what is the answer for (1 - 4r) to the power of 4?

Question

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Answer 1

use Binomial theorem.

the 4th row of the pascal triangle is 1,4,6,4, and 1.

(1-4r)^4

= 1^4 (-4r)^0 + (4) 1^3 (-4r)^1 + (6) 1^2 (-4r)^2 + (4) 1^1 (-4r)^3 + 1 (1)^0 (-4r)^4

= 1 + 4(-4r) + 6)(16r^2) + 4(-64r^3) + 1(256r^4)

= 1 - 16r + 96r^2 - 384r^3 + 256r^4

Answer 2

(1 - 4r)^2(1 - 4r)^2

So when (1 - 4r)^2, it equals to 16r^2 - 8r + 1

But it also multiplies with the same value again, so square 16r^2 - 8r + 1.

(16r^2 - 8r + 1)^2

The answer will be 256r^4 + 64r^2 + 1

Answer 3

Expand and distribute...
(1-4r)(1-4r)(1-4r)(1-4r) = (1-8r+16r²)(1-8r+16r²) = 1-16r+96r²-256r³+256r^4
Or use binomial theorem...
1(1^4)(-4r)º+4(1³)(-4r)¹+6(1²)(-4r)²+4(1¹)(-4r)³+1(1º)(-4r)^4 = 1-16r+96r²-256r³+256r^4

Answer 4

1-16r+96r^2-256r^3+256r^4