 81n^4

Answer 2

The coefficients go like this:
1
1 2 1
1 3 3 1
1 4 6 4 1

Answer 3

The coefficients are

1 4 6 4 1

625m^4 + 4500*m^3*n + 1350*m^2*n^2 + 540*m*n^3 + 81*n^4

Answer 4

(a+b)^n= a^n + n*a^(n-1)*b +(n(n-1)/(1*2))*a^n-2*b^2 + ... +b^n

I'll write just the coefficients.
1 corresponds to (5m)^4
4 corresponds to (5m)^3 and (3n)
6 corresponds to (5m)^2 and (3n)^2
4 ..........."............ (5m) and (3n)^3
1 ..........."............ (3n)^4
Always keep in mind that the coefficients start from 1 go up, come down with symmetrical values (if both terms are positive) and stop again at 1.

Answer 5

(a + b)^n = ∑ C(n, k) · a^(n - k) · b^k,
for k going from 0 to n.

(5m)^4 + 4(5m)³(3n) + 6(5m)²(3n)² + 4(5m)(3n)³ + (3n)^4
= 625m^4 + 4(125m³)(3n) + 6(25m²)(9n²) + 4(5m)(27n³) + 81n^4
= 625m^4 + 1500m³n + 1350m²n² + 540mn³ + 81n^4

Answer 6

(5m + 3n)^4 = 625m^4 + 1500m^3n + 1350m^2n^2 + 540mn^3 + 81n^4

Answer 7

Multiply each and each and every time period in the first set of brackets with suggestions from each and every time period in the second one. as an social gathering m^2*(2m^2-5m-4) -10m*(2m^2-5m-4)-3*(2m^2-5m-4) when you do each and each and every of the enlargement and sequence 2*m^4-25*m^3+40*m^2+fifty 5*m+12

Answer 8

use binomial expansion

ans : 625m^4 + 1500m^3n + 1350m^2n^2 + 540mn^3 + 81n^4.

Answer 9

No trick here. This is expanded by the binomial theorem. See examples already given.

Answer 10

81n^4 + 540mn³ + 1350m²n² + 500m³n + 625m^4