X 5 32 - How To Discuss
X 5 32
w factor (x
(aõ bà µ) = (a b) (aà + a³b + a²b² + ab³ + bô)
(xà µ 32) = (x 2) (xô + x³ (2) + x² (2)  + x (2) ³ + (2) Â)
(xà  32) = (x 2) (xô + 2x³ + 4x² + 8x + 16)
X 5 32
X 5 32
This is a very good question. Here are some helpful things you should know:
x ny n = (xy) (x (n 1) + x (n 2) y + x (n 3) y 2 + ... + x 2 y (n 3) + xy (n 2) + y (n 1))
This sounds scary, I know, but it helps when you look at some examples. Consider cases n = 2, 3, 4 and 5:
x 2 y 2 = (x y) (x + y)
x 3 y 3 = (x y) (x 2 + xy + y 2)
x 4 y 4 = (x y) (x 3 + x 2 y + xy 2 + y 3)
x 5 y 5 = (x y) (x 4 + x 3 y + x 2 y 2 + x y 3 + y 4)
Etc.
So in this case we have y = 2, then:
x 5 32 = (x2) (x 4 + 2x 3 + 4x 2 + 8x + 16)
Quartic factor is irreparable. It can be shown that a major factor from x n to a n, where a is a number, cannot be reduced if and only if n is a prime number (e.g. n = 5). ۔ Tasting is trivial, but it's true, so I won't spend much time considering it in any other way.
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Another unconventional way of looking at things is the geometric series. Catch:
x 4 + x 3 y + x 2 y 2 + x y 3 + y 4
Or maybe imagine a geometric series with a long equation and a general ratio of the initial term x 4 and y / x. The first term is x 4, the second term is:
x 4 * y / x = x 3y
The third term is:
x 3 y * y / x = x 2 y 2
Etc. We know the formula of the geometric series combination which gives us:
x 4 + x 3 y + x 2 y 2 + x y 3 + y 4 = x 4 * (1 (y / x) 5) / (1 y / x)
= x 5 * (1 (y / x) 5) / (xy)
= (x 5 y 5) / (x y)
x 5 y 5 = (x y) (x 4 + x 3 y + x 2 y 2 + x y 3 + y 4)
Chewing gum.
X5 32
x 5 32 = x 5 2 5, so we need to add (x2) (x 4 + ... + 2 4). If you multiply what we get.
x 5 2x 4 + 16x 32 y ... We need to add the term in ... to cancel 2x 4 + 16x.
To cancel 2x 4, we need to add 2x 3 ... so if (x 2) we get (x 4 + 2x 3 + ... + 16)
x 5 + 2x 4 + 16x 2x 4 4x 3 32
x 5 + 16x 4x 3 32
To cancel 4x 3, we need to add 4x 2 to ..., so if (x2) (x 4 + 2x 3 + 4x 2 + ... + 16)
x 5 + 2x 4 + 4x 3 + 16x 2x 4 4x 3 8x 2 32
x 5 + 16x 8x 2 32
To cancel 8x 2, we have to add 8x to ... if (x 2) we get (x 4 + 2x 3 + 4x 2 + 8x + ... + 16)
x 5 + 2x 4 + 4x 3 + 8x 2 + 16x 2x 4 4x 3 8x 2 16x 32
x 5 32
So we don't need more terms ...
x 5 32 = (x2) (x 4 + 2x 3 + 4x 2 + 8x + 16)
X 5 32
X 5 32
(x 5 32) = (x2) (x 4 + 2x³ + 4x² + 8x + 16)
